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Essential Mathematics for Economics

Student’s Manual Essential Mathematics for Economic Analysis
4 edition th Knut Sydsæter Peter Hammond Arne Strøm

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Preface
This student’s solutions manual accompanies Essential Mathematics for Economic Analysis (4th edition, FT SM ⊃ Prentice Hall, 2012). Its main purpose is to provide more detailed solutions to the problems marked ⊂ in the text. The answers provided in this Manual should be used in combination with any shorter answers provided in the main text. There are a few cases where only part of the answer is set out in detail, because the rest follows the same pattern. We would appreciate suggestions for improvements from our readers, as well as help in weeding out inaccuracies and errors. Oslo and Coventry, July 2012 Knut Sydsæter (knutsy@econ.uio.no) Peter Hammond (hammond@stanford.edu) Arne Strøm (arne.strom@econ.uio.no)

Contents
1 2 3 4 5 6 7 8 9 10 11 Introductory Topics I: Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Introductory Topics II: Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Introductory Topics III: Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Functions of One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Properties of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Derivatives in Use . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Single-Variable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Interest Rates and Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Functions of Many Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

12 Tools for Comparative Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 13 14 15 16 17 Multivariable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 Constrained Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Matrix and Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Determinants and Inverse Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 1

INTRODUCTORY TOPICS I: ALGEBRA

1

Chapter 1 Introductory Topics I: Algebra
1.3
5. (a) (2t −1)(t 2 −2t +1) = 2t (t 2 −2t +1)−(t 2 −2t +1) = 2t 3 −4t 2 +2t −t 2 +2t −1 = 2t 3 −5t 2 +4t −1 (b) (a + 1)2 + (a − 1)2 − 2(a + 1)(a − 1) = a 2 + 2a + 1 + a 2 − 2a + 1 − 2a 2 + 2 = 4. Alternatively, apply the quadratic identity x 2 + y 2 − 2xy = (x − y)2 with x = a + 1 and y = a − 1 to obtain (a + 1)2 + (a − 1)2 − 2(a + 1)(a − 1) = [(a + 1) − (a − 1)]2 = 22 = 4. (c) (x + y + z)2 = (x + y + z)(x + y + z) = x(x + y + z) + y(x + y + z) + z(x + y + z) = x 2 + xy + xz + yx + y 2 + yz + zx + zy + z2 = x 2 + y 2 + z2 + 2xy + 2xz + 2yz (d) With a = x + y + z and b = x − y − z, (x + y + z)2 − (x − y − z)2 = a 2 − b2 = (a + b)(a − b) = 2x(2y + 2z) = 4x(y + z). 13. (a) a 2 + 4ab + 4b2 = (a + 2b)2 by the ﬁrst quadratic identity. (d) 9z2 − 16w2 = (3z − 4w)(3z + 4w), according to the difference-of-squares formula. (e) − 1 x 2 + 2xy − 5y 2 = − 1 (x 2 − 10xy + 25y 2 ) = 5 5 − 1 (x − 5y)2 (f) a 4 − b4 = (a 2 − b2 )(a 2 + b2 ), using the difference-of-squares formula. Since 5 a 2 − b2 = (a − b)(a + b), the answer in the book follows.

1.4

5. (a)

1 1 x+2 x−2 x+2−x+2 4 − = − = = 2 x−2 x+2 (x − 2)(x + 2) (x + 2)(x − 2) (x − 2)(x + 2) x −4 (b) Since 4x + 2 = 2(2x + 1) and 4x 2 − 1 = (2x + 1)(2x − 1), the lowest common denominator (LCD) is 2(2x + 1)(2x − 1). Then (6x + 25)(2x − 1) − 2(6x 2 + x − 2) 42x − 21 21 6x + 25 6x 2 + x − 2 = − = = 2−1 4x + 2 4x 2(2x + 1)(2x − 1) 2(2x + 1)(2x − 1) 2(2x + 1) 18b2 − a(a − 3b) + 2(a 2 − 9b2 ) a(a + 3b) a a 18b2 +2= = = − (c) 2 2 a − 9b a + 3b (a + 3b)(a − 3b) (a + 3b)(a − 3b) a − 3b 1 1 (a + 2) − a 2 1 (d) − = = = 8ab 8b(a + 2) 8ab(a + 2) 8ab(a + 2) 4ab(a + 2) 2 5t 2t t (2 − t) 3t −t (t − 2) 3t −3t 2 2t − t · − = · = · = (e) t +2 t −2 t −2 t +2 t −2 t +2 t −2 t +2 1 1 1 a 1 − 2a a 1 − 2a a− = 2 − (4a − 2) = 4 − 4a = 4(1 − a) (f) = 1 2 = 4a − 2, so 2 − 0.25 0.25 4 2 1 2(x + 1) + x − 3x(x + 1) 2 − 3x 2 + −3= = x x+1 x(x + 1) x(x + 1) t t t (2t − 1) − t (2t + 1) −2t − = = 2 2t + 1 2t − 1 (2t + 1)(2t − 1) 4t − 1 4x 2x − 1 3x(x − 2) + 4x(x + 2) − (2x − 1) 7x 2 + 1 3x − − = = 2 x+2 2−x (x − 2)(x + 2) (x − 2)(x + 2) x −4 1 1 1 1 1 1 1 1 + xy − 2 · x2y2 − 2 + y+x y2 − x2 x y x2 y x y x2 y = = = x + y (e) = = 2 1 1 1 1 1 1 1 y + x2 · xy + 2 + 2 · x2y2 xy xy x2 y x2 y To clear the fractions within both the numerator and denominator, multiply both by xy to get y−x a(y − x) = a(y + x) y+x

6. (a) (b) (c)

(d) (f)

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

2

CHAPTER 1
1 4

INTRODUCTORY TOPICS I: ALGEBRA
5 20

8. (a)

−

1 5

= n

−

4 20

=

1 20 ,

so

1 4

−

1 −2 5

=

(b) n −

(c)

(d)

(e) (f)

n·n n(n − 1) − n2 −n =n− = = 1 1 n−1 n−1 n−1 ·n 1− 1− n n 1 1 u 1 1 1 + = + =1 + = Let u = x p−q . Then 1 + x p−q 1 + x q−p 1 + u 1 + 1/u 1+u 1+u 1 1 + 2 (x 2 − 1) x+2 1 (x + 1) + 1 x−1 x −1 = = = 3 2 − 2x + 1) 2 x − x − 2x + 2 (x + 2)(x (x − 1)2 (x 2 − 1) x− x+1 1 1 − 2 2 − (x + h)2 2 2 −2x − h 1 x −2xh − h 1 (x + h) x = 2 − 2 = = 2 , so 2 (x + h)2 2 2 x x x (x + h) h x (x + h)2 (x + h) 2x 10x 2 = . Multiplying denominator and numerator by x 2 − 1 = (x + 1)(x − 1) yields 5x(x − 1) x−1 =n−

1 −2 20 n2

= 202 = 400

1.5
5. The answers given in the main text for each respective part emerge after multiplying both numerator √ √ √ √ √ √ √ and denominator by the following: (a) 7 − 5 (b) 5 − 3 (c) 3 + 2 (d) x y − y x √ √ √ (e) x + h + x (f) 1 − x + 1. 12. (a) (2x )2 = 22x = 2x if and only if 2x = x 2 , or if and only if x = 0 or x = 2. (b) Correct because a p−q = a p /a q . (c) Correct because a −p = 1/a p . (d) 51/x = 1/5x = 5−x if and only if 1/x = −x or −x 2 = 1, so there is no real x that satisﬁes the equation. (e) Put u = a x and v = a y , which reduces the equation to uv = u + v, or 0 = uv − u − v = (u − 1)(v − 1) − 1. This is true only for special values of u and v and so for special values of x and y. In particular, the equation is false when x = y = 1. √ √ (f) Putting u = x and v = y reduces the equation to 2u · 2v = 2uv , which holds if and only if uv = u + v, as in (e) above.
2

1.6
4. (a) 2 < 3x + 1 3x + 1 − 2(2x + 4) −x − 7 3x + 1 has the same solutions as − 2 > 0, or > 0, or >0 2x + 4 2x + 4 2x + 4 2x + 4 A sign diagram reveals that the inequality is satisﬁed for −7 < x < −2. A serious error is to multiply the inequality by 2x + 4, without checking the sign of 2x + 4. If 2x + 4 < 0, mulitiplying by this number will reverse the inequality sign. (It might be a good idea to test the inequality for some values of x. For example, for x = 0 it is not true. What about x = −5?) 480 − 3n 120 ≤ 0.75, or ≤ 0. A sign diagram reveals that the (b) The inequality is equivalent to n 4n inequality is satisﬁed for n < 0 and for n ≥ 160. (Note that for n = 0 the inequality makes no sense. For n = 160, we have equality.) (c) Easy: g(g − 2) ≤ 0 etc. (d) Note that p2 − 4p + 4 = (p − 2)2 , and p+1 ≥ 0. The fraction makes no sense if p = 2. The conclusion follows. the inequality reduces to (p − 2)2 −n − 2 − 2n − 8 −3n − 10 −n − 2 − 2 > 0, i.e. > 0, or > 0, etc. (e) The inequality is equivalent to n+4 n+4 n+4 (f) See the text and use a sign diagram. (Don’t cancel x 2 . If you do, x = 0 appears as a false solution.)

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 1

INTRODUCTORY TOPICS I: ALGEBRA

3

5. (a) Use a sign diagram. (b) The inequality is not satisﬁed for x = 1. If x = 1, it is obviously satisﬁed if and only x + 4 > 0, i.e. x > −4 (because (x − 1)2 is positive when x = 1). (c) Use a sign diagram. (d) The inequality is not satisﬁed for x = 1/5. If x = 1/5, it is obviously satisﬁed for x < 1. (e) Use a sign diagram. (Note that (5x − 1)11 has the same sign as 5x − 1.) 3x − 1 −(1 + x 2 ) 3x − 1 > x + 3 if and only if − (x + 3) > 0, i.e. > 0, so x < 0. (1 + x 2 is (f) x x x x−3 x−3 −2x(x + 2) always positive.) (g) > 2x − 1 if and only if − (2x − 1) < 0, i.e. < 0. x+3 x+3 x+3 Then use a sign diagram. (h) x 2 − 4x + 4 = (x − 2)2 , which is 0 for x = 2, and strictly positive for x = 2. (i) x 3 + 2x 2 + x = x(x 2 + 2x + 1) = x(x + 1)2 . Since (x + 1)2 is always ≥ 0, we see that x 3 + 2x 2 + x ≤ 0 if and only if x ≤ 0.

Review Problems for Chapter 1
5. (a) (2x)4 = 24 x 4 = 16x 4 (b) 2−1 − 4−1 =
1 2

−

1 4

1 = 4 , so (2−1 − 4−1 )−1 = 4.

(d) −(−ab3 )−3 = −(−1)−3 a −3 b−9 = a −3 b−9 , so a 5 · a 3 · a −2 a6 = 3 = a3 [−(−ab3 )−3 (a 6 b6 )2 ]3 = [a −3 b−9 a 12 b12 ]3 = [a 9 b3 ]3 = a 27 b9 (e) a −3 · a 6 a −3 −3 3 3 −3 8 x 3 8 x x · (f) · −2 = = = (x 5 )−3 = x −15 2 x 8 x −2 x −2 (c) Cancel the common factor 4x 2 yz2 . √ √ √ √ √ √ √ √ 9. All are straightforward, except (c), (g), and (h): (c) − 3 3 − 6 = −3 + 3 6 = −3 + 3 3 2 √ = −3 + 3 2 (g) (1 + x + x 2 + x 3 )(1 − x) = (1 + x + x 2 + x 3 ) − (1 + x + x 2 + x 3 )x = 1 − x 4 (h) (1 + x)4 = (1 + x)2 (1 + x)2 = (1 + 2x + x 2 )(1 + 2x + x 2 ) and so on. 12. (a) and (b) are easy. (c) ax + ay + 2x + 2y = a(x + y) + 2(x + y) = (a + 2)(x + y) (d) 2x 2 − 5yz + 10xz − xy = 2x 2 + 10xz − (xy + 5yz) = 2x(x + 5z) − y(x + 5z) = (2x − y)(x + 5z) (e) p2 − q 2 + p − q = (p − q)(p + q) + (p − q) = (p − q)(p + q + 1) (f) u3 + v 3 − u2 v − v 2 u = u2 (u − v) + v 2 (v − u) = (u2 − v 2 )(u − v) = (u + v)(u − v)(u − v) = (u + v)(u − v)2 . 16. (a) s s s(2s + 1) − s(2s − 1) 2s − = = 2 2s − 1 2s + 1 (2s − 1)(2s + 1) 4s − 1 −x(x + 3) − (1 − x)(x − 3) − 24 1−x 24 −7(x + 3) −7 x = − − = = (b) 3−x x + 3 x2 − 9 (x − 3)(x + 3) (x − 3)(x + 3) x−3 1 y−x y−x = . = (c) Multiplying numerator and denominator by x 2 y 2 yields 2 2 y −x (y − x)(y + x) x+y

17. (a) Cancel the factor 25ab. (b) x 2 − y 2 = (x + y)(x − y). Cancel x + y. (c) The fraction can be 2a − 3b 4x − x 3 x(2 − x)(2 + x) x(2 + x) (2a − 3b)2 = . (d) = = written as 2 2 (2a − 3b)(2a + 3b) 2a + 3b 4 − 4x + x (2 − x) 2−x 25. Let each side have length s, and let the area be K. Then K is the sum of the areas of the triangles 1 1 1 ABP , BCP , and CAP in Fig. SM1.R.25, which equals 2 sh1 + 2 sh2 + 2 sh3 = K. It follows that h1 + h2 + h3 = 2K/s, which is independent of where P is placed.
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

4

CHAPTER 2

INTRODUCTORY TOPICS II: EQUATIONS C

s

h3

h2 P h1

s

A

s

B

Figure SM1.R.25

Chapter 2 Introductory Topics II: Equations
2.1
3. (a) We note ﬁrst that x = −3 and x = −4 both make the equation absurd. Multiplying the equation by the common denominator (x + 3)(x + 4) yields (x − 3)(x + 4) = (x + 3)(x − 4), i.e. x 2 + x − 12 = x 2 − x − 12, and thus x = 0. (b) Multiplying by the common denominator (x − 3)(x + 3) yields 3(x + 3) − 2(x − 3) = 9, from which we get x = −6. (c) Multiplying by the common denominator 15x (assuming that x = 0) yields 18x 2 − 75 = 10x 2 − 15x + 8x 2 , from which we get x = 5. 5. (a) Multiplying by the common denominator 12 yields 9y − 3 − 4 + 4y + 24 = 36y, and so y = 17/23. (b) Multiplying by 2x(x + 2) yields 8(x + 2) + 6x = 2(2x + 2) + 7x, from which we ﬁnd x = −4. 2 − 2z − z = (c) Multiplying both numerator and denominator in the ﬁrst fraction by 1 − z leads to (1 − z)(1 + z) 6 . Multiplying each side by (1 − z2 )(2z + 1) yields (2 − 3z)(2z + 1) = 6 − 6z2 , and so z = 4. 2z + 1 p 1 (d) Expanding the parentheses we get p − 3 − 4 + 12 − 1 + p = − 1 . Multiplying by the common 4 8 3 3 3 denominator 24 gives the equation 6p − 9 − 6 + 2p − 8 + 8p = −8, whose solution is p = 15/16.

2.2
2. (a) Multiply both sides by abx to obtain b+a = 2abx. Hence, x = b a 1 1 1 b+a = + = + . 2ab 2ab 2ab 2 a b (b) Multiply the equation by cx + d to obtain ax + b = cAx + dA, or (a − cA)x = dA − b, and thus 1 x = (dA − b)/(a − cA). (c) Multiply the equation by x 1/2 to obtain 2 p = wx 1/2 , thus x 1/2 = p/2w, √ 2 /4w 2 . (d) Multiply each side by so, by squaring each side, x = p 1 + x to obtain 1 + x + ax = 0, 2 = b2 /a 2 , so x = ±b/a. (f) We see immediately that x = 0. so x = −1/(1 + a). (e) x αx − a = βx − b if and only if (α − β)x = a − b, so x = (a − b)/(α − β). √ Squaring each side of pq = 3q + 5 yields pq = (3q + 5)2 , so p = (3q + 5)2 /q. Y = 94 + 0.2(Y − (20 + 0.5Y )) = 94 + 0.2Y − 4 − 0.1Y , so 0.9Y = 90, implying that Y = 100. r 1/3 . K = Q4 , so K 3 = 2wQ4 /r, and hence K = 2wQ4 /r Raise each side to the 4th power: K 2 2w 1/2 L3/4 leads to 2L/K = r/w, (e) Multiplying numerator and denominator in the left-hand fraction by 4K 1 from which we get L = rK/2w. (f) Raise each side to the 4th power: 16 p4 K −1 (r/2w) = r 4 . It follows −1 = 32r 3 w/p 4 , so K = 1 p 4 r −3 w −1 . that K 32

4. (a) (b) (c) (d)

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 2

INTRODUCTORY TOPICS II: EQUATIONS

5

5. (a)

√ 1 1 T −t tT 1 = − = , so s = . (b) KLM = B + αL, so KLM = (B + αL)2 , and s t T tT T −t so M = (B + αL)2 /KL. (c) Multiplying each side by x − z yields x − 2y + xz = 4xy − 4yz, or (x + 4y)z = 4xy − x + 2y, and so z = (4xy − x + 2y)/(x + 4y). (d) V = C − CT /N, so CT /N = C − V and thus T = N(1 − V /C).

2.3
5. (a) See the text. (b) If the smaller of the two natural numbers is n, then the larger is n + 1, so the requirement is that n2 + (n + 1)2 = 13. This reduces to 2n2 + 2n − 12 = 0, i.e. n2 + n − 6 = 0, with solutions n = −3 and n = 2, so the two numbers are 2 and 3. (If we had asked for integer solutions, we would have −3 and −2 in addition.) (c) If the shortest side is x, the other is x+14, so according to Pythagoras’s Theorem x 2 +(x+14)2 = 342 , or x 2 + 14x − 480 = 0. The only positive solution is x = 16, and then x + 14 = 30. (d) If the usual driving speed is x km/h and the usual time spent is t hours, then xt = 80. 16 minutes is 16/60 = 4/15 hours, so driving at the speed x +10 for t −4/15 hours gives (x +10)(t −4/15) = 80. From the ﬁrst equation, t = 80/x. Inserting this into the second equation, we get (x + 10)(80/x − 4/15) = 80. Rearranging, we obtain x 2 + 10x − 3000 = 0, whose positive solution is x = 50. So his usual driving speed is 50 km/h.

2.4
4. (a) If the two numbers are x and y, then x + y = 52 and x − y = 26. Adding the two equations gives 2x = 78, so x = 39, and then y = 52 − 39 = 13. (b) Let the cost of one table be $x and the cost of one chair $y. Then 5x + 20y = 1800 and 2x + 3y = 420. Solving this system yields x = 120, y = 60. (c) Let x and y be the number of units produced of A and B, respectively. This gives the equations 3 x = 2 y and 300x + 200y = 13 000. If we use the expression for x from the ﬁrst equation and insert it in the second, we get 450y + 200y = 13 000, which yields y = 20, and then x = 30. Thus, 30 units of quality A and 20 of quality B should be produced. (d) If the person invested $x at 5% and $y at 7.2%, then x + y = 10 000 and 0.05x + 0.072y = 676. The solution is x = 2000 and y = 8000.

2.5
2. (a) The numerator 5 + x 2 is never 0, so there are no solutions. (b) The equation is obviously equivalent (x + 1)2 x 2 + 1 + 2x = 0, or 2 = 0, so x = −1. (c) x = −1 is clearly no solution. The given to x2 + 1 x +1 1/3 − 1 x(x + 1)−2/3 = 0. Multiplying this equation by (x + 1)2/3 yields equation is equivalent to (x + 1) 3 x + 1 − 1 x = 0, whose solution is x = −3/2. (d) Multiplying by x − 1 yields x + 2x(x − 1) = 0, or 3 x(2x − 1) = 0. Hence x = 0 or x = 1/2. 3. (a) z = 0 satisﬁes the equation. If z = 0, then z − a = za + zb, or (1 − a − b)z = a. If a + b = 1 we have a contradiction. If a + b = 1, then z = a/(1 − a − b). (b) The equation is equivalent to (1 + λ)μ(x − y) = 0, so λ = −1, μ = 0, or x = y. (c) μ = ±1 makes the equation meaningless. Otherwise, multiplying the equation by 1 − μ2 yields λ(1 − μ) = −λ, or λ(2 − μ) = 0, so λ = 0 or μ = 2. (d) The equation is equivalent to b(1 + λ)(a − 2) = 0, so b = 0, λ = −1, or a = 2.
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

6

CHAPTER 3

INTRODUCTORY TOPICS II: MISCELLANEOUS

Review Problems for Chapter 2
2. (a) Assuming x = ±4, multiplying by the common denominator (x − 4)(x + 4) reduces the equation to x = −x, so x = 0. (b) The given equation makes sense only if x = ±3. If we multiply the equation by the common denominator (x + 3)(x − 3) we get 3(x + 3)2 − 2(x 2 − 9) = 9x + 27 or x 2 + 9x + 18 = 0, with the solutions x = −6 and x = −3. The only solution of the given equation is therefore x = −6. (c) Subtracting 2x/3 from each side simpliﬁes the equation to 0 = −1 + 5/x, whose only solution is x = 5. (d) Assuming x = 0 and x = ±5, multiply by the common denominator x(x − 5)(x + 5) to get x(x − 5)2 − x(x 2 − 25) = x 2 − 25 − (11x + 20)(x + 5). Expanding each side of this equation gives x 3 − 10x 2 + 25x − x 3 + 25x = x 2 − 25 − 11x 2 − 75x − 100, which simpliﬁes to 50x = −125 − 75x with solution x = −1. 5. (a) Multiply the equation by 5K 1/2 to obtain 15L1/3 = K 1/2 . Squaring each side gives K = 225L2/3 . (b) Raise each side to the power 1/t to obtain 1 + r/100 = 21/t , and so r = 100(21/t − 1). b−1 b−1 (c) abx0 = p, so x0 = p/ab. Now raise each side to the power 1/(b − 1). (d) Raise each side to the power −ρ to get (1 − λ)a −ρ + λb−ρ = c−ρ , or b−ρ = λ−1 (c−ρ − (1 − λ)a −ρ ). Now raise each side to the power −1/ρ.

Chapter 3 Introductory Topics II: Miscellaneous
3.1
3. (a)–(d): In each case, look at the last term in the sum and replace n by k to get an expression for the kth term. Call it sk . Then the sum is n sk . (e) The coefﬁcients are the powers 3n for n = 1, 2, 3, 4, 5, k=1 so the general term is 3n x n . (f) and (g) see answers in the text. (h) This is tricky. One has to see that each term is 198 larger that the previous term. (The problem is related to the story about Gauss on page 56.) 7. (a) n ck 2 = c · 12 + c · 22 + · · · + c · n2 = c(12 + 22 + · · · + n2 ) = c n k 2 k=1 k=1 2 2 (b) Wrong even for n = 2: The left-hand side is (a1 + a2 )2 = a1 + 2a1 a2 + a2 , but the right-hand side 2 2 is a1 + a2 . (c) Both sides equal b1 + b2 + · · · + bN . (d) Both sides equal 51 + 52 + 53 + 54 + 55 . 2 2 (e) Both sides equal a0,j + · · · + an−1,j . (f) Wrong even for n = 2. Then left-hand side is a1 + a2 /2, but the right-hand side is (1/k)(a1 + a2 ).

3.3
1. (a) See the text.
2 4

(b) s=0 r=2

rs r +s

2

2

= s=0 2s 2+s 2 3
2

2

+ 3 4 n 2

3s 3+s + 4 5

2

+ +

4s 4+s 4 4
2

2

=0+ (after considerable arithmetic). m n n m

+ m 2

+

6 5

2

+

8 6

2

=5+

3113 3600

(c) i=1 j =1

(i + j ) =
2 j =1 i=1

i + i=1 j =1

j 2 . After using formulae (3.2.4) and (3.2.5), we can

n 1 1 1 write this as j =1 2 m(m + 1) + m 6 n(n + 1)(2n + 1) = n 2 m(m + 1) + m 1 n(n + 1)(2n + 1) = i=1 6 p 1 mn(2n2 + 3n + 3m + 4). (Note that k=1 a = pa.) 6

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 3 m 2 m m m

INTRODUCTORY TOPICS II: MISCELLANEOUS

7

(d) i=1 j =1

i = j i=1

(i + i ) =
2 i=1

i+ i=1 1 i 2 = 2 m(m + 1) + 1 m(m + 1)(2m + 1) = 1 m(m + 1)(m + 2) 6 3 n n j =1 m m n

4. a is the mean of the column means aj because ¯ ¯

1 n

aj = ¯ j =1

1 n

1 m

arj r=1 =

1 mn

arj = a. ¯ r=1 j =1

To prove (∗), note that because arj − a is independent of the summation index s, it is a common factor ¯ m when we sum over s, so s=1 (arj − a)(asj − a) = (arj − a) m (asj − a) for each r. Next, summing ¯ ¯ ¯ ¯ s=1 over r gives m m m m

(arj − a)(asj − a) = ¯ ¯ r=1 s=1 m s=1 (arj r=1

(arj − a) ¯ s=1 (asj − a) ¯

(∗∗)

because − a) is a common factor when we sum over r. Using the properties of sums and the ¯ deﬁnition of aj , we have ¯ m m m

(arj − a) = ¯ r=1 r=1

arj − r=1 a = maj − ma = m(aj − a) ¯ ¯ ¯ ¯ ¯ m s=1 (asj

Similarly, replacing r with s as the index of summation, one also has Substituting these values into (∗∗) then conﬁrms (∗).

− a) = m(aj − a). ¯ ¯ ¯

3.4
√ √ √ 6. (a) If (i) x − 4 = x + 5 − 9, then squaring each side gives (ii) x −√ = ( x + 5 − 9)2 . Expanding 4 the square on the right-hand side of (ii) gives x − 4 = x + 5 − 18 x + 5 + 81, which reduces to √ √ 18 x + 5 = 90 or x + 5 = 5, implying that x + 5 = 25 and so x = 20. This shows that if x is a solution of (i), then x = 20. No other value of x can satisfy (i). But if we √ check this solution, we ﬁnd that √ with x = 20 the LHS of (i) becomes 16 = 4, and the RHS becomes 25 − 9 = 5 − 9 = −4. Thus the LHS and the RHS are different. This means that equation (i) actually has no solutions at all. (But note that 42 = (−4)2 , i.e. the square of the LHS equals the square of the RHS. That is how the spurious solution x = 20 managed to sneak in.) √ √ + as (b) If x is a solution of√ (iii) x − 4 = 9 − x√ 5, then just √ in part (a) we ﬁnd that x must be a solution 2 . Now, (9 − x + 5 )2 = ( x + 5 − 9)2 , so equation (iv) is equivalent of (iv) x − 4 = (9 − x + 5 ) to equation (ii) in part (a). This means that (iv) has exactly one solution, namely x = 20. Inserting this value of x into equation (iii), we ﬁnd that x = 20 is a solution of (iii). A geometric explanation of the results can be given with reference to Figure SM3.4.6. √ We see that the two solid curves in the ﬁgure have no point in common, that is, the expressions x − 4 √ √ √ and x + 5 − 9 are not equal for any value of x. (In fact, the difference x − 4 − ( x + 5 − 9) increases with x, so there is no point of intersection farther to the right, either.) This explains why the equation √ √ in (a) has no solution. The dashed curve y = 9 − x + 5, on the other hand, intersects y = x + 5 for x = 20 (and only there), and this corresponds to the solution in part (b). Comment: In part (a) it was necessary to check the result, because the transition from (i) to (ii) is only an implication, not an equivalence. Similarly, it was necessary to check the result in part (b), since the transition from (iii) to (iv) also is only an implication — at least, it is not clear that it is an equivalence. (Afterwards, it turned out to be an equivalence, but we could not know that until we had solved the equation.)
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

8

CHAPTER 3

INTRODUCTORY TOPICS II: MISCELLANEOUS y y =9− 5 y= 5 10 √ x−4 15 20 25 x √ x+5

-5 y=
Figure SM3.4.6

√

x+5−9

√ 7. (a) Here we have “iff” since 4 = 2. (b) It is easy to see by means of a sign diagram that x(x + 3) < 0 precisely when x lies in the open interval (−3, 0). Therefore we have an implication from left to right (that is, “only if”), but not in the other direction. (For example, if x = 10, then x(x + 3) = 130.) (c) x 2 < 9 ⇐⇒ −3 < x < 3, so x 2 < 9 only if x < 3. If x = −5, for instance, we have x < 3 but x 2 > 9. Hence we cannot have “if” here. (d) x 2 + 1 is never 0, so we have “iff” here. (e) If x > 0, then x 2 > 0, but x 2 > 0 also when x < 0. (f) x 4 + y 4 = 0 ⇐⇒ x = 0 and y = 0, which implies that we have “only if”. If x = 0 and, say, y = 1, then x 4 + y 4 = 1, so we cannot have “if” here. 9. (a) If x and y are not both nonnegative, at leat one of them must be negative, i.e. x < 0 or y < 0. (b) If not all x are greater than or equal to a, at least one x must be less than a. (c) At least one of them is less than 5. (Would it be easier if the statement to negate were “Neither John nor Diana is less than 5 years old”?) (d)–(f) See the answers in the text.

3.7
3. For n = 1, both sides are 1/2. As the induction hypothesis, suppose that (∗) is true for n = k. Then the sum of the ﬁrst k + 1 terms is 1 1 1 1 1 k 1 + + + ··· + + = + 1·2 2·3 3·4 k(k + 1) (k + 1)(k + 2) k + 1 (k + 1)(k + 2) But 1 k(k + 2) + 1 (k + 1)2 k+1 k + = = = k + 1 (k + 1)(k + 2) (k + 1)(k + 2) (k + 1)(k + 2) k+2 which is (∗) for n = k + 1. Thus, by induction, (∗) is true for all n. 4. The claim is true for n = 1. As the induction hypothesis, suppose k 3 + (k + 1)3 + (k + 2)3 is divisible by 9. Note that (k + 1)3 + (k + 2)3 + (k + 3)3 = (k + 1)3 + (k + 2)3 + k 3 + 9k 2 + 27k + 27 = k 3 + (k + 1)3 + (k + 2)3 + 9(k 2 + 3k + 3). This is divisible by 9 because the induction hypothesis implies that the sum of the ﬁrst three terms is divisible by 9, whereas the last term is also obviously divisible by 9.

Review Problems for Chapter 3
6. (b) ⇒ false (because x 2 = 16 also has the solution x = −4), ⇐ true, because if x = 4, then x 2 = 16. (c) ⇒ true because (x − 3)2 ≥ 0; ⇐ false because with y > −2 and x = 3, one has (x − 3)2 (y + 2) = 0. (d) ⇒ and ⇐ both true, since the equation x 3 = 8 has the solution x = 2 and no others. (In the terminology of Section 6.3, the function f (x) = x 3 is strictly increasing. See Problem 6.3.3 and see the graph in Fig. 7, page 88.)
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 4

FUNCTIONS OF ONE VARIABLE

9

9. Consider Fig. A3.6.8 on page 657 in the book. Let nk denote the number of students in the set marked Sk , for k = 1, 2, . . . , 8. Suppose the sets A, B, and C refer to those who study English, French, and Spanish, respectively. Since 10 students take all three languages, n7 = 10. There are 15 who take French and Spanish, so 15 = n2 + n7 , and thus n2 = 5. Furthermore, 32 = n3 + n7 , so n3 = 22. Also, 110 = n1 + n7 , so n1 = 100. The rest of the information implies that 52 = n2 + n3 + n6 + n7 , so n6 = 52 − 5 − 22 − 10 = 15. Moreover, 220 = n1 + n2 + n5 + n7 , so n5 = 220 − 100 − 5 − 10 = 105. Finally, 780 = n1 + n3 + n4 + n7 , so n4 = 780 − 100 − 22 − 10 = 648. The answers are therefore: (a) n1 = 100, (b) n3 + n4 = 648 + 22 = 670, (c) 1000 − 8 ni = 1000 − 905 = 95. i=1 10. According to Problem 3.R.3(a) there are 100 terms. Using the trick that led to (3.2.4): R= 3+ 5+ 7 + · · · + 197 + 199 + 201 7+ 5+ 3 R = 201 + 199 + 197 + · · · +

Summing vertically term by term gives 2R = 204 + 204 + 204 + · · · + 204 + 204 + 204 = 100 × 204 = 20400, and thus R = 10200. (b) S = 1001 + 2002 + 3003 + · · · + 8008 + 9009 + 10010 = 1001(1 + 2 + 3 + · · · + 8 + 9 + 10) = 1001 · 55 = 55055. 11. For (a) and (b) see text. (c) For n = 1, the inequality is correct by part (a) (and for n = 2, it is correct by part (b)). As the induction hypothesis, suppose that (1 + x)n ≥ 1 + nx when n equals the arbitrary natural number k. Because 1 + x ≥ 0, we have (1 + x)k+1 = (1 + x)k (1 + x) ≥ (1 + kx)(1 + x) = 1 + (k + 1)x + kx 2 ≥ 1 + (k + 1)x, where the last inequality holds because k > 0. Thus, the induction hypothesis holds for n = k + 1. Therefore, by induction, Bernoulli’s inequality is true for all natural numbers n.

Chapter 4 Functions of One Variable
4.2
1. (a) f (0) = 02 + 1 = 1, f (−1) = (−1)2 + 1 = 2, f (1/2) = (1/2)2 + 1 = 1/4 + 1 = 5/4, √ √ and f ( 2) = ( 2)2 + 1 = 2 + 1 = 3. (b) (i) Since (−x)2 = x 2 , f (x) = f (−x) for all x. (ii) f (x +1) = (x +1)2 +1 = x 2 +2x +1+1 = x 2 +2x +2 and f (x)+f (1) = x 2 +1+2 = x 2 +3. Thus 2 equality holds if and only if x 2 + 2x + 2 = x 2 + 3, i.e. if and only if x = 1/2. (iii) f (2x) = (2x)√+ 1 = √ 1 4x 2 + 1 and 2f (x) = 2x 2 + 2. Now, 4x 2 + 1 = 2x 2 + 2 ⇔ x 2 = 1/2 ⇔ x = ± 1/2 = ± 2 2. 13. (a) We require 5 − x ≥ 0, so x ≤ 5. (b) The denominator x 2 − x = x(x − 1) must be different from 0, so x = 0 and x = 1. (c) To begin with, the denominator must be nonzero, so we require x = 2 and x = −3. Moreover, since we can only take the square root of a nonnegative number, the fraction (x − 1)/(x − 2)(x + 3) must be ≥ 0. A sign diagram reveals that Df = (−3, 1] ∪ (2, ∞). Note in particular that the function is deﬁned with value 0 at x = 1.

4.4
10. The points that satisfy the inequality 3x + 4y ≤ 12 are those that lie on or below the straight line 3x + 4y = 12, as explained in Example 6 for a similar inequality. The points that satisfy the inequality x − y ≤ 1, or equivalently, y ≥ x − 1, are those on or above the straight line x − y = 1. Finally, the points that satisfy the inequality 3x + y ≥ 3, or equivalently, y ≥ 3 − 3x, are those on or above the straight line 3x + y = 3. The set of points that satisfy all these three inequalities simultaneously is the shaded set shown in Fig. A4.4.10 of the text.
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

10

CHAPTER 4

FUNCTIONS OF ONE VARIABLE

4.6
2 2 2 9. (b) We ﬁnd that f (x) = Ax 2 + Bx + C, where A = a1 + a2 + · · · + an , B = 2(a1 b1 + a2 b2 + · · · + an bn ), 2 2 2 . Now, if B 2 − 4AC > 0, then according to formula (2.3.4), the equation and C = b1 + b2 + · · · + bn f (x) = Ax 2 + Bx + C = 0 would have two distinct solutions, which contradicts f (x) ≥ 0 for all x. Hence B 2 − 4AC ≤ 0 and the conclusion follows.

4.7
1. (a) Following Note 2 and (6), all integer roots must divide 6. Thus ±1, ±2, ±3, and ±6 are the only possible integer roots. Of these 8 different candidates, we ﬁnd that −2, −1, 1, and 3 all are roots, and since there can be no more than 4 roots in a polynomial equation of degree 4, we have found them all. In fact, the equation can be written as (x + 2)(x + 1)(x − 1)(x − 3) = 0. (b) This has the same possible integer roots as (a). But only −6 and 1 are integer solutions. (The third root is −1/2.) (c) Neither 1 nor −1 satisﬁes the equation, so there are no integer roots. (d) First multiply the equation by 4 to have integer coefﬁcients. Then ±1, ±2, and ±4 are seen to be the only possible integer solutions. In fact, 1, 2, −2 are all solutions. 3. (a) The answer is 2x 2 + 2x + 4 + 3/(x − 1), because + 2x − 1) ÷ (x − 1) = 2x 2 + 2x + 4 (2x 3 3 − 2x 2 2x 2x 2 + 2x − 1 2x 2 − 2x 4x − 1 4x − 4 3 (b) The answer is x2 + 1, because (x 4 + x 3 + x 2 + x) ÷ (x 2 + x) = x 2 + 1 x4 + x3 x2 + x x2 + x 0 (c) The answer is x3 − 4x 2 + 3x + 1 − 4x/(x 2 no remainder + x + 1), because remainder

(x 5 − 3x 4 x5 + x4 + x3 − 4x 4 − x 3 − 4x 4 − 4x 3 − 4x 2

+ 1) ÷ (x 2 + x + 1) = x 3 − 4x 2 + 3x + 1 +1

3x 3 + 4x 2 +1 3x 3 + 3x 2 + 3x x 2 − 3x + 1 x2 + x + 1 − 4x
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

remainder

CHAPTER 4

FUNCTIONS OF ONE VARIABLE

11

(d) The answer is 3x 5 + 6x 3 − 3x 2 + 12x − 12 + (28x 2 − 36x + 13)/(x 3 − 2x + 1), because (3x 8 3x 8 − 6x 6 + 3x 5 6x 6 − 3x 5 + 6 4 + 6x 3 6x − 12x x2 x2 + 1) ÷ (x 3 − 2x + 1) = 3x 5 + 6x 3 − 3x 2 + 12x − 12 + 1

− 3x 5 + 12x 4 − 6x 3 + x 2 + 6x 3 − 3x 2 − 3x 5

+ 1

12x 4 − 12x 3 + 4x 2 + 1 4 2 + 12x 12x − 24x − 12x 3 + 28x 2 − 12x + 1 − 12x 3 + 24x − 12 28x 2 − 36x + 13 remainder

4. (a) Since the graph intersects the x-axis at the two points x = −1 and x = 3, we try the quadratic function f (x) = a(x + 1)(x − 3), for some constant a > 0. But the graph passes through 1 the point (1, −2), so we need f (1) = −2. Since f (1) = −4a for our chosen function, a = 2 . 1 This leads to the formula y = 2 (x + 1)(x − 3). (b) Because the equation f (x) = 0 must have roots x = −3, 1, 2, we try the cubic function f (x) = b(x + 3)(x − 1)(x − 2). Then f (0) = 6b. According to the graph, f (0) = −12. So b = −2, and hence y = −2(x + 3)(x − 1)(x − 2). (c) Here we try a cubic polynomial of the form y = c(x + 3)(x − 2)2 , with x = 2 as a double 1 root. Then f (0) = 12c. rom the graph we see that f (0) = 6, and so c = 2 . This leads to the formula 1 2. y = 2 (x + 3)(x − 2) 8. Polynomial division gives (x 2 x2 − γx + βx − (β + γ )x − (β + γ )x − β(β + γ ) β(β + γ ) and so E = α x − (β + γ ) + ) ÷ (x + β) = x − (β + γ )

remainder

αβ(β + γ ) β(β + γ ) = αx − α(β + γ ) + . x+β x+β

4.8
4. (a) C. The graph is a parabola and since the coefﬁcient in front of x 2 is positive, it has a minimum point. √ (b) D. The function is deﬁned for x ≤ 2 and crosses the y-axis at y = 2 2 ≈ 2.8. (c) E. The graph is a parabola and since the coefﬁcient in front of x 2 is negative, it has a maximum point. (d) B. When x increases, y decreases, and y becomes close to −2 when x is large. (e) A. The function is deﬁned for x ≥ 2 and increases as x increases. 1 (f) F. Let y = 2 − ( 2 )x . Then y increases as x increases. For large values of x, one has y close to 2.
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

12

CHAPTER 4

FUNCTIONS OF ONE VARIABLE

4.10
3. (a) 3x 4x+2 = 8 when 3x 4x 42 = 8 or (12)x 42 = 8, and so 12x = 1/2. Then x ln 12 = ln 1−ln 2 = − ln 2, so x = − ln 2/ ln 12. (b) Since ln x 2 = 2 ln x, the equation reduces to 7 ln x = 6, so ln x = 6/7, and thus x = e6/7 . (c) One possible method is to write the equation as 4x (1 − 4−1 ) = 3x (3 − 1), or 4x · (3/4) = 3x · 2, so (4/3)x = 8/3, implying that x = ln(8/3)/ ln(4/3). Alternatively, start by dividing both sides by 3x to obtain (4/3)x (1−1/4) = 3−1 = 2, so (4/3)x = 8/3 as before. In (d)–(f) use the deﬁnition a loga x = x. 2 (d) log2 x = 2 implies that 2log2 x = 22 or x = 4. (e) logx e2 = 2 implies that x logx e = x 2 or e2 = x 2 . Hence x = e. (f) log3 x = −3 implies that 3log3 x = 3−3 or x = 1/27. 4. (a) Directly, the equation Aert = Best implies that ert /est = B/A, so e(r−s)t = B/A. Applying ln to B 1 ln . Alternatively, one can apply ln to each side of each side gives (r − s)t = ln(B/A), so t = r −s A the original equation, leading to ln A + rt = ln B + st, and then solve for t. (b) Let t denote the number of years after 1990. Assuming continuous exponential growth, when the GNP of the two nations is the same, one must have 1.2 · 1012 · e0.09t = 5.6 · 1012 · e0.02t . Applying the answer found in part (a), we obtain t= 1 5.6 · 1012 14 1 ln ln ≈ 22 = 12 0.09 − 0.02 1.2 · 10 0.07 3

According to this, the two countries would have the same GNP approximately 22 years after 1990, so in 2012. (For annual GDP, which is not quite the same as GNP, the latest (June 2012) International Monetary Fund estimates for the year 2012, in current US dollars, are 15.6 · 1012 for the USA, and 8.0 · 1012 for China.)

Review Problems for Chapter 4
14. (a) p(x) = x(x 2 + x − 12) = x(x − 3)(x + 4), because x 2 + x − 12 = 0 for x = 3 and x = −4. (b) ±1, ±2, ±4, ±8 are the only possible integer zeros. By trial and error we ﬁnd that q(2) = q(−4) = 0, so 2(x − 2)(x + 4) = 2x 2 + 4x − 16 is a factor for q(x). By polynomial division we ﬁnd that q(x) ÷ (2x 2 + 4x − 16) = x − 1/2, so q(x) = 2(x − 2)(x + 4)(x − 1/2).
1 17. Check by direct calculation that p(2) = 4 23 −22 − 11 2+ 15 = 2−4− 11 + 15 = 0, so x−2 must be a factor 4 2 2 2 1 1 of p(x). By direct division, we ﬁnd that p(x) ÷ (x − 2) = 4 (x 2 − 2x − 15) = 4 (x + 3)(x − 5), so x = −3 and x = 5 are the two other zeros. (Alternative: q(x) has the same zeros as 4p(x) = x 3 −4x 2 −11x +30. This polynomial can only have ±1, ±2, ±3, ±5, ±10, ±15, and ±30 as integer zeros. It is tedious work to ﬁnd the zeros in this way.) a + b/x 19. For the left-hand graph, note that for x = 0, one has y = f (x) = , so that y tends to a as 1 + c/x x becomes large positive or negative. The graph shows that a > 0. There is a break point at x = −c, and −c > 0, so c < 0. f (0) = b/c > 0, so b < 0. The right-hand graph of the quadratic function g is a parabola which is convex, so p > 0. Moreover r = g(0) < 0. Finally, g(x) has its minimum at x = x ∗ = −q/2p. Since x ∗ > 0 and p > 0, we conclude that q < 0.

22. (a) ln(x/e2 ) = ln x − ln e2 = ln x − 2 for x > 0. (b) ln(xz/y) = ln(xz) − ln y = ln x + ln z − ln y for x, y, z > 0 (c) ln(e3 x 2 ) = ln e3 + ln x 2 = 3 + 2 ln x for x > 0. (In general, ln x 2 = 2 ln |x|.) 1 3 1 3 (d) When x > 0, note that 2 ln x − 2 ln(1/x) − ln(x + 1) = 2 ln x − 2 (− ln x) − ln(x + 1) = 2 ln x − 2 − ln(x + 1) = ln[x 2 /(x + 1)]. ln(x + 1) = ln x
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 5

PROPERTIES OF FUNCTIONS

13

Chapter 5 Properties of Functions
5.3
4. (a) f does have an inverse since it is one-to-one. This is shown in the table by the fact that all the numbers in the second row, the domain of f −1 , are different. The inverse assigns to each number in the second row, the corresponding number in the ﬁrst row. In particular, f −1 (2) = −1. (b) Since f (x) increases by 2 for each unit increase in x, one has f (x) = 2x + a for a suitable constant a. 1 But then a = f (0) = 4, so f (x) = 2x + 4. Solving y = 2x + 4 for x yields x = 2 y − 2, so interchanging −1 (x) = 1 x − 2. x and y gives y = f 2 9. (a) (x 3 − 1)1/3 = y ⇐⇒ x 3 − 1 = y 3 ⇐⇒ x = (y 3 + 1)1/3 . If we use x as the independent variable, f −1 (x) = (x 3 + 1)1/3 . is the domain and range for both f and f −1 . x+1 (b) The domain is all x = 2, and for all such x one has = y ⇐⇒ x + 1 = y(x − 2) ⇐⇒ x−2 −2y − 1 2y + 1 (1 − y)x = −2y − 1 ⇐⇒ x = = . Using x as the independent variable, f −1 (x) = 1−y y−1 (2x + 1)/(x − 1). The domain of the inverse is all x = 1. (c) Here y = (1 − x 3 )1/5 + 2 ⇐⇒ y − 2 = (1 − x 3 )1/5 ⇐⇒ (y − 2)5 = 1 − x 3 ⇐⇒ x 3 = 1 − (y − 2)5 ⇐⇒ x = [1 − (y − 2)5 ]1/3 . With x as the free variable, f −1 (x) = [1 − (x − 2)5 ]1/3 . is the domain and range for both f and f −1 . 10. (a) The domain is and the range is (0, ∞), so the inverse is deﬁned on (0, ∞). From y = ex+4 , ln y = x + 4, so x = ln y − 4, y > 0. (b) The range is , which is the domain of the inverse. From y = ln x − 4, one has ln x = y + 4, and so x = ey+4 . (c) The domain is . On this domain the function is increasing, with y → ln 2 as x → −∞ and y → ∞ as x → ∞. So the range of the function is (ln 2, ∞). From y = ln 2 + ex−3 one has ey = 2 + ex−3 , so ex−3 = ey − 2. Hence, x = 3 + ln(ey − 2) for y > ln 2.

5.4

√ √ and 1. (a) The curve intersects the axes x = 0 and y = 0 at the points (0, √ 3) √ (± 6, 0) respectively. It is ± also entirely bounded by the rectangle whose four corners are (± 6, ± 3). Moreover, it is symmetric √ √ any about both axes, since all its points take the form (± ξ , ± η), where ξ, η are √ pair of positive real √ 2 + 2η2 = 6. Putting x = y yields the four points (± 2, ± 2) on the curve. More numbers satisfying ξ points can be found by ﬁxing any x satisfying −6 < x < 6, then solving for y. (The curve is called an ellipse. See the next section.) (b) The same argument as in (a) shows that the curve intersects only the axis x = 0, at (0, ±1). There are no points on the graph where y 2 < 1. As in (a), it is symmetric about both axes. It comes in two separate parts: below y = −1; above y = 1. Putting x 2 = 1 and then √ √ x 2 = 9 yields the additional points (±1, ± 2) and (±3, ± 10). (The graph is a hyperbola. See the next section.)

5.5
8. The method of completing the square used in problem 5 shows that x 2 + y 2 + Ax + By + C = 0 ⇐⇒ 2 2 1 1 1 x 2 + Ax + y 2 + By + C = 0 ⇐⇒ x 2 + Ax + 2 A + y 2 + By + 2 B = 4 (A2 + B 2 − 4C) ⇐⇒
1 1 1 Provided that A2 + B 2 > 4C, the last equation is that x + 2 A + y + 2 B = 4 (A2 + B 2 − 4C). √ 1 1 1 of a circle centred at − 2 A, − 2 B with radius 2 A2 + B 2 − 4C. If A2 + B 2 = 4C, the graph consists 1 1 2 + B 2 < 4C, the solution set is empty. only of the point − 2 A, − 2 B . For A 2 2

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5.6
1. In each case, except (c), the rule deﬁnes a function because it associates with each member of the original set a unique member in the target set. For instance, in (d), if the volume V of a sphere is given, the formula V = 4 πr 3 in the appendix implies that the radius is r = (3V /4π)1/3 . But then 3 the formula S = 4πr 2 in the appendix gives the surface area. Substituting for r in this formula gives S = 4π(3V /4π)2/3 = (36π)1/3 V 3/2 that expresses the surface area of a sphere as a function of its volume.

Review Problems for Chapter 5
7. (a) The function f is deﬁned and strictly increasing for ex > 2, i.e. x > ln 2. Its range is because f (x) → −∞ as x → ln 2+ , and f (x) → ∞ as x → ∞. From y = 3 + ln(ex − 2), we get ln(ex − 2) = y − 3, and so ex − 2 = ey−3 , or ex = 2 + ey−3 , so x = ln(2 + ey−3 ). Hence f −1 (x) = ln(2 + ex−3 ), x∈ . (b) Note that f is strictly increasing. Moreover, e−λx → ∞ as x → −∞, and e−λx → 0 as x → ∞. Therefore, f (x) → 0 as x → −∞, and f (x) → 1 as x → ∞. So the range of f , and therefore the a we get e−λx + a = a/y, so e−λx = a(1/y − 1), or domain of f −1 , is (0, 1). From y = −λx e +a −λx = ln a + ln(1/y − 1). Thus x = −(1/λ) ln a − (1/λ) ln(1/y − 1), and therefore the inverse is f −1 (x) = −(1/λ) ln a − (1/λ) ln(1/x − 1), with x ∈ (0, 1).

Chapter 6 Differentiation
6.2
5. For parts (a)–(c) we set out the explicit steps of the recipe in (6.2.3). (a) (A): f (a + h) = f (0 + h) = 3h + 2 (B): f (a + h) − f (a) = f (h) − f (0) = 3h + 2 − 2 = 3h (C)–(D): [f (h) − f (0)]/ h = 3 (E): [f (h) − f (0)]/ h = 3 → 3 as h → 0, so f (0) = 3. The slope of the tangent at (0, 2) is 3. (b) (A): f (a+h) = f (1+h) = (1+h)2 −1 = 1+2h+h2 −1 = 2h+h2 (B): f (1+h)−f (1) = 2h+h2 (C)–(D): [f (1+h)−f (1)]/ h = 2+h (E): [f (1+h)−f (1)]/ h = 2+h → 2 as h → 0, so f (1) = 2. (c) (A): f (3 + h) = 2 + 3/(3 + h) (B): f (3 + h) − f (3) = 2 + 3/(3 + h) − 3 = −h/(3 + h) (C)–(D): [f (3 + h) − f (3)]/ h = −1/(3 + h) (E): [f (3 + h) − f (3)]/ h = −1/(3 + h) → −1/3 as h → 0, so f (3) = −1/3. For parts (d)–(f) we still follow the recipe in (6.2.3), but express the steps more concisely. (d) [f (h) − f (0)]/ h = (h3 − 2h)/ h = h2 − 2 → −2 as h → 0, so f (0) = −2. (e) f (−1 + h) − f (−1) −1 + h + 1/(−1 + h) + 2 h2 − 1 + 1 h = , which simpliﬁes to = →0 h h h(h − 1) h−1 as h → 0, so f (0) = 0. f (1 + h) − f (1) (1 + h)4 − 1 h4 + 4h3 + 6h2 + 4h + 1 − 1 = = = h3 + 4h2 + 6h + 4 → 4 as h h h h → 0, so f (1) = 4.

(f)

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15

√ √ x + h and b = x gives 8. (a) Applying the formula (a − b)(a + b) = a 2 − b2 with a = √ √ √ √ ( x + h − x)( x + h + x) = (x + h) − x = h. √ √ √ √ f (x + h) − f (x) ( x + h − x)( x + h + x) 1 (b) = =√ √ √ √ h h( x + h + x) x+h+ x 1 1 f (x + h) − f (x) 1 =√ (c) From part (b), √ −→ √ = x −1/2 . h 2 x + h + x h→0 2 x

6.5
5. (a) 3h 1/3 − 2/3h h−2 1 1 1/3 − 2/3h = = = → as h → 2 h−2 3h(h − 2) 3h(h − 2) 3h 6 (b) When x → 0, then (x 2 − 1)/x 2 = 1 − 1/x 2 → −∞ as x → 0. √ 32(t − 3) 32 32t − 96 32t − 96 3 = = → 8, as t → 3, so 3 2 → 8 = 2 as t → 3. (c) 2 t − 2t − 3 (t − 3)(t + 1) √ t + 1 t − 2t − 3 √ √ √ √ √ 1 ( h + 3 − 3)( h + 3 + 3) h+3− 3 1 (d) = =√ √ √ −→ √ . √ h→0 2 3 h h( h + 3 + 3) h+3+ 3 (t + 2)(t − 2) t −2 2 t2 − 4 = = → − as t → −2. (e) 2 t + 10t + 16 (t + 2)(t + 8) t +8 3 √ √ √ 2− x 1 1 (f) Observe that 4 − x = (2 + x)(2 − x), so lim = lim √ = . x→4 4 − x x→4 2 + x 4 f (x) − f (1) x 2 + 2x − 3 (x − 1)(x + 3) = = = x + 3 → 4 as x → 1 x−1 x−1 x−1 f (x) − f (1) = x + 3 → 5 as x → 2. (b) x−1 (2 + h)2 + 2(2 + h) − 8 h2 + 6h f (2 + h) − f (2) = = = h + 6 → 6 as h → 0. (c) h h h x 2 + 2x − a 2 − 2a x 2 − a 2 + 2(x − a) (x − a)(x + a) + 2(x − a) f (x) − f (a) = = = (d) x−a x−a x−a x−a (e) Same answer as in (d), putting x − a = h.

6. (a)

= x + a + 2 → 2a + 2 as x → a. (f)

(a + h)2 + 2a + 2h − (a − h)2 − 2a + 2h f (a + h) − f (a − h) = = 4a + 4 → 4a + 4 as h → 0. h h

6.7
3. (a) y = 1 = x −6 ⇒ y = −6x −7 , using the power rule (6.6.4). x6 √ 3 1 (b) y = x −1 (x 2 + 1) x = x −1 x 2 x 1/2 + x −1 x 1/2 = x 3/2 + x −1/2 ⇒ y = 2 x 1/2 − 2 x −3/2 x+1 1 · (x − 1) − (x + 1) · 1 −2 3 ⇒ y = (c) y = x −3/2 ⇒ y = − 2 x −5/2 (d) y = = 2 x−1 (x − 1) (x − 1)2 x 1 4 5 (e) y = 5 + 5 = x −4 + x −5 ⇒ y = − 5 − 6 x x x x 3x − 5 3(2x + 8) − 2(3x − 5) 34 ⇒ y = (f) y = = (g) y = 3x −11 ⇒ y = −33x −12 2 2x + 8 (2x + 8) (2x + 8)2 3x − 1 3(x 2 + x + 1) − (3x − 1)(2x + 1) −3x 2 + 2x + 4 ⇒ y = (h) y = 2 = x +x+1 (x 2 + x + 1)2 (x 2 + x + 1)2

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6. (a) y = 6x − 12 = 6(x − 2) ≥ 0 ⇐⇒ x ≥ 2, so y is increasing in [2, ∞). √ √ (b)√ ) = x 3 − 3x√ x(x 2 − 3) = x(x − 3)(x + 3), so (using a sign diagram) y is increasing in y = − 3, 0 and in 3, ∞ . √ √ √ √ 2(2 + x 2 ) − (2x)(2x) 2(2 − x 2 ) 2( 2 − x)( 2 + x) = 2 = , so y is increasing in [− 2, 2]. (c) y = (2 + x 2 )2 (x + 2)2 (x 2 + 2)2 (2x − 3x 2 )(x + 1) − (x 2 − x 3 ) −2x 3 − 2x 2 + 2x −x(x − x1 )(x − x2 ) = = , where x1,2 = (d) y = 2 2 2(x + 1) 2(x + 1) (x + 1)2 √ 1 1 − 2 ∓ 2 5. Then y is increasing in (−∞, x1 ] and in [0, x2 ]. 7. (a) y = −1 − 2x = −3 when x = 1, so the slope of the tangent is −3. Since y = 1 when x = 1, the point–slope formula gives y − 1 = −3(x − 1), or y = −3x + 4. (b) y = 1 − 2(x 2 + 1)−1 , so y = 4x/(x 2 + 1)2 = 1 and y = 0 when x = 1. The tangent is y = x − 1. (c) y = x 2 − x −2 , so y = 2x + 2x −3 = 17/4 and y = 15/4 when x = 2, hence y = (17/4)x − 19/4. 1 4x 3 (x 3 + 3x 2 + x + 3) − (x 4 + 1)(3x 2 + 6x + 1) 1 when x = 0, so = − and y = 2 + 1)(x + 3)]2 [(x 9 3 y = −(x − 3)/9. (d) y = at + b a(ct + d) − (at + b)c ad − bc 9. (a) We use the quotient rule: y = ⇒ y = = 2 ct + d (ct + d) (ct + d)2 √ (b) y = t n a t + b = at n+1/2 + bt n ⇒ y = (n + 1/2)at n−1/2 + nbt n−1 1 0 · (at 2 + bt + c) − 1 · (2at + b) −2at − b (c) y = 2 = ⇒ y = 2 + bt + c)2 at + bt + c (at (at 2 + bt + c)2

6.8
3. (a) y = 1 = (x 2 + x + 1)−5 = u−5 , where u = x 2 + x + 1. By the chain rule, x + 1)5 = −5(2x + 1)(x 2 + x + 1)−6 . y = √ √ 1 (b) With u = x + x + x, y = u = u1/2 , so y = 2 u−1/2 u . Now, u = x + v 1/2 , with 1 1 1 v = x + x 1/2 . Then u = 1 + 2 v −1/2 v , where v = 1 + 2 x −1/2 . Thus, in the end, y = 2 u−1/2 u = (x 2 + (−5)u−6 u
1 2

x + (x + x 1/2 )1/2

−1/2

1 1 1 + ( 2 (x + x 1/2 )−1/2 (1 + 2 x −1/2 ) . (c) See the text.

6.10
4. (a) y = 3x 2 + 2e2x is obviously positive everywhere, so y increases in (−∞, ∞). (b) y = 10xe−4x + 5x 2 (−4)e−4x = 10x(1 − 2x)e−4x . A sign diagram shows that y increases in [0, 1/2]. 2 2 2 (c) y = 2xe−x + x 2 (−2x)e−x = 2x(1 − x)(1 + x)e−x . A sign diagram shows that y increases in (−∞, −1] and in [0, 1].

6.11
3. For most of these problems we need the chain rule. That is important in itself! But it implies in particular d u 1 that if u = f (x) is a differentiable function of x that satisﬁes f (x) > 0, then ln u = u = . u u dx 1 1 1 1 (a) y = ln(ln x) = ln u with u = ln x implies that y = u = = . u ln x x x ln x
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17

√ −2x 1 1 −x 1 − x 2 = ln u with u = 1 − x 2 implies that y = u = √ = . √ 2 2 1 − x2 u 1 − x2 1−x √ 1 (Alternatively: 1 − x 2 = (1 − x 2 )1/2 ⇒ y = 2 ln(1 − x 2 ), and so on.) 1 1 (c) y = ex ln x ⇒ y = ex ln x + ex = ex ln x + x x 2 3 x3 2 2 x3 2 x3 1 ⇒ y = 3x e ln x + e 2 2x = ex 3x 2 ln x 2 + (d) y = e ln x x x ex 2x + 3 x 2 (e) y = ln(e + 1) ⇒ y = x (f) y = ln(x + 3x − 1) ⇒ y = 2 e +1 x + 3x − 1 2 x − 1)−1 x (ex − 1)−2 (h) y = e2x 2 −x ⇒ y = −2e ⇒ y = (4x − 1)e2x −x (g) y = 2(e (b) y = ln 5. (a) One must have x 2 > 1, i.e. x > 1 or x < −1. (b) ln(ln x) is deﬁned when ln x is deﬁned and 1 positive, that is, for x > 1. (c) The fraction is deﬁned when ln(ln x) is deﬁned and different ln(ln x) − 1 from 1. From (b), ln(ln x) is deﬁned when x > 1. Further, ln(ln x) = 1 ⇐⇒ ln x = e ⇐⇒ x = ee . 1 Conclusion: is deﬁned ⇐⇒ x > 1 and x = ee . ln(ln x) − 1 6. (a) The function is deﬁned for 4 − x 2 > 0, that is in (−2, 2). f (x) = −2x/(4 − x 2 ) ≥ 0 in (−2, 0], so this is where y is increasing. (b) The function is deﬁned for x > 0. f (x) = x 2 (3 ln x + 1) ≥ 0 for ln x ≥ −1/3, or x ≥ e−1/3 , so y is increasing in [e−1/3 , ∞). (c) The function is deﬁned for x > 0, 2(1 − ln x)(−1/x)2x − 2(1 − ln x)2 (1 − ln x)(ln x − 3) = . A sign diagram reveals that y is and y = 2 4x 2x 2 increasing in x when 1 ≤ ln x ≤ 3 and so for x in [e, e3 ]. 9. In these problems we can use logarithmic differentiation. Alternatively we can write the functions in the form f (x) = eg(x) and then use the fact that f (x) = eg(x) g (x) = f (x)g (x). 1 f (x) (a) Let f (x) = (2x)x . Then ln f (x) = x ln(2x), so = 1 · ln(2x) + x · · 2 = ln(2x) + 1. f (x) 2x √ √ √ x = e x ln x , Hence, f (x) = f (x)(ln(2x) + 1) = (2x)x (ln x + ln 2 + 1). (b) f (x) = x x = eln x √ √ √ √ 1 x ln x d √ x ln x x 1 so f (x) = e ( x ln x) = x = x x− 2 ( 2 ln x + 1). · √ + dx x 2 x √ 1 1 1 √ (c) ln f (x) = x ln x = 2 x ln x, so f (x)/f (x) = 2 (ln x + 1), which gives f (x) = 2 ( x )x (ln x + 1).
1 11. (a) See the answer in the text. (b) Let f (x) = ln(1 + x) − 2 x. Then f (0) = 0 and moreover f (x) = 1 1/(x + 1) − 2 = (1 − x)/2(x + 1), which is positive in (0, 1), so f (x) > 0 in (0, 1), and the left-hand inequality is established. To prove the other inequality, put g(x) = x − ln(1 + x). Then g(0) = 0 and g (x) = 1 − 1/(x + 1) = x/(x + 1) > 0 in (0, 1), so the conclusion follows. √ √ √ (c) Let f (x) = 2( x − 1) − ln x. Then f (1) = 0 and f (x) = (1/ x) − 1/x = ( x − 1)/x, which is positive for x > 1. The conclusion follows.

Review Problems for Chapter 6
15. (a) y = 2 ex − e−x ln x ≥ 0 if x ≥ 1. (b) y = x ≥ 0 ⇐⇒ ex ≥ e−x ⇐⇒ e2x ≥ 1 ⇐⇒ x ≥ 0 x e + e−x (x − 1)(x − 2) 3x = ≥ 0 ⇐⇒ x ≤ 1 or x ≥ 2. (Use a sign diagram.) (c) y = 1 − 2 x +2 x2 + 2

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Chapter 7 Derivatives in Use
7.1
3. (a) Implicit differentiation w.r.t. x yields (∗) 1 − y + 3y + 3xy = 0. Solving for y yields y = (1+3y)/(1−3x). The deﬁnition of the function implies that y = (x −2)/(1−3x). Substituting this in the expression for y gives y = −5/(1 − 3x)2 . Differentiating (∗) w.r.t. x gives −y + 3y + 3y + 3xy = 0. Inserting y = (1 + 3y)/(1 − 3x) and solving for y gives y = 6y /(1 − 3x) = −30/(1 − 3x)3 . (b) Implicit differentiation w.r.t. x yields (∗) 5y 4 y = 6x 5 , so y = 6x 5 /5y 4 = (6/5)x 1/5 . Differentiating (∗) w.r.t. x gives 20y 3 (y )2 + 5y 4 y = 30x 4 . Inserting y = 6x 5 /5y 4 and solving for y yields y = 6x 4 y −4 − 4y −1 (y )2 = 6x 4 y −4 − (144/25)x 10 y −9 = (6/25)x −4/5 . 8. (a) y + xy = g (x) + 3y 2 y , and solve for y . (b) g (x + y)(1 + y ) = 2x + 2yy , and solve for y . (c) 2(xy + 1)(y + xy ) = g (x 2 y)(2xy + x 2 y ), and solve for y . (How did we differentiate g(x 2 y) w.r.t. x? Well, if z = g(u) and u = x 2 y, then z = g (u)u where u is a product of two functions that both depend on x. So u = 2xy + x 2 y .) 10. (a) Differentiating w.r.t. x, keeping in mind that y depends on x, yields 2(x 2 + y 2 )(2x + 2yy ) = a 2 (2x − 2yy ). Then solve for y . (b) Note that x = 0 would imply that y = 0. Excluding this possibility, we see that y = 0 when √ 1 1 x 2 + y 2 = a 2 /2, or y 2 = 2 a 2 − x 2 . Inserting this into the given equation yields x = ± 4 a 6 and so √ 1 y = ± 2 a 2. This yields the four points on the graph at which the tangent is horizontal.

7.2
4. (a) Using (ii) and (iii) to substitute for C and Y respectively in equation (i), one has Y = f (Y ) + I + ¯ X − g(Y ). Differentiating w.r.t. I yields dY/dI = f (Y )(dY/dI ) + 1 − g (Y )(dY/dI ) = (f (Y ) − g (Y ))(dY/dI ) + 1 (∗)

Thus, dY/dI = 1/ 1 − f (Y ) + g (Y ) . Imports should increase when income increases, so g (Y ) > 0. It follows that dY/dI > 0. (b) Differentiating (∗) w.r.t. I yields, in simpliﬁed notation, d 2 Y /dI 2 = (f − g )(dY/dI ) + (f − g )(d 2 Y/dI 2 ), so d 2 Y/dI 2 = (f − g )(dY/dI )/(1 − f + g )2 = (f − g )/(1 − f + g )3 .

7.3
5. (a) dy/dx = −e−x−5 , so dx/dy = 1/(dy/dx) = 1/ − e−x−5 = −ex+5 . (b) dy/dx = −e−x /(e−x + 3), so dx/dy = −(e−x + 3)/e−x = −1 − 3ex (c) Implicit differentiation w.r.t. x yields y 3 + x(3y 2 )(dy/dx) − 3x 2 y − x 3 (dy/dx) = 2. Solve for dy/dx, and then invert.

7.4
3. (a) f (0) = 1 and f (x) = −(1 + x)−2 , so f (0) = −1. Then f (x) ≈ f (0) + f (0)x = 1 − x. (b) f (0) = 1 and f (x) = 5(1 + x)4 , so f (0) = 5. Then f (x) ≈ f (0) + f (0)x = 1 + 5x. 1 1 1 (c) f (0) = 1 and f (x) = − 4 (1 − x)−3/4 , so f (0) = − 4 . Then f (x) ≈ f (0) + f (0)x = 1 − 4 x.
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19
2

8. Implicit differentiation yields: 3exy + 3xexy (y 2 + x2yy ) − 2y = 6x + 2yy . (To differentiate 3xexy 2 2 2 2 use the product rule to get 3exy + 3x(d/dx)exy . The chain rule gives (d/dx)exy = exy (y 2 + x2yy ).) 3 For x = 1, y = 0 we get 3 − 2y = 6, so y = −3/2. (b) y(x) ≈ y(1) + y (1)(x − 1) = − 2 (x − 1)

7.5
2. f (x) = (1 + x)−1 , f (x) = −(1 + x)−2 , f (x) = 2(1 + x)−3 , f (4) (x) = −6(1 + x)−4 , f (5) (x) = 24(1 + x)−5 . Then f (0) = 0, f (0) = 1, f (0) = −1, f (0) = 2, f (4) (0) = −6, f (5) (0) = 24, and so 1 1 1 1 1 1 f (x) ≈ f (0) + 1! f (0)x + 2! f (0)x + 3! f (0)x 3 + 4! f (4) (0)x 4 + 5! f (5) (0)x 5 = x − 2 x 2 + 1 x 3 − 3 1 4 x + 1 x5. 4 5 √ 5 3. With f (x) = 5(ln(1+x)− 1 + x ) = 5 ln(1+x)−5(1+x)1/2 we get f (x) = 5(1+x)−1 − 2 (1+x)−1/2 , −2 + 5 (1 + x)−3/2 , and so f (0) = −5, f (0) = 5 , f (0) = − 15 . and the Taylor f (x) = −5(1 + x) 4 2 4 1 5 polynomial of order 2 about x = 0 is f (0) + f (0)x + 2 f (0)x 2 = −5 + 2 x − 15 x 2 . 8

7.6
4. (a) We use Taylor’s formula (3) with g(x) = (1 + x)1/3 and n = 2. Then g (x) = 1 (1 + x)−2/3 , g (x) = 3 − 2 (1+x)−5/3 , and g (x) = 10 (1+x)−8/3 , so g(0) = 1, g (0) = 1 , g (0) = − 2 , g (c) = 10 (1+c)−8/3 . 9 27 3 9 27 5 1 It follows that g(x) = 1 + 3 x − 1 x 2 + R3 (x), where R3 (x) = 1 10 (1 + c)−8/3 x 3 = 81 (1 + c)−8/3 x 3 . 9 6 27 (b) c ∈ (0, x)√ x ≥ 0, so (1 + c)−8/3 ≤ 1, and the inequality follows. and (c) Note that 3 1003 = 10(1 + 3 · 10−3 )1/3 . Using the approximation in part (a) gives (1 + 3 · 10−3 )1/3 ≈ √ 1+ 1 3·10−3 − 1 (3·10−3 )2 = 1+10−3 −10−6 ) = 1.000999, and so 3 1003 ≈ 10.00999. By part (b), the 9 3 5 error R3 (x) in the approximation (1+3·10−3 )1/3 ≈ 1.000999 satisﬁes |R3 (x)| ≤ 81 (3·10−3 )3 = 5 10−9 . 3 √ Hence the error in the approximation 3 1003 ≈ 10.00999 is 10|R3 (x)| ≤ 50 10−9 < 2 · 10−8 , implying 3 that the answer is correct to 7 decimal places.

7.7
9. (a) Elx A = xg x dA x x xf = 0 (b) Elx (fg) = (fg) = (f g + fg ) = + = Elx f + Elx g A dx fg fg f g x f xg xf xg gf − fg f (c) Elx = − = Elx f − Elx g = = 2 g (f/g) g f g f g (d) See the answer in the text. (e) Is like (d), but with +g replaced by −g, and +g by −g . x u dz du x dz = = Elu f (u) Elx u (f) z = f (g(u)), u = g(x) ⇒ Elx z = z dx u z du dx

7.8
3. By the results in (7.8.3), all the functions are continuous wherever they are deﬁned. So (a) and (d) are deﬁned everywhere. In (b) we must exclude x = 1; in (c) the function is deﬁned for x < 2. Next, in (e) we must exclude values of x that make the denominator 0. These values satisfy x 2 + 2x − 2 = 0, or √ (x + 1)2 = 3, so they are x = ± 3 − 1. Finally, in (f), the ﬁrst fraction requires x > 0, and then the other fraction is also deﬁned.

7.9
1. (a) limx→0+ (x 2 + 3x − 4) = 02 + 3 · 0 − 4 = −4 x−x lim− = lim− 0 = 0. x→0 x→0 x x + |x| x+x = lim+ = lim+ 2 = 2. (c) |x| = x for x > 0. Hence, lim+ x→0 x→0 x→0 x x (b) |x| = −x for x < 0. Hence, lim− x→0 x + |x| = x

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CHAPTER 7

DERIVATIVES IN USE

√ √ (d) As x → 0+ one has x → 0 and so −1/ x → −∞. (e) As x → 3+ one has x − 3 → 0+ and so x/(x − 3) → ∞. (f) As x → 3− one has x − 3 → 0− , and so x/(x − 3) → −∞. 4. (a) Vertical asymptote, x = −1. Moreover, x 2 ÷ (x + 1) = x − 1 + 1/(x + 1), so y = x − 1 is an asymptote as x → ±∞. (b) No vertical asymptote. Moreover. (2x 3 − 3x 2 + 3x − 6) ÷ (x 2 + 1) = 2x − 3 + (x − 3)/(x 2 + 1), so y = 2x − 3 is an asymptote as x → ±∞. (c) Vertical asymptote, x = 1. Moreover, (3x 2 + 2x) ÷ (x − 1) = 3x + 5 + 5/(x − 1), so y = 3x + 5 is an asymptote as x → ±∞. (d) Vertical asymptote, x = 1. Moreover, (5x 4 − 3x 2 + 1) ÷ (x 3 − 1) = 5x + (−3x 2 + 5x + 1)/(x 3 − 1), so y = 5x is an asymptote as x → ±∞. √ 7. This is rather tricky because the denominator is 0 at x1,2 = 2 ± 3. A sign diagram shows that f (x) > 0 only in (−∞, 0) and in (x1 , x2 ). The text explains where f increases. See also Fig. SM7.9.7. y 6 4 2 −6 −4 −2 −2 −4
Figure SM7.9.7

f (x) =

3x −x 2 + 4x − 1

2

4

6

8

x

7.10
4. Recall from (4.7.6) that any integer root of the equation f (x) = x 4 + 3x 3 − 3x 2 − 8x + 3 = 0 must be a factor of the constant term 3. The way to see this directly is to notice that we must have 3 = −x 4 − 3x 3 + 3x 2 + 8x = x(−x 3 − 3x 2 + 3x + 8) and if x is an integer then the bracketed expression is also an integer. Thus, the only possible integer solutions are ±1 and ±3. Trying each of these possibilities, we ﬁnd that only −3 is an integer solution. There are three other real roots, with approximate values x0 = −1.9, y0 = 0.4, and z0 = 1.5. If we use Newton’s method once for each of these roots we get the more accurate approximations −0.1749 f (−1.9) x1 = −1.9 − = −1.9 − ≈ −1.9 + 0.021 = −1.879 f (−1.9) 8.454 −0.4624 f (0.4) = 0.4 − ≈ 0.4 − 0.053 = 0.347 y1 = 0.4 − f (0.4) −8.704 −0.5625 f (1.5) = 1.5 − ≈ 1.5 + 0.034 = 1.534 z1 = 1.5 − f (1.5) 16.75

7.12

3. (a) lim

1 x−1 1 “0” = = lim = (or use x 2 − 1 = (x + 1)(x − 1)). 2−1 x→1 x x→1 2x 0 2 x 3 + 3x 2 − 4 3x 2 + 6x 6x + 6 “0” “0” = = lim = = lim =3 (b) lim 3 2 + 8x + 4 2 + 10x + 8 x→−2 x + 5x x→−2 3x x→−2 6x + 10 0 0

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 7

DERIVATIVES IN USE

21

(c) lim

x 4 − 4x 3 + 6x 2 − 8x + 8 “0” 4x 3 − 12x 2 + 12x − 8 “0” = = lim = = x→2 x→2 x 3 − 3x 2 + 4 0 3x 2 − 6x 0 12x 2 − 24x + 12 =2 lim x→2 6x − 6 1 ln x − x + 1 “0” (1/x) − 1 “0” (−1/x 2 ) = lim = = lim =− = 2 x→1 (x − 1) x→1 2(x − 1) x→1 0 0 2 2

(d) lim

4 7 − 7x + 1 1 ln(7x + 1) − ln(4x + 4) “0” 7x + 1 4x + 4 = 3 (e) lim ln = lim = = lim x→1 x − 1 x→1 x→1 4x + 4 x−1 0 1 8 x −x x (ln x + 1) − 1 x (ln x + 1)2 + x x (1/x) x x x “0” “0” = = lim = = lim = −2 (f) lim x→1 1 − x + ln x x→1 x→1 0 −1 + 1/x 0 −1/x 2 (using Example 6.11.4 to differentiate x x ). 8. L = lim f (x) 1/g(x) “0” −1/(g(x))2 g (x) (f (x))2 g (x) = lim = = lim = lim = · · x→a g(x) x→a 1/f (x) x→a −1/(f (x))2 f (x) x→a (g(x))2 f (x) 0 g (x) 1 L2 lim = L2 lim . The conclusion follows. (This argument ignores problems with x→a f (x) x→a f (x)/g (x) “division by 0”, when either f (x) or g (x) tends to 0 as x tends to a.)

Review Problems for Chapter 7
10. (a) We must have 1+x > 0, so the domain of f is the interval −1 < x < 1. As x → 1− , one has 1−x f (x) → ∞; as x → −1− , one has f (x) → −∞. Since f (x) = 1/(1 − x 2 ) > 0 when −1 < x < 1, 1+x 1 1+x one has ln = 2y, so f is strictly increasing and the range of f is . (b) From y = ln 1−x 2 1−x 1+x = e2y . Then solve for x. 1−x

12. (a) f (x) = 2/(2x + 4) = (x + 2)−1 and f (x) = −(x + 2)−2 . We get f (0) = ln 4, f (0) = 1/2, and 1 f (0) = −1/4, so f (x) ≈ f (0) + f (0)x + 2 f (0)x 2 = ln 4 + x/2 − x 2 /8. (b) g (x) = −(1/2)(1 + x)−3/2 and g (x) = (3/4)(1 + x)−5/2 . We get g(0) = 1, g (0) = −1/2, and g (0) = 3/4, so g(x) ≈ 1 − x/2 + 3x 2 /8. (c) h (x) = e2x + 2xe2x and h (x) = 4e2x + 4xe2x . We get h(0) = 0, h (0) = 1, and h (0) = 4, so h(x) ≈ x + 2x 2 . 15. With x =
1 2
2 and n = 5, formula (7.6.6) yields e 2 = 1 + 1! + 1 ( 2 )6 c 6! e 1 ( 2 )6 6! 2 1 1 1 ( 2 )2 2!

+

1 ( 2 )3 3!

+

1 ( 2 )4 4!

+

1 ( 2 )5 5!

+

1 ( 2 )6 c 6! e ,

where c

1 1 < is some number between 0 and 2 . Now, R6 ( 2 ) = 1 1 c < e 2 < 2. Thus it follows that fact that c < 2 implies e
1 1

=

1 23040

≈ 0.00004340, where we used the

1 1 1 1 2 2 2 2 2 e 2 ≈ 1 + 1! + 2! + 3! + 4! + 5! = 1 + 2 + 1 + 48 + 384 + 3840 ≈ 1.6486979. Because the error is 8 1 less than 0.000043, the approximation e 2 ≈ 1.649 is correct to 3 decimal places.

( 1 )2

( 1 )3

( 1 )4

( 1 )5

23. (a) lim− (x 2 − 3x + 2) = 9 − 9 + 2 = 2 (b) Tends to +∞. x→3 √ 3 − x + 17 tends to +∞ as x → −1− , but to −∞ as x → −1+ , so there is no limit as x → −1. (c) x+1
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

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SINGLE-VARIABLE OPTIMIZATION

(2 − x)ex − x − 2 “0” −ex + (2 − x)ex − 1 “0” −ex − ex + (2 − x)ex = lim = lim = = x→0 x→0 x→0 x3 0 3x 2 0 6x 1 −xex −ex = lim = lim =− . x→0 6x x→0 6 6 (Cancelling x at the penultimate step avoids using l’Hôpital’s rule a third time.) 5 x 2 − 6x + 9 2x − 6 1 “0” “0” − 2 = lim 3 = = lim 2 = = (e) lim 2 − 3x + 18 x→3 x − 3 x→3 x − 4x x→3 3x − 8x − 3 x −x−6 0 0 1 1 2 x−4 1 “0” lim = (f) lim 2 = = lim = . (Can you ﬁnd another way?) x→3 6x − 8 x→4 2x − 32 x→4 4x 5 0 16 (x − 2)(x − 1) x 2 − 3x + 2 = = x − 1, which tends to 1 as x → 2. (g) If x = 2, then x−2 x−2 √ √ √ (4 − x + 17 )(4 + x + 17 ) 4 − x + 17 16 − x − 17 = (h) If x = −1, then = = √ √ 2x + 2 (2x + 2)(4 + x + 17 ) (2x + 2)(4 + x + 17 ) −1 1 which tends to − as x → −1. √ 16 2(4 + x + 17 ) (ln x)2 ln x 2 1 (i) lim = lim = 0, because of (7.12.3). x→∞ 3x 2 3 x→∞ x √ √ 24. When x → 0, the numerator tends to b − d and the denominator to 0, so the limit does not exist when d = b. If d = b, however, then √ √ 1 −1/2 − 1 c(cx + b)−1/2 a−c ax + b − cx + b “0” 2 a(ax + b) 2 = = lim = √ lim x→0 x→0 x 0 1 2 b (d) lim

Chapter 8 Single-Variable Optimization
8.2
√ √ √ 8(3x 2 + 4) − (8x)(6x) 8(2 − 3x)(2 + 3x) 2. h (x) = = , so h has stationary points at x1 = −2 3/3 2 2 (3x 2 + 4)2 √ (3x + 4) and x2 = 2 3/3. A sign diagram shows that h (x) < 0 in (−∞, x1 ) and in (x2 , ∞), whereas h (x) > 0 in (x1 , x2 ). Therefore h is strictly decreasing in (−∞, x1 ], strictly increasing in [x1 , x2 ], and strictly decreasing again√ [x2 , ∞). Then, because h(x) →√ as x → ±∞, it follows that the maximum of h in 0 occurs at x2 = 2 3/3 and the minimum at x1 = −2 3/3. 8. (a) y = ex − 2e−2x and y = ex + 4e−2x . Hence y = 0 when ex = 2e−2x , or e3x = 2, i.e. x = 1 ln 2. 3 Since y > 0 everywhere, the function is convex and this is a minimum point. (b) y = −2(x − a) − 4(x − b) = 0 when x = 1 (a + 2b). This is a maximum point since y = −6. 3 (c) y = 1/x − 5 = 0 when x = 1 . This is a maximum point since y = −1/x 2 < 0 for all x > 0. 5 10. (a) f (x) = k − Aαe−αx = 0 when x = x0 = (1/α) ln(Aα/k). Note that x0 > 0 if and only if Aα > k. Moreover f (x) = Aα 2 e−αx > 0 for all x ≥ 0, so x0 solves the minimization problem. (b) Substituting for A in the answer to (a) gives the expression for the optimal height x0 . Its value increases as p0 (probability of ﬂooding) or V (cost of ﬂooding) increases, but decreases as δ (interest rate) or k (marginal construction cost) increases. The signs of these responses are obviously what an economist would expect. (Not only an economist, actually.)
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 8

SINGLE-VARIABLE OPTIMIZATION

23

8.3
1 2. (a) π(Q) = Q(a−Q)−kQ = (a−k)Q−Q2 so π (Q) = (a−k)−2Q = 0 for Q = Q∗ = 2 (a−k). This ∗ ) = −( 1 (a − k))2 + (a − k) 1 (a − k) = maximizes π because π (Q) < 0. The monopoly proﬁt is π(Q 2 2 1 1 (a − k)2 . (b) dπ(Q∗ )/dk = − 2 (a − k) = −Q∗ , as in Example 3. 4 (c) The new proﬁt function is π(Q) = π(Q) + sQ = (a − k)Q − Q2 + sQ. Then π (Q) = ˆ ˆ 1 1 ˆ ˆ a − k − 2Q + s = 0 when Q = 2 (a − k + s). Evidently Q = 2 (a − k + s) = a − k provided s = a − k, which is the subsidy required to induce the monopolist to produce a − k units.

8.4
2. In all cases the maximum and minimum exist by the extreme value theorem. Follow the recipe in (8.4.1). (a) f (x) = −2 for all x in [0, 3], so the recipe tells us that both the maximum and minimum points are at the ends of the interval [0, 3]. Since f (0) = −1 and f (3) = −7, the maximum is at x = 0, the minimum at x = 3. (Actually the sign of f (x) alone implies that the maximum is at the lower end of the interval, and the minimum at the upper end.) (b) f (−1) = f (2) = 10 and f (x) = 3x 2 − 3 = 0 at x = ±1. The only stationary point in the interval [−1, 2] is x = 1, where f (1) = 6. There are two maxima at the endpoints, and a minimum at x = 1. (c) f (x) = x + 1/x, so f (1/2) = f (2) = 5/2 at the endpoints. Also, f (x) = 1 − 1/x 2 = 0 at x = ±1. 1 The only stationary point in the interval [ 2 , 2] is x = 1, where f (1) = 2. There are two maxima at the endpoints, and a minimum at x = 1. √ (d) At the endpoints one has f (−1) = 4 and f ( 5) = 0. Because f (x) = 5x 2 (x 2 − 3), there are two √ √ stationary points √ the interval [−1, 5] at x = 0 and x = 3. The values at these stationary points are in √ √ f (0) = 0 and f ( 3) = −6 3. The maximum is at x = −1 and the minimum is at x = 3. (e) f (x) = 3x 2 − 9000x + 6 · 106 = 3(x − 1000)(x − 2000) = 0 when x = 1000 and x = 2000. At these stationary points f (1000) = 2.5·109 and f (2000) = 2·109 . There is a minimum at the endpoint x = 0 and a maximum at x = 3000. 6. (a) (f (2) − f (1))/(2 − 1) = (4 − 1)/1 = 3 and f (x) = 2x, so√ (x) = 3 when x = x ∗ = 3/2. f √ (b) (f (1) − f (0))/1 = (0 − 1)/1 = −1 and f (x) = −x/ 1 − x 2 = −1 when x = 1 − x 2 . 1 Squaring each side of the last equation gives x 2 = 1 − x 2 √ so x 2 = 2 . This has two solutions and √ √ 1 1 x = ± 2 2, of which only the positive solution satisﬁes x = 1 − x 2 . So we require x = x ∗ = 2 2. (c) (f (6) − f (2))/4 = −1/6 and f (x) = −2/x 2 = −1/6 when −12/x 2 = −1 or x 2 = 12, and so √ √ √ x = ± 12. The required solution√ [2, 6] is x = x ∗ = 12 = 2 3. in √ √ 1 (d)√ (4) − f (0))/4 = 1/4 = ( 25 − 9)/4 = (5 − 3)/4 = 1/2 and f (x) = 2 2x/ 9 + x 2 = (f √ 2. x/ 9 + x 2 = 1/2 when 2x = 9 + x√ Squaring each side of the last equation gives 4x 2 = 9+x 2 and so √ 2 = 9. This has two solutions x = ± 3, of which only the positive solution satisﬁes x/ 9 + x 2 = 1/2. 3x √ So we require x = x ∗ = 3.

8.5
4. (i) π(Q) = 1840Q − (2Q2 + 40Q + 5000) = 1800Q − 2Q2 − 5000. Since π (Q) = 1800 − 4Q = 0 for Q = 450, and π (Q) = −4 < 0, it follows that Q = 450 maximizes proﬁts. (ii) π(Q) = 2200Q − 2Q2 − 5000. Since π (Q) = 2200 − 4Q = 0 for Q = 550, and π (Q) = −4 < 0, it follows that Q = 550 maximizes proﬁts. (iii) π(Q) = −2Q2 − 100Q − 5000. Here π (Q) = −4Q − 100 < 0 for all Q ≥ 0, so the endpoint Q = 0 maximizes proﬁts.
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

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CHAPTER 8

SINGLE-VARIABLE OPTIMIZATION

8.6
2. (a) Strictly decreasing, so no extreme points. (Actually the sign of f (x) alone implies that the maximum is at the lower end of the interval, and the minimum at the upper end.) (b) f (x) = 3x 2 − 3 = 0 for x = ±1. Because f (x) = 6x, we have f (−1) = −6 and f (1) = 6, so x = −1 is a local maximum point, and x = 1 is a local minimum point. (c) f (x) = 1 − 1/x 2 = 0 for x = ±1. With f (x) = 2/x 3 , we have f (−1) = −2 and f (1) = 2, so x = −1 is a local maximum point, and x = 1 is a local minimum point. (d) f (x) = 5x 4 − 15x 2 = 5x 2 (x 2 − 3) and f (x) = 20x 3 − 30x. There √ three√ are stationary points at √ √ √ x = 0 and x = ±√ 3. Because f (0) = 0, whereas f (− 3) = −20 · 3 3 + 30 3 = −30√3 < 0 √ √ and f ( 3) = 30 3 > 0, there is a local maximum at x = − 3 and a local minimum at x = 3. (e) This parabola has a local (and global) minimum at x = 3. (f) f (x) = 3x 2 + 6x = 3x(x + 2) and f (x) = 6x + 6. There are two stationary points at x = 0 and x = −2. Because f (0) = 6 and f (−2) = −6, there is a local maximum at x = −2 and a local minimum at x = 0. 3. See the graph in Fig. SM8.6.3. (a) The function f (x) is deﬁned if and only if x = 0 and x ≥ −6. f (x) = 0 at x = −6 and at x = −2. At any other point x in the domain, f (x) has the same sign as (x + 2)/x, so f (x) > 0 if x ∈ (−6, −2) or x ∈ (0, ∞). (b) We ﬁrst ﬁnd the derivative of f : f (x) = − 1 2√ x+2 −4x − 24 + x 2 + 2x (x + 4)(x − 6) x+6+ = = √ √ √ 2 2 x+6 x x 2 x+6 2x 2x 2 x + 6

By means of a sign diagram we see that f (x) > 0 if −6 < x < −4, f (x) < 0 if −4 < x < 0, f (x) < 0 if 0 < x < 6, f (x) > 0 if 6 < x. Hence f is strictly decreasing in [−4, 0) and in (0, 6], strictly increasing in [−6, −4] and in [6, ∞). It follows from the ﬁrst-derivative test (Thm. 8.6.1) that the two√ points x = −4 and x =√ are respectively a local maximum and a local minimum, with 6 √ 1 f (−4) = 2 2 and f (6) = 4 8 = 8 2/3. Also, according to the deﬁnition (8.6.1), the point x = −6 3 is another local minimum point. √ (c) Since limx→0 x + 6 = 6 > 0, while limx→0− (1 + 2/x) = −∞ and limx→0+ (1 + 2/x) = ∞, we see that limx→0− f (x) = −∞ and limx→0+ f (x) = ∞. Furthermore, lim f (x) = lim 1 x 2 − 2x − 24 ·√ = lim 2 x→∞ 2x x+6 y 8 6 4 2 −6 −4 −2 −2 −4
Figure SM8.6.3

x→∞

x→∞

1 1 12 − − 2 2 x x

1 ·√ x+6

=

1 ·0=0 2

2

4

6

8 10

x

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 8

SINGLE-VARIABLE OPTIMIZATION

25

7. f (x) = x 3 + ax + b → ∞ as x → ∞, and f (x) → −∞ as x → −∞. By the intermediate value theorem, the continuous function f has at least one real root. We have f (x) = 3x 2 + a. We consider two cases. First, in case a ≥ 0, one has f (x) > 0 for all x = 0, so f is strictly increasing, and there is only one real root. Note that 4a 3 + 27b2 ≥ 0 in this case. √ √ Second, in case a < 0, one has f (x) = 0 for x = ± −a/3 = ± p, where p = −a/3. Because √ √ f (x) = 3x, the function f has a local maximum at x = − p, where y = b + 2p p, and a local √ √ minimum at x = p, where y = b − 2p p. If either of these local extreme values is 0, the equation has a double root, which is the case if and only if 4p 3 = b2 , that is, if and only if 4a 3 + 27b2 = 0. Otherwise, the equation has three real roots if and only if the local maximum value is positive and the √ √ √ local minimum value is negative. This occurs ⇐⇒ b > −2p p and b < 2p p ⇐⇒ |b| < 2p p ⇐⇒ b2 < 4p3 ⇐⇒ 4a 3 + 27b2 < 0.

8.7
3. The answers given in the text can be found in a straightforward way by considering the signs of the following derivatives: (a) y = −e−x (1 + x), y = xe−x (b) y = (x − 1)/x 2 , y = (2 − x)/x 3 1 − 2 ln x 6 ln x − 5 (c) y = x 2 e−x (3 − x), y = xe−x (x 2 − 6x + 6) (d) y = ,y = 3 x x4 (e) y = 2ex (ex − 1), y = 2ex (2ex − 1) (f) y = e−x (2 − x 2 ), y = e−x (x 2 − 2x − 2)

Review Problems for Chapter 8
8. (a) The answer given in the text is easily found from the derivative h (x) = ex (2 − e2x ) . (2 + e2x )2 1 (b) In fact, h is strictly increasing in (−∞, 2 ln 2], which includes (−∞, 0]. Also lim h(x) = 0, and x→−∞ h(0) = 1/3. Thus, h deﬁned on (−∞, 0] has an inverse deﬁned on (0, 1/3] with values in (−∞, 0]. ex To ﬁnd the inverse, note that = y ⇐⇒ y(ex )2 − ex + 2y = 0. This quadratic equation in ex 2 + e2x has the roots ex = [1 ± 1 − 8y 2 ]/2y. We require the solution to satisfy x ≤ 0 and so ex ≤ 1 when 0 < y < 1/3. Now, taking the positive square root would give ex > 1/2y > 6 when 0 < y < 1/3. So we must have ex = [1 − √1 − 8y 2 ]/2y, and so x = ln(1 − 1 − 8y 2 ) − ln(2y). Using x as the free variable, h−1 (x) = ln(1− 1 − 8x 2 )−ln(2x). The function and its inverse are graphed in Fig. SM8.R.8. y 0.5 h −1.5 −1.0 −0.5 −0.5 h−1 −1.0 −1.5
Figure SM8.R.8

0.5

x

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

26

CHAPTER 9

INTEGRATION

√ 10. (a) Because 3 u is deﬁned for all real u, the only points x not in the domain are the ones for which √ √ √ x 2 −a = 0, or x = ± a. So the domain of f √ consists of all x = ± a. Since 3 u > 0 if and only if u > 0, 3 the denominator in the expression for f (x), x 2 − a, is positive if and only if x 2 > a, i.e. if and only if √ √ x < − a or x > a. The numerator in the expression for f (x) is x, and a sign diagram then reveals that √ √ x −x f (x) is positive in (− a, 0) and in ( a, ∞). Since f (−x) = 3 = −√ = −f (x), 3 2−a 2−a x (−x) the graph of√ is symmetric about the origin. f 3 (b) Writing x 2 − a as (x 2 − a)1/3 and then differentiating yields f (x) =
1 2 1 · (x 2 − a)1/3 − x · 1 (x 2 − a)−2/3 · 2x x 2 − a − x · 1 2x (x − 3a) 3 3 = = 3 2 2 − a)2/3 2 − a)4/3 (x (x (x − a)4/3

Here the second equality was obtained by multiplying both denominator and numerator by (x 2 − a)2/3 . √ Of course, f (x) is not deﬁned at ± a. Except at these points, the denominator is always positive (since √ √ √ 1 (x 2 − a)4/3 = ((x 2 − a)1/3 )4 ). The numerator, 3 (x 2 − 3a) = 1 (x + 3a)(x − 3a), is 0 at x = ± 3a, 3 √ √ √ and nonnegative in (−∞, − 3a]√ in [ 3a, ∞). Since f and f are not deﬁned at ± a, we ﬁnd that √ √ √ √ √ f (x) is increasing in (−∞, − 3a] and in [ 3a, ∞), decreasing in [− 3a, −√a), in (− a, a), and √ √ √ in ( a, 3a]. It follows that x = − 3a is a local maximum point and x = 3a is a local minimum point. (c) Differentiating once more, we ﬁnd that f (x) =
2 2 3 x(x 2 2 − a)4/3 − 1 (x 2 − 3a) · 4 (x 2 − a)1/3 · 2x 3 3 9 x(9a − x ) = (x 2 − a)8/3 (x 2 − a)7/3

Here the second equality was obtained by dividing each term in the denominator and the numerator by (x 2 − a)1/3 , then simplifying the numerator. The resulting expression for f (x) shows that there are √ √ inﬂection points where x equals −3 a, 0, and 3 a. (f (x) is 0 at these points, and changes sign around each of them.) 11. Note ﬁrst that f (x) → 0 as x → ±∞ (divide denominator and numerator by x 3 ). Differentiation yields 18x 2 (x 4 + x 2 + 2) − 6x 3 (4x 3 + 2x) −6x 2 (x 4 − x 2 − 6) −6x 2 (x 2 − 3)(x 2 + 2) = = (x 4 + x 2 + 2)2 (x 4 + x 2 + 2)2 (x 4 + x 2 + 2)2 √ so f is stationary when x√ 0 and when x = ± 3. Moreover, f changes sign from negative to positive = √ as x increases through − 3, then it switches back to negative as x increases through 3. It follows that √ √ x = 3 is a local (and global) maximum point, that x = − 3 is a local (and global) minimum point, and x = 0 is neither. (It is an inﬂection point.) Note moreover that f (−x) = −f (x) for all x, so the graph is symmetric about the origin. The graph of f is shown in Fig. A8.R.11 in the text. f (x) =

Chapter 9 Integration
9.1
4. (a) (b) (t 3 + 2t − 3) dt = (x −1)2 dx = t 3 dt + 2t dt −
1 3 dt = 4 t 4 + t 2 − 3t + C

(x 2 −2x +1) dx = 1 x 3 −x 2 +x +C. Alternative: Since 3

d (x −1)3 = 3(x −1)2 , dx

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 9

INTEGRATION

27

we have (c)

(x − 1)2 dx = 1 (x − 1)3 + C1 . This agrees with the ﬁrst answer, with C1 = C + 1/3. 3
1 (x 2 + x − 2) dx = 1 x 3 + 2 x 2 − 2x + C 3 1 (x + 2)3 dx = 4 x 4 + 2x 3 + 6x 2 + 8x + C,

(x − 1)(x + 2) dx =

(d) Either ﬁrst evaluate (x + 2)3 = x 3 + 6x 2 + 12x + 8, to get or: (f)
1 (x + 2)3 = 4 (x + 2)4 + C1 .

(e)

1 (e3x − e2x + ex ) dx = 1 e3x − 2 e2x + ex + C 3

x 3 − 3x + 4 dx = x

x2 − 3 +

4 dx = 1 x 3 − 3x + 4 ln |x| + C 3 x

5. (a) First simplify the integrand: (y − 2)2 dy = √ y

(y − 2)2 y 2 − 4y + 4 = = y 3/2 − 4y 1/2 + 4y −1/2 . From this we get √ √ y y

(y 3/2 − 4y 1/2 + 4y −1/2 ) dy = 2 y 5/2 − 8 y 3/2 + 8y 1/2 + C. 5 3 x3 1 = x2 − x + 1 − , so x+1 x+1 x3 x3 x2 dx = − + x − ln |x + 1| + C. x+1 3 2 x(1 + x 2 )15 dx =
1 32 (1

(b) Polynomial division: (c)

d (1 + x 2 )16 = 16(1 + x 2 )15 · 2x = 32x(1 + x 2 )15 , so dx

+ x 2 )16 + C.

13. f (x) =

1 1 1 1 (x −2 + x 3 + 2) dx = −x −1 + x 4 + 2x + C. With f (1) = we have = −1 + + 2 + C, 4 4 4 4 1 4 1 5 −1 so C = −1. Now integration yields f (x) = (−x + x +2x −1) dx = − ln x + x +x 2 −x +D. 4 20 1 1 + 1 − 1 + D, so D = − . With f (1) = 0 we have 0 = − ln 1 + 20 20

9.2
3

5. We do only (c) and (f):
3

(f)
2

√ √ 6. (a) f (x) = x 3 − 3x 2 + 2x and so f (x) = 3x 2 − 6x + 2 = 0 for x0 = 1 − 3/3 and x1 = 1 + 3/3. We see that f (x) > 0 ⇐⇒ x < x0 or x > x1 . Also, f (x) < 0 ⇐⇒ x0 < x < x1 . So f is (strictly) increasing in (−∞, x0 ] and in [x1 , ∞), and (strictly) decreasing in [x0 , x1 ].
1

1 + t dt = t −1

−2 3 2

1 2 2x

− 1 x 3 dx = 3

3 −2

( 1 x3 − 6
9 2

1 4 12 x ) 4 2

=

3 −2

1 3 12 x (2 5 2

27 − x) = − 12 +

32 12

=

5 12 .

1 ln(t − 1) + 2 t 2 = ln 2 +

−

= ln 2 +

(b) See the graph in the text.
0

f (x) dx =
0

1

(x 3 − 3x 2 + 2x) dx =

1 0

1 x4 1 − x3 + x2 = − 0 = . 4 4 4

9.3
1

4. (a)
0

(x p+q + x p+r ) dx =

1 0

x p+r+1 1 1 x p+q+1 + = + p+q +1 p+r +1 p+q +1 p+r +1 (12x 2 − 6x) dx = 4x 3 − 3x 2 + C,

(b) Equality (i) implies a +b = 6. Also, f (x) = 2ax +b, so equality (ii) implies 2a +b = 18. It follows that a = 12 and b = 6, so f (x) = 12x 2 − 6x. But then f (x) =
2

and since we want
0

(4x 3 − 3x 2 + C) = 18, we must have 16 − 8 + 2C = 18, hence C = 5.

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

28

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INTEGRATION
1 1 1

5. (a) See the text.

(b) (x 2 + 2)2 dx = (x 4 + 4x 2 + 4) dx = 0 √ 0 1 2 1 x +x+ x+1 x(x + 1) + (x + 1)1/2 (c) dx = dx = x+1 x+1 0 0 1 √ √ 1 1 3 ( 2 x 2 + 2(x + 1)1/2 ) = 2 + 2 2 − 2 = 2 2 − 2
0

0 1 0

( 1 x 5 + 4 x 3 + 4x) = 83/15 5 3 (x + (x + 1)−1/2 ) dx =

(d) A

x+c+b−c d A(b − c) d x+b d + =A + =A+ + . Now integrate. x+c x x+c x x+c x

9. W (T ) = K(1 − e− T )/ T . Here W (T ) → 0 as T → ∞, and by l’Hôpital’s rule, W (T ) → K as T → 0+ . For T > 0, we ﬁnd W (T ) = Ke− T (1 + T − e T )/ T 2 < 0 because e T > 1 + T (see Problem 6.11.11). We conclude that W (T ) is strictly decreasing and that W (T ) ∈ (0, K). 2 10. (a) f (x) = √ > 0 for all x > 0. Also, f (x) → −∞ as x → 0, whereas √ x + 4 ( x + 4 − 2) f (x) → ∞ as x → ∞. It follows that f is strictly increasing on (0, ∞), with range equal to . Hence f √ √ has an inverse deﬁned on . To ﬁnd the inverse, note that y = 4 ln( x + 4 −2) ⇐⇒ ln( x + 4 −2) = √ y/4 ⇐⇒ x + 4 = ey/4 + 2 ⇐⇒ x + 4 = (ey/4 + 2)2 ⇐⇒ x = ey/2 + 4ey/4 . It follows that the inverse is g(x) = ex/2 + 4ex/4 . (b) See Fig. A9.3.10. (c) In Fig. A9.3.10 the graphs of f and g are symmetric about the line y = x, so area A = area B. But area B is the difference between the area of a rectangle with base a and height 10, and the area below the graph of g over the interval [0, a]. Therefore, A = B = 10a −
0 a

(ex/2 + 4ex/4 ) dx = 10a − 2ea/2 − 16ea/4 + 2 + 16

√ √ √ √ Because a = f (10) = 4 ln( 14 − 2), we have ea/2 = ( 14 − 2)2 = 14 − 4 14 + 4 = 18 − 4 14 and √ also ea/4 = 14 − 2. Hence, √ √ √ √ A = B = 10a − 2(18 − 4 14) − 16( 14 − 2) + 18 = 40 ln( 14 − 2) + 14 − 8 14 ≈ 6.26

9.4
2. (a) Let n be the total number of individuals. The number of individuals with income in the interval [b, 2b] 2b 2b 2b nB is then N = n . Their total income is M = n Br −2 dr = n −Br −1 = Br −2 r dr = 2b b b b
2b

n b Br −1 dr = n

2b b

B ln r = nB ln 2. Hence the mean income is m = M/N = 2b ln 2.
2b b

(b) Total demand is x(p) =
2b

nD(p, r)f (r) dr = b 2b

nApγ r δ Br −2 dr = nABp γ b 2b

r δ−2 dr =

nABpγ b r δ−1 δ−1

= nABpγ bδ−1

2δ−1

−1 . δ−1

9.5
1. (a) See the text. (c) (b)
1 3xe4x dx = 3x · 4 e4x − 1 3 3 · 4 e4x dx = 4 xe4x − 3 4x 16 e

+C xe−x dx.

(1 + x 2 )e−x dx = (1 + x 2 )(−e−x ) −

2x(−e−x ) dx = −(1 + x 2 )e−x + 2

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 9

INTEGRATION

29

Using the answer to (a) to evaluate the last integral, we get (1 + x 2 )e−x dx = −(1 + x 2 )e−x − 2xe−x − 2e−x + C = −(x 2 + 2x + 3)e−x + C (d)
1 xln x dx = 2 x 2 ln x − 1 21 2x x 1 dx = 2 x 2 ln x − 1 2x 1 1 dx = 2 x 2 ln x − 4 x 2 + C

2. (a) See the text. If follows that
2 0

d (b) Recall that dx 2x = 2x ln 2, and therefore 2x / ln 2 is the indeﬁnite integral of 2x . x 2x − ln 2
2 0

x2x dx =

2 0

8 2x dx = − ln 2 ln 2

2 0

4 3 2x 8 1 8 − − = − = 2 2 2 (ln 2) ln 2 (ln 2) (ln 2) ln 2 (ln 2)2

(c) First use integration by parts on the indeﬁnite integral. By (9.5.1) with f (x) = x 2 and g(x) = ex , (∗) x 2 ex dx = x 2 ex − 2xex dx. To evaluate the last integral we must use integration by parts once 2xex dx = 2xex − 2ex dx = 2xex − (2ex + C).
1 0

more. With f (x) = 2x and g(x) = ex , we get Inserted into (∗) this gives
1

x 2 ex dx = x 2 ex − 2xex + 2ex + C, and hence,
1

x 2 ex dx =
1 0

1 0

(x 2 ex −

2xex + 2ex ) = (e − 2e + 2e) − (0 − 0 + 2) = e − 2. Alternatively, and more compactly, using formula (9.5.2):
0

x 2 ex dx =

1 0

x 2 ex − 2
0

xex dx = e − 2

1 0

xex −
0

1

ex dx = e − 2 e −

ex = e − 2.

√ (d) We must write the integrand in the form f (x)g (x). If we let f (x) = x and g (x) = 1 + x = (1 + x)1/2 , then what is g? A certain amount of reﬂection should suggest choosing g(x) = 2 (1 + x)3/2 . 3 Using (9.5.2) then gives
3 0

√ x 1 + x dx =

3 0

x · 2 (1 + x)3/2 − 3
2 3

3 0

1 · 2 (1 + x)3/2 dx 3

4 4 + x)5/2 = 16 − 15 (45/2 − 1) = 16 − 15 · 31 = 7 11 15 √ Alternatively, we could have found the indeﬁnite integral of x 1 + x ﬁrst, and then evaluated the deﬁnite integral by using deﬁnition (9.2.3) of the deﬁnite integral. Figure SM9.5.2(d) shows the area under the √ graph of y = x 1 + x over the interval [0, 3], and you should ask yourself if 7 11 is a reasonable estimate 15 of the area of A.

=3·

· 43/2 −

2 3 2 3 0 5 (1

y 6 5 4 3 2 1 1 2 3 A

√ y =x 1+x

4

5

x

Figure SM9.5.2(d)

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

30

CHAPTER 9

INTEGRATION
T T

6. (a) By formula (9.5.2), −T −rT 1 e + r r
T T 0 0

te−rt dt =

t
0

−1 −rt 1 e = 2 (1 − (1 + rT )e−rT ). Multiply this expression by b. r r
T 0

−1 −rt e − r

T 0

−1 −rt −T −rT 1 e dt = e + r r r

T 0

e−rt dt =

(b)
0 T

(a + bt)e−rt dt = a (a − bt + ct )e
2 −rt 2

e−rt dt + b
T 0

T

te−rt dt, and so on using (a).
T 0

(c)
0

dt = a
0

e
−rt

−rt

dt − b
T 0

te−rt dt + c
0 −rt

T

t 2 e−rt dt. Use the previous results
T 0

T 0

and

t e

2 −rt

dt =

T

t (−1/r)e
0

−

2t (−1/r)e

dt = −(1/r)T 2 e−rT + (2/r)

te−rt dt.

9.6
2. (a) See the text.
1 x 3 +2 +C. 3e

(b) With u = x 3 +2 we get du = 3x 2 dx and

x 2 ex

3 +2

dx =

1 u 3 e du

1 = 3 eu +C =

ln u ln(x + 2) dx = du. 2x + 4 2u This does not look simpler than the original integral. A better idea is to substitute u = ln(x + 2). Then dx ln(x + 2) 2 2 1 1 1 du = and dx = 2 u du = 4 (u) + C = 4 (ln(x + 2)) + C. x+2 2x + 4 √ √ (d) First attempt: u = 1+x. Then, du = dx, and x 1 + x dx = (u−1) u du = (u3/2 −u1/2 ) du (c) First one might try u = x +2, which gives du = dx and √ 2 5/2 2 3/2 2 2 u − u + C = (1 + x)5/2 − (1 + x)3/2 + C. Second attempt: u = 1 + x. Then u2 = 1 + x 5 3 5 3√ and 2udu = dx. Then the integral is x 1 + x dx = (u2 − 1)u2u du = (2u4 − 2u3 ) du, and so on. Check that you get the same answer. Actually, even integration by parts works in this case. See Problem 9.5.2(d). x3 x2 · x dx = dx = (e) With u = 1 + x 2 one has x 2 = u − 1, and du = 2x dx, so (1 + x 2 )3 (1 + x 2 )3 1 u−1 −1 1 1 1 1 + du = 2 (u−2 − u−3 ) du = − 2 u−1 + 4 u−2 + C = + C. 2 3 2) u 2(1 + x 4(1 + x 2 )2 √ (f) With u = 4 − x 3 , u2 = 4−x 3 , and 2u du = −3x 2 dx, so x 5 4 − x 3 dx = x 3 4 − x 3 x 2 dx = = (4−u2 ) u (− 2 )u du = 3 6. (a) I =
0 1 8 2 2 (− 8 u2 + 2 u4 ) du = − 9 u3 + 15 u5 +C = − 8 (4−x 3 )3/2 + 15 (4−x 3 )5/2 +C. 3 3 9 1 0

(x 4 − x 9 )(x 5 − 1)12 dx =
0 0

−x 4 (x 5 − 1)13 dx. Introduce u = x 5 − 1. Then du = 5x 4 dx.

Now we use (9.6.2) along with the facts that u = −1 when x = 0 and u = 0 when x = 1. The integral
1 13 1 14 1 becomes I = − 5 u du = − 70 u = 70 . −1 √ −1 ln x (b) With u = x one has u2 = x and 2u du = dx. Then √ dx = 2 ln u2 du = 4 ln u du = √ √ √ √x √ 4(u ln u − u) + C = 4 x ln x − 4 x + C = 2 x ln x − 4 x + C. (Integration by parts also works √ in this case, with f (x) = ln x and g (x) = 1/ x.) √ (c) With u = 1 + x one has (u − 1)2 = x, so 2(u − 1) du = dx. Again we use (9.6.2) along with the facts that u = 1 when x = 0 and u = 3 when x = 4. The speciﬁed integral becomes 3 3 3 √ 2(u − 1) (u1/2 − u−1/2 ) du = 2 ( 2 u3/2 − 2u1/2 ) = 8 . (The substitution u = 1 + x du = 2 √ 3 3 u 1 1 1 also works.)

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 9

INTEGRATION

31

7. (a) With u = 1 + e

√ 1 √ · e x dx. Now x = 1 gives u = 1 + e and x = 4 gives 2 x √ 4 1+e2 1+e2 e x 2 du 2 . Thus √ =2 u=1+e ln u = 2 ln(1 + e2 ) − 2 ln(1 + e). dx = √ u x (1 + e x ) 1 1+e 1+e (b) A natural substitution is u = ex + 1 leading to du = ex dx and so dx = du/ex = du/(u − 1). 1/3 e1/3 +1 dx 1 1/3 + 1. Thus, When x = 0, u = 2, when x = 1/3, u = e = du = x +1 e u(u − 1) 0 2 1/3 +1 1/3 +1 e e 1 1 − du = ln |u − 1| − ln |u| = 1 − ln(e1/3 + 1) + ln 2 = ln 2 − ln(e−1/3 + 1) 3 u−1 u 2 2 1 1/3 + 1) = ln[e1/3 /(e1/3 + 1)] = − ln(e−1/3 + 1). because 3 − ln(e e−x , the suggested substitution t = e−x (or even better u = 1+e−x ), Rewriting the integrand as 1 + e−x dt = −e−x dx works well. Verify that you get the same answer. √ x

one has du =

(c) With z4 = 2x − 1 one has 4z3 dz = 2 dx. Also x = 8.5 gives z = 2 and x = 41 gives z = 3. The
3

integral becomes 7 + 2 ln 2.
2

2z3 dz =2 z2 − z

3 2

z2 dz =2 z−1

3

z+1+
2

1 dz = 2 z−1

3 2

1 2 2 z +z+ln(z−1)

dz =

9.7
3. (a) See the text.
∞ 0

(b) Using a simpliﬁed notation and the result in Example 1(a), we have
∞ 0

(x − 1/λ)2 λe−λx dx = −
2 ∞

(x − 1/λ)2 e−λx +
∞ 0

∞

2 (x − 1/λ) e−λx dx

= 1/λ + 2
0 ∞

xe

−λx

dx − (2/λ)
0 ∞ 0

e

−λx

dx = 1/λ2 + 2/λ2 − 2/λ2 = 1/λ2 , where we have used
∞ 0

both the result of Example 1(a) and part (a) in order to derive the penultimate equality. (c)
0

(x − 1/λ)3 λe−λx dx = −
3 ∞ 0 2

(x − 1/λ)3 e−λx +
3

3 (x − 1/λ)2 e−λx dx

= −1/λ + (3/λ)

(x − 1/λ) λe

−λx

dx = −1/λ + (3/λ)(1/λ2 ) = 2/λ3 , where we have used the

result of part (b) in order to derive the penultimate equality. 5. (a) f (x) = (1 − 3 ln x)/x 4 = 0 at x = e1/3 , with f (x) > 0 for x < e1/3 and f (x) < 0 for x > e1/3 . Hence f has a maximum at (e1/3 , 1/3e). Since f (x) → −∞ as x → 0+ , there is no minimum. Note that f (x) → 0 as x → ∞. (Use l’Hôpital’s rule.) b (b) a x −3 ln x dx = −

b

b = 1 and a → 0. But

a ∞ −3 ln x 1 x

1 −2 x ln x + 2

b a

1 −3 x dx = 2

b a

1 1 − x −2 ln x − x −2 . This diverges when 2 4 √ 1/ x + 2 dx and I2 = √ = 2 5, and

dx = 1/4.
3 −2+ε

7. Provided that both limits exist, the integral is the sum of I1 = lim+
3−ε ε→0

lim+

−2 ε→0

1/ 3 − x dx. Here I1 = lim+ ε→0 3−ε −2 ε→0

√

3 −2+ε

√ 2 x+2

√ √ = lim+ 2 5 − 2 ε ε→0 ε→0

I2 = lim+

√ √ √ √ −2 3 − x = lim+ −2 ε + 2 5 = 2 5.

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

32

CHAPTER 9

INTEGRATION

√ √ √ 12. The substitution u = (x − μ)/ 2σ gives du = dx/σ 2, and so dx = σ 2 du. It is used in (a)–(c). +∞ +∞ 1 2 f (x) dx = √ e−u du = 1, by (9.7.8). (a) π −∞ −∞ +∞ +∞ √ 1 2 xf (x) dx = √ (μ + 2σ u)e−u du = μ, using part (a) and Example 3. (b) π −∞ −∞ +∞ +∞ +∞ 1 2 2 √ 2 (c) (x − μ)2 f (x) dx = 2σ 2 u2 √ e−u σ 2 du = σ 2 √ u2 e−u du. Now inteπ −∞ σ 2π −∞ −∞ +∞ √ 2 2 2 2 1 1 1 u2 e−u du = 2 π. Hence the gration by parts yields u2 e−u du = − 2 ue−u + 2 e−u du, so integral equals σ 2 .
−∞

9.8
10. (a) If f = r, the equation can be rewritten as x = (r − f )x 1 − ˙ x . There are two constant (1 − f/r)K solutions x ≡ 0 and x ≡ (1 − f/r)K, though the latter is negative, so biologically meaningless, unless x ¯ . Using (9.8.7), the ¯ ¯ f ≤ r. With r = r − f and K = (1 − f/r)K, the equation is x = r x 1 − ¯ K solution is ¯ (1 − f/r)K K = x(t) = ¯ (1 − f/r)K − x0 −(r−f )t K − x0 −¯ t 1+ e 1+ e r x0 x0 In the special case when f = r, the equation reduces to x = −rx 2 /K. Separating the variables gives ˙ 2 = (r/K) dt, and integration gives 1/x = rt/K + C. If x(0) = x , we get C = 1/x , so the −dx/x 0 0 1 solution is x = → 0 as t → ∞. rt/K + 1/x0 (b) When f > r, the solution to the differential equation given in part (a) is still valid even though both ¯ K and r are negative. Because the ﬁsh rate f exceeds the replenishment rate r, however, the ﬁsh stock ¯ steadily declines. Indeed, as t → ∞ one has e−(r−f )t → ∞ and in fact the solution of the equation satisﬁes x(t) → 0. That is, the ﬁsh stock tends to extinction.

9.9
2. (a) dx/dt = e2t /x 2 . Separate: x 2 dx =
1 e2t dt. Integrate: 1 x 3 = 2 e2t + C1 . Solve for x: 3
3

3 3 x 3 = 2 e2t + 3C1 = 2 e2t + C, with C = 3C1 . Hence, x =

3 2t 2e

+ C. (It is important to put in the
3

1 3 constant at the integration step. Adding it later leads to an error: 1 x 3 = 2 e2t , x 3 = 2 e2t , x = 3 This is a solution only if C = 0, and is not the general solution.)

3 2t 2 e +C.

(b) dx/dt = e−t ex , so

e−x dx =

e−t dt. Integrate: −e−x = −e−t +C1 . Solve for x: e−x = e−t +C,

with C = −C1 . Hence, −x = ln(e−t + C), so x = − ln(e−t + C). (c) Directly from (9.9.3). (d) Similar to (a). (e) By (9.9.5), x = Ce2t + e2t
1 2

(−t)e−2t dt = Ce2t − e2t

te−2t dt. Here

1 te−2t dt = t (− 2 )e−2t +

1 1 1 1 1 1 e−2t dt = (− 2 t − 4 )e−2t and thus x = Ce2t − e2t (− 2 t − 4 )e−2t = Ce2t + 2 t + 4 .

(f) By (9.9.5), x = Ce−3t + e−3t

e3t tet

2 −3t

dt = Ce−3t + e−3t

1 tet dt = Ce−3t + 2 et
2

2 −3t

.

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

CHAPTER 9

INTEGRATION

33

5. (a) See the text. Hence, K 1−α = K 1−α

(b)

K −α dK =

γ L0 (1 − α) βt e β (1 − α)γ L0 βt 1−α (e − 1) + K0 , from which we ﬁnd K. = β

1 γ L0 βt K 1−α = e + C1 . 1−α β γ L0 (1 − α) 1−α + (1 − α)C1 , so + (1 − α)C1 . At t = 0 one has K0 = β γ L0 eβt dt, so

Review Problems for Chapter 9
1

4. (a) 5/4. (Example 9.7.2.) e (b)
0 2

1 (1 + x 4 )5 = 31/20 20 e 1 e 1

∞

(c)
0

5te−t −
0

∞

5e−t dt = 5

∞ 0

e−t = −5

(d)
1 2

(ln x) dx =
2

e 1

x(ln x) − 2

ln x dx = e − 2 (x ln x − x) = e − 2

0 1 2 3 1 2 1 (e) (x + 1)3/2 = (93/2 − 1) = 52/9 (f) ln(e3z + 5) = (ln 6 − ln 5) = ln(6/5) 9 3 3 0 9 −∞ 3 e/2 e/2 1 1 1 1 (g) x 4 ln(2x) − x 3 dx = (e/2)4 − [(e/2)4 − (1/2)4 ] = (1/256)(3e4 + 1) 4 1/2 4 1/2 4 16

(h) Introduce u =
∞

√

x. Then

u2

= x, so 2u du = dx, and
1

∞

e− x √ dx = x

√

∞ 1

e−u 2u du = u

2
1

e−u du = 2e−1 .

5. (a) With u = 9 +

√ x one has x = (u − 9)2 and so dx = 2(u − 9) du. Also, u = 9 when x = 0 and 25 14 14 14 18 1 2(u − 9) du = du = 10 − 18 ln . 2− u = 14 when x = 25. Thus √ dx = u u 9 9+ x 0 9 9 √ 2 − 2 and so dt = 2u du. Also, u = 2 when t = 2 and u = 3 when (b) With u = t + 2 one has t = u 7 √ 3 3 3 1 5 2 3 = t t + 2 dt = (u2 − 2)u · 2u du = 2 (u4 − 2u2 ) du = 2 t = 7. Hence 5u − 3u 2 − 32 − 16 = 422 − 76 = 886/15. 3 5 3 √ 5 3 (c) With u = 19x 3 + 8 one has u3 = 19x 3 + 8 and so 3u2 du = 57x 2 dx. Also, x = 0 gives u = 2 and
243 5 54 3

−

2

2

2

2

x = 1 gives u = 3. Therefore
0

1

57x 2 19x 3 + 8 dx =
3

3

3u3 du =

3 2

From the equilibrium condition f (Q∗ ) = 10. (a) As in Example 9.4.3, ﬁrst we need to ﬁnd P ∗ and 100 − 0.05Q∗ = g(Q∗ ) = 0.1Q∗ + 10, we obtain 0.15Q∗ = 90, and so Q∗ = 600. Then P ∗ = g(Q∗ ) = 0.1Q∗ + 10 = 70. Moreover, 600 600 600 0.05 2 (f (Q) − P ∗ ) dQ = (30 − 0.05q) dq = CS = 30Q − Q = 9000 2 0 0 0 PS =
0 600

2 Q∗ .

3 4 4u

= 195/4.

(P ∗ − g(Q)) dQ =
0

600

(60 − 0.1Q) dQ = 0.1Q∗ .

600 0

60Q −

0.1 2 Q = 18000 2

(b) Equilibrium occurs when + 5) = 4.5 + Clearing fractions and then simplifying, we ∗ )2 + 50Q∗ − 275 = 0. The only positive solution is Q∗ = 5, and then P ∗ = 5. obtain (Q CS =
0 5

50/(Q∗

50 − 5 dQ = Q+5

5 0

[50 ln(Q + 5) − 5Q] = 50 ln 2 − 25
5 0

PS =
0

5

(5 − 4.5 − 0.1Q) dQ =

(0.5Q − 0.05Q2 ) = 2.5 − 1.25 = 1.25

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11. (a) f (t) = 4

2 ln t · (1/t) · t − (ln t)2 · 1 (2 − ln t) ln t =4 , and 2 t t2 [2 · (1/t) − 2 ln t · (1/t)] t 2 − [2 ln t − (ln t)2 ] 2t (ln t)2 − 3 ln t + 1 =8 f (t) = 4 t4 t3 (b) f (t) = 0 ⇐⇒ ln t (2 − ln t) = 0 ⇐⇒ ln t = 2 or ln t = 0 ⇐⇒ t = e2 or t = 1. But f (1) = 8 > 0 and f (e2 ) = −8e−6 , so t = 1 is a local minimum point and t = e2 ≈ 7.4 is a local maximum point. We ﬁnd f (1) = 0 and f (e2 ) = 16e−2 ≈ 2.2. e2 (ln t)2 2 ], so the area is A = 4 (c) The function is positive on [1, e dt. With u = ln t as a new variable, t 1 2 2 1 du = 1 dt. When t = 1 then u = 0, and t = e2 implies u = 2. Hence, A = 4 0 u2 du = 4 0 3 u3 = 32 . t 3 x −2 dx =
1 1 t dt, and so −1/x = 2 t 2 + C1 , or x = 1/(C − 2 t 2 ) (with C = −C1 ).

13. (a) Separable.

(b) and (c): Direct use of (9.9.3). 1 1 t e5t − 5 5 e5t dt =

(d) Using (9.9.5), x = Ce−5t + 10e−5t

te5t dt. Here
1 5t 25 e )

te5t dt =

1 5t 1 te − e5t . Thus x = Ce−5t + 10e−5t ( 1 te5t − 5 5 25 et/2 et dt = Ce−t/2 +e−t/2 t 2 e3t dt = Ce−3t + e−3t
1 3t 3 te

= Ce−5t + 2t − 2 . 5

(e) x = Ce−t/2 +e−t/2 (f) x = Ce−3t + e−3t

2 2 e3t/2 dt = Ce−t/2 +e−t/2 e3t/2 = Ce−t/2 + et . 3 3 1 2 3t 2 t e − te3t dt 3 3 = Ce−3t + 1 t 2 − 2 t + 3 9
2 27 .

= Ce−3t + 1 t 2 − 2 e−3t 3 3

−

1 3

e3t dt

16. (a) and (b), see the text. (c) F (x) = f (x) = −λ2 ae−λx (e−λx + a)−2 + 2λ2 ae−2λx (e−λx + a)−3 = aλ2 e−λx (e−λx −a)(e−λx +a)−3 . Note that F (x) = 0 for e−λx = a, i.e. for x0 = −(ln a/λ). Since F (x) changes sign about x0 = − ln(a/λ), this is an inﬂection point. F (x0 ) = F (−(ln a/λ)) = a/(a + a) =
∞

1/2. See the graph in Fig. A9.R.16. a→−∞ b→∞

(d)

lim [F (0) − F (a)] + lim [F (b) − F (0)] = 1, by (a).

−∞

f (x) dx =

0

a→−∞ a

lim

f (x)dx + lim

b

b→∞ 0

f (x) dx =

Chapter 10 Interest Rates and Present Values
10.2
6. With g(x) = (1 + r/x)x for all x > 0 one has ln g(x) = x ln(1+r/x). Differentiating gives g (x)/g(x) = ln(1 + r/x) + x(−r/x 2 )/(1 + r/x) = ln(1 + r/x) − (r/x)/(1 + r/x), as claimed in the problem. Putting h(u) = ln(1 + u) − u/(1 + u), one has h (u) = u/(1 + u)2 > 0 for u > 0, so h(u) > 0 for all u > 0, implying that g (x)/g(x) = h(r/x) > 0 for all x > 0. So g(x) is strictly increasing for x > 0. Because g(x) → er as x → ∞, it follows that g(x) < er for all x > 0. Continuous compounding of interest is best for the lender.

10.4
6. Let x denote the number of years beyond 1971 that the extractable resources of iron will last. We need to solve the equation 794 + 794 · 1.05 + · · · + 794 · (1.05)x = 249 · 103 . Using (10.4.3), 794[1 − (1.05)x+1 ]/(1 − 1.05) = 249 · 103 or (1.05)x+1 = 249 · 103 · 0.05/794 + 1 = 12450/794 + 1 ≈
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35

16.68. Using a calculator, we ﬁnd x ≈ (ln 16.68/ ln 1.05)−1 ≈ 56.68, so the resources will be exhausted part way through the year 2028. 8. (a) The quotient of this inﬁnite series is e−rt , so the sum is f (t) = (b) f (t) = P (t)e−rt P (t) . = rt 1 − e−rt e −1

P (t)(ert − 1) − P (t)rert , and t ∗ > 0 can only maximize f (t) if f (t ∗ ) = 0, that is, if (ert − 1)2 r P (t ∗ ) ∗ ∗ P (t ∗ )(ert − 1) = rP (t ∗ )ert , which implies that = . ∗) P (t 1 − e−rt ∗ r 1 1 “0” = lim ∗ −rt ∗ = ∗ = (c) lim −rt ∗ r→0 1 − e r→0 t e 0 t 11. See Fig. SM10.4.11. If p > 1, then ∞ (1/np ) = 1+ ∞ (1/np ) is ﬁnite because ∞ (1/np ) is the n=1 n=2 n=2 sum of the shaded rectangles, and this sum is certainly less than the area under the curve y = 1/x p over [1, ∞), which is equal to 1/(p − 1). If p ≤ 1, the sum ∞ (1/np ) is the sum of the larger rectangles in n=1 the ﬁgure, and this sum is larger than the area under the curve y = 1/x p over [1, ∞), which is unbounded when p ≤ 1. Hence, ∞ (1/np ) diverges in this case. n=1 y 1 y = 1/x p

1
Figure SM10.4.11

2

3

4

x

10.7
5. According to (10.7.1), the internal rate must satisfy −100 000 + 10 000 10 000 10 000 + ··· + =0 + (1 + r)2 (1 + r)20 1+r

After dividing all the terms by 10 000, and putting s = 1/(1 + r), we have to show that the equation f (s) = s 20 + s 19 + · · · + s 2 + s − 10 = 0 has a unique positive solution. Since f (0) = −10 and f (1) = 10, by the intermediate value theorem (Theorem 7.10.1), there exists a number s ∗ between 0 and 1 such that f (s ∗ ) = 0. This s ∗ is the unique positive root because f (s) > 0 for all s > 0. In fact, from (10.4.3), f (s) = −10 + (s − s 21 )/(1 − s), and f (s ∗ ) = 0 ⇐⇒ (s ∗ )21 − 11s ∗ + 10 = 0. Problem 7.R.26 asks for an approximation to the unique root of this equation in the interval (0, 1). The answer is 3 s = s ∗ = 0.928, so r ∗ = 1/s ∗ − 1 ≈ 0.0775, which means that the internal rate of return is about 7 4 %.

Review Problems for Chapter 10
8. (a) See the text. (b) We use formula (10.5.3) on the future value of an annuity: (5000/0.04)[(1.04)4 − 1] = 21 232.32 (c) The last of seven payments will be on 1st January 2006, when the initial balance of 10 000 will have earned interest for 10 years. So K must solve 10 000 · (1.04)10 + K[(1.04)8 − 1]/0.04 = 70 000. We ﬁnd that K ≈ 5990.49.
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

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CHAPTER 11

FUNCTIONS OF MANY VARIABLES
√ t/2 −rt

11. (a) f (t) = 100e

1 √ − r . We see that f (t) = 0 for t = t ∗ = 1/16r 2 . Since f (t) > 0 for 4 t t < t ∗ and f (t) < 0 for t > t ∗ , it follows that t ∗ maximizes f (t). √ 1 (b) f (t) = 200e−1/t e−rt 2 − r . We see that f (t) = 0 for t = t ∗ = 1/ r. Since f (t) > 0 for t < t ∗ t and f (t) < 0 for t > t ∗ , it follows that t ∗ maximizes f (t). e

Chapter 11 Functions of Many Variables
11.2
5. (a)–(c) are easy. (d) z = x y = (eln x )y = ey ln x = eu with u = y ln x. Then zx = eu ux = x y (y/x) = yx y−1 . Similarly zy = eu uy = x y ln x. Moreover, zxx = (∂/∂x)(yx y−1 ) = y(y − 1)x y−2 . (When differentiating x y−1 partially w.r.t. x, one treats y as a constant, so the rule dx a /dx = ax a−1 applies.) Similarly zyy = (∂/∂y)(x y ln x) = x y (ln x)2 and zxy = (∂/∂y)(yx y−1 ) = x y−1 + yx y−1 ln x.

11.3
9. (a) It might help to regard the ﬁgure as a contour map of a mountain, whose level curves join points at the same height above mean sea level. Near P the terrain is rising in the direction of the positive x-axis, so fx (P ) > 0, and it is also sloping down in the direction of the positive y-axis so fy (P ) < 0. Near Q, the terrain slopes in the opposite directions. Hence fx (Q) < 0 and fy (Q) > 0. (b) (i) The line x = 1 has no point in common with any of the given level curves. (ii) The line y = 2 intersects the level curve z = 2 at x = 2 and x = 6 (approximately). (c) If you start at the point (6, 0) and move up along the line 2x + 3y = 12, the ﬁrst marked level curve you meet is z = f (x, y) = 1. Moving further you meet level curves with higher z-values. The level curve with the highest z-value you meet is z = 3, which is the level curve that just touches the straight line. 10. The stated inequalities on the partial derivatives imply that F (1, 0) − F (0, 0) =
0 1 1 0

F1 (x, 0) dx ≥

2 dx = 2 F (0, 1) − F (0, 0) =
1 0 1 0

F (2, 0) − F (1, 0) =
1

2

F1 (x, 0) dx ≥ 2;

F2 (0, y) dy ≤ 1

F (1, 1) − F (0, 1) =
0

1

F1 (x, 1) dx ≥ 2; F (1, 1) − F (1, 0) =

F2 (1, y) dy ≤ 1

11.5
3. (a) In the ﬁrst week it buys 120/50 = 2.4 million shares, followed successively by 120/60 = 2 million shares, 120/45 = 2.667 million, 120/40 = 3 million, 120/75 = 1.6 million, and ﬁnally 120/80 = 1.5 million in the sixth week. The total is 13.167 million shares. (b) The arithmetic mean price is $350/6 = 58.33. But the total cost at that price of the 13.167 million shares which the fund has acquired would be $13.167×58.33 = 768.031 million. So using the arithmetic mean price would overstate the fund’s cost by $48.031 million. A more accurate statement of the mean price is $720 / 13.167 = 54.68 per share. Routine arithmetic shows that this is the harmonic mean of the six prices deﬁned by formula (c) of Example 2.
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37

11.6
2. (a)–(d) are routine. (e) f (x, y, z) = (x 2 + y 3 + z4 )6 = u6 , with u = x 2 + y 3 + z4 . Then f1 = 6u5 u1 = 6(x 2 + y 3 + z4 )5 2x = 12x(x 2 + y 3 + z4 )5 , f2 = 6u5 u2 = 6(x 2 + y 3 + z4 )5 3y 2 = 18y 2 (x 2 + y 3 + z4 )5 , f3 = 6u5 u3 = 6(x 2 + y 3 + z4 )5 4z3 = 24z3 (x 2 + y 3 + z4 )5 (f) f (x, y, z) = exyz = eu , with u = xyz, gives f1 = eu u1 = exyz yz. Similarly, f2 = eu u2 = exyz xz, and f3 = eu u3 = exyz xy. 10. From f = x y we get (∗) ln f = y z ln x. Differentiating (∗) w.r.t x yields fx /f = y z /x, and so z z fx = fy z /x = x y y z /x = y z x y −1 . Differentiating (∗) w.r.t y yields fy /f = zy z−1 ln x, and so fy = z z zy z−1 (ln x)x y . Differentiating (∗) w.r.t z yields fz /f = y z (ln y)(ln x), and so fz = y z (ln x)(ln y)x y . 11. For (x, y) = (0, 0), f1 = y(x 4 +4x 2 y 2 −y 4 )(x 2 +y 2 )−2 and f2 (x, y) = x(x 4 −4x 2 y 2 −y 4 )(x 2 +y 2 )−2 . Thus, for y = 0, f1 (0, y) = −y. This is also correct for y = 0, because f1 (0, 0) = limh→0 [f (h, 0) − f (0, 0)]/ h = 0. Similarly, f2 (x, 0) = x for all x. It follows that f12 (0, y) = (∂/∂y)f1 (0, y) = −1 for all y. In particular, f12 (0, 0) = −1. Similarly, f21 (x, 0) = (∂/∂x)f2 (x, 0) = 1 for all x, and so f21 (0, 0) = 1. Straightforward differentiation shows that, for (x, y) = (0, 0), x 6 + 9x 4 y 2 − 9x 2 y 4 − y 6 (∗) (x 2 + y 2 )3 Thus, outside the origin, the two cross partials are equal, in accordance with Young’s theorem. At the origin, however, we have seen that f12 (0, 0) = −1 and f21 (0, 0) = 1. Therefore, at least one of f12 and f21 must be discontinuous there. Indeed, it follows from (∗) that f12 (x, 0) = 1 for all x = 0 and f12 (0, y) = −1 for all y = 0. Thus, as close to (0, 0) as we want, we can ﬁnd points where f12 equals 1 and also points where f12 equals −1. Therefore f12 cannot be continuous at (0, 0). Exactly the same argument shows that f21 cannot be continuous at (0, 0). f12 (x, y) = f21 (x, y) = z 11.7
2. (a) YK = aAK a−1 and YK = aBLa−1 , so KYK + LYL = aAK a + aBLa = a(AK a + BLa ) = aY (b) KYK + LYL = KaAK a−1 Lb + LAK a bLb−1 = aAK a Lb + bAK a La = (a + b)AK a Lb = (a + b)Y 2aKL5 − bK 4 L2 2bK 5 L − aK 2 L4 and YL = , so (c) YK = (aL3 + bK 3 )2 (aL3 + bK 3 )2 K 2 L2 (aL3 + bK 3 ) K 2 L2 2aK 2 L5 − bK 5 L2 + 2bK 5 L2 − aK 2 L5 = = = Y. KYK + LYL = (aL3 + bK 3 )2 aL3 + bK 3 (aL3 + bK 3 )2 (According to Section 12.6 the functions in (a), (b), and (c) are homogeneous of degrees a, a + b, and 1, respectively, so the results we obtained are immediate consequences of Euler’s Theorem, (12.6.2).) 7. YK = (−μ/ )a(− )K − YL = (−μ/ )b(− )L−
−1 Aeλt −1 Aeλt

aK − + bL− aK − + bL− bL− )Aeλt

−(μ/ )−1 −(μ/ )−1

= μaK − = μbL−

−1 Aeλt

aK − + bL−

−(μ/ )−1 −(μ/ )−1

,

−1 Aeλt

Thus, KYK + LYL = + + = μY . (This function is homogeneous of degree μ, so the result is an immediate consequence of Euler’s Theorem, (12.6.2).)

μ(aK −

aK −

bL−

−(μ/ )−1

aK − + bL−

.

11.8
4. mDm − D ∂ pD p pD =p = 2 (mDm − D) = 2 [Elm D − 1] > 0 if and only if Elm D > 1, 2 ∂m m m m m so pD/m increases with m if Elm D > 1. (Using the formulas in Problem 7.7.9, the result also follows from the fact that Elm (pD/m) = Elm p + Elm D − Elm m = Elm D − 1.)

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Review Problems for Chapter 11
12. (a) See the text. (b) We want to ﬁnd all (x, y) that satisfy both equations (i) 4x 3 − 8xy = 0 and (ii) 4y − 4x 2 + 4 = 0. From (i), 4x(x 2 − 2y) = 0, which implies that x = 0, or x 2 = 2y. For x = 0, (ii) yields y = −1, so (x, y) = (0, −1) is one solution. For x 2 = 2y, (ii) reduces to 4y − 8y + 4 = 0, √ √ or y = 1. But then x 2 = 2, so x = ± 2. Hence, two additional solutions are (x, y) = (± 2, 1).

Chapter 12 Tools for Comparative Statics
12.1
5. (a) If z = F (x, y) = x + y with x = f (t) and y = g(t), then F1 = F2 = 1, so the chain rule formula (12.1.1) gives dz/dt = 1 · f (t) + 1 · g (t) = f (t) + g (t). (b) is like (a), except that F2 = −1 so the chain rule gives dz/dt = f (t) − g (t). (c) If z = F (x, y) = xy with x = f (t) and y = g(t), then F1 (x, y) = y, F2 (x, y) = x, dx/dt = f (t), and dy/dt = g (t), so formula (12.1.1) gives dz/dt = F1 (x, y)(dx/dt) + F2 (x, y)(dy/dt) = yf (t) + xg (t) = f (t)g(t) + f (t)g (t). 1 x dx x = f (t), (d) If z = F (x, y) = with x = f (t) and y = g(t), then F1 (x, y) = , F2 (x, y) = − 2 , y y y dt dy 1 dz dx dy x and = g (t), so formula (12.1.1) gives = F1 (x, y) + F2 (x, y) = f (t) − 2 g (t) = dt dt dt dt y y yf (t) − xg (t) f (t)g(t) − f (t)g (t) = . y2 (g(t))2 (e) If z = F (x, y) = G(x), independent of y, with x = f (t), then F1 = G and F2 = 0, so (12.1.1) gives dz/dt = G (x) · f (t), which is the chain rule for one variable. 6. Let U (x) = u(x, h(x)). Then U (x) = u1 + u2 h (x) = αx α−1 + x α + zα α αzα−1 − α + zα x z 4a 3 x (ax 4 + b)−2/3 3

Because the term in large parentheses equals −αx α /z(x α + zα ), simplifying gives U (x) = 4ax 3 αx α−1 αx α αx α−1 (3z3 − 4ax 4 ) − = x α + zα z(x α + zα ) 3z2 3(x α + zα )z3 √ 4 3b/a, whereas

But z3 = ax 4 + b so 3z3 − 4ax 4 = 3b − ax 4 . It follows that U (x) = 0 when x = x ∗ = U (x) > 0 for x < x ∗ and U (x) < 0 for x > x ∗ . Hence x ∗ maximizes U .

7. Differentiating (12.1.1) w.r.t. t yields, d 2 z/dt 2 = (d/dt)[F1 (x, y) dx/dt] + (d/dt)[F2 (x, y) dy/dt]. Here (d/dt)[F1 (x, y) dx/dt] = [F11 (x, y) dx/dt + F12 (x, y) dy/dt]dx/dt + F1 (x, y) d 2 x/dt 2 , (d/dt)[F2 (x, y) dy/dt] = [F21 (x, y) dx/dt + F22 (x, y) dy/dt] dy/dt + F2 (x, y) d 2 y/dt 2 . Summing these two while assuming that F12 = F21 , the conclusion follows.

12.2
2. (a) Let z = F (x, y) = xy 2 with x = t + s 2 and y = t 2 s. Then F1 (x, y) = y 2 , F2 (x, y) = 2xy, ∂x/∂t = 1, and ∂y/∂t = 2ts. Then (12.2.1) gives ∂z/∂t = F1 (x, y)(∂x/∂t) + F2 (x, y)(∂y/∂t) = y 2 + 2xy2ts = (t 2 s)2 + 2(t + s 2 )t 2 s2ts = t 3 s 2 (5t + 4s 2 ). ∂z/∂s is found in the same way. 2y ∂x ∂y −2sx ts ∂z = F1 (x, y) + F2 (x, y) = et+s + e , etc. (b) 2 ∂t ∂t ∂t (x + y) (x + y)2
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39

8. (a) Let v = x 3 + y 3 + z3 − 3xyz, so that u = ln v. Then ∂u/∂x = (1/v)(∂v/∂x) = (3x 2 − 3yz)/v. Similarly, ∂u/∂y = (3y 2 − 3xz)/v, and ∂u/∂z = (3z2 − 3xy)/v. Hence, x ∂u ∂u ∂u 1 1 1 3v +y +z = (3x 3 − 3xyz) + (3y 3 − 3xyz) + (3z3 − 3xyz) = =3 ∂x ∂y ∂z v v v v

which proves (i). Equation (ii) is then proved by elementary algebra. (b) Note that f is here a function of one variable. With z = f (u) where u = x 2 y, we get ∂z/∂x = f (u)ux = 2xyf (x 2 y). Likewise, ∂z/∂y = x 2 f (x 2 y), so x∂z/∂x = 2x 2 yf (x 2 y) = 2y∂z/∂y.

12.3
2. (a) See the text. (b) Put F (x, y) = x − y + 3xy. Then F1 = 1 + 3y, F2 = −1 + 3x, F11 = 0, F12 = 3, and F22 = 0. So y = −F1 /F2 = −(1 + 3y)/(−1 + 3x). Moreover, using equation (12.3.3), 1 6(1 + 3y) 6(1 + 3y)(−1 + 3x) = . F11 (F2 )2 − 2F12 F1 F2 + F22 (F1 )2 = y =− (F2 )3 (−1 + 3x)3 (−1 + 3x)2 (c) Put F (x, y) = y 5 − x 6 . Then F1 = −6x 5 , F2 = 5y 4 , F11 = −30x 4 , F12 = 0, F22 = 20y 3 , so y = −F1 /F2 = −(−6x 5 /5y 4 ) = 6x 5 /5y 4 . Moreover, using equation (12.3.3), y =− 144x 10 6x 4 1 . (−30x 4 )(5y 4 )2 + 20y 3 (−6x 5 )2 = 4 − y 25y 9 (5y 4 )3

3. (a) With F (x, y) = 2x 2 + xy + y 2 , one has y = −F1 /F2 = −(4x + y)/(x + 2y) = −4 at (2, 0). Moreover, using (12.3.3) gives y = −(28x 2 + 14y 2 + 14xy)/(x + 2y)3 . At (2, 0) this gives y = −14. The point–slope formula for the tangent gives y = −4x + 8. (b) y = 0 requires y = −4x. Inserting this into the original equation gives a quadratic equation for x. Along with the corresponding values of y, this gives the two points indicated in the text.

12.4
√ 3. (a) Here equation (∗) is P /2 L∗ = w. Solve for L∗ . The rest is routine. (b) The ﬁrst-order condition is now Pf (L∗ ) − CL (L∗ , w) = 0 (∗)

To ﬁnd the partial derivatives of L∗ , we will differentiate (∗) partially w.r.t. P and w. First, we ﬁnd the partial derivative of Pf (L∗ ) w.r.t. P using the product rule. The result is f (L∗ )+Pf (L∗ )(∂L∗ /∂P ). Then the partial derivative of CL (L∗ , w) w.r.t. P is CLL (L∗ , w)(∂L∗ /∂P ). So differentiating (∗) w.r.t. P yields f (L∗ )+Pf (L∗ )(∂L∗ /∂P )−CLL (L∗ , w)(∂L∗ /∂P ) = 0. Solving for ∂L∗ /∂P gives the answer. Second, differentiating (∗) w.r.t. w yields [Pf (L∗ ) − CLL (L∗ , w)](∂L∗ /∂w) − CLw (L∗ , w) = 0. Solving for ∂L∗ /∂w gives the answer. 6. (a) F1 (x, y) = ey−3 + y 2 and F2 (x, y) = xey−3 + 2xy − 2. Hence, the slope of the tangent to the level curve F (x, y) = 4 at the point (1, 3) is y = −F1 (1, 3)/F2 (1, 3) = −10/5 = −2. Then use the point–slope formula.
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(b) Taking the logarithm of both sides, we get (1 + c ln y) ln y = ln A + α ln K + β ln L. Differentiating c ∂y 1 ∂y α partially with respect to K gives ln y + (1 + c ln y) = . Solving for ∂y/∂K yields the y ∂K y ∂K K given answer. Then ∂y/∂L is found in the same way.

12.5
3. With F (K, L) = AK a Lb , the partial derivatives are FK = aF /K, FL = bF /L, FKK = a(a − 1)F /K 2 , FKL = abF /KL, and FLL = b(b − 1)F /L2 . But then the numerator of the expression for σyx is −FK FL (KFK + LFL ) = −(aF /K)(bF /L)(a + b)F = −ab(a + b)F 3 /KL, whereas the denominator is KL (FL )2 FKK − 2FK FL FKL + (FK )2 FLL = KLF 3 [b2 a(a − 1) − 2a 2 b2 + a 2 b(b − 1)]/K 2 L2 = −ab(a + b)F 3 /KL. It follows that σKL = 1.

12.6
3. (2): (3): (4): (5): xf1 (x, y) + yf2 (x, y) = x(y 2 + 3x 2 ) + y(2xy) = 3(x 3 + xy 2 ) = 3f (x, y) It is easy to see that f1 (x, y) = y 2 + 3x 2 and f2 (x, y) = 2xy are homogeneous of degree 2. f (x, y) = x 3 + xy 2 = x 3 [1 + (y/x)2 ] = x 3 f (1, y/x) = y 3 [(x/y)3 + x/y] = y 3 f (x/y, 1) x 2 f11 + 2xyf12 + y 2 f22 = x 2 (6x) + 2xy(2y) + y 2 (2x) = 6x 3 + 6xy 2 = 3 · 2f (x, y)

9. Let C and D denote the the numerator and the denominator in the expression for σyx in Problem 12.5.3. Because F is homogeneous of degree one, Euler’s theorem implies that C = −F1 F2 F , and (12.6.6) implies that xF11 = −yF12 and yF22 = −xF21 = −xF12 . Hence, D = xy (F2 )2 F11 − 2F1 F2 F12 + (F1 )2 F22 = −F12 y 2 (F2 )2 + 2xyF1 F2 + x 2 (F1 )2 = −F12 (xF1 + yF2 )2 = −F12 F 2 using Euler’s theorem again. It follows that σxy = C/D = (−F1 F2 F )/(−F12 F 2 ) = F1 F2 /F F12 .

12.7
1. (a) and (f) are easy. For (b), note that xgx + ygy + zgz = g(x, y, z) + 2. Because this is not equal to kg(x, y, z) for any k, Euler’s theorem implies that g is not homogeneous of any degree. √ √ √ √ √ √ √ t ( x + y + z) tx + ty + tz = = t −1/2 h(x, y, z) for all t > 0, so (c) h(tx, ty, tz) = tx + ty + tz t (x + y + z) √ (tx)2 + (ty)2 t 2 (x 2 + y 2 ) √ = t xy ln = h is homogeneous of degree −1/2. (d) G(tx, ty) = txty ln txty t 2 xy tG(x, y) for all t > 0, so G is homogeneous of degree 1. (e) xHx + yHy = x(1/x) + y(1/y) = 2. Since 2 is not equal to k(ln x +ln y) for any constant k, Euler’s theorem implies that H is not homogeneous of any degree. 2. (a) We ﬁnd that f (tx1 , tx2 , tx3 ) = 1 (tx1 tx2 tx3 )2 1 1 + + 4 + (tx )4 + (tx )4 (tx1 ) tx1 tx2 tx3 2 3 6 (x x x )2 1 1 1 1 t 1 2 3 = tf (x1 , x2 , x3 ) + + = 4 4 4 4 x1 x2 x3 t (x1 + x2 + x3 ) t

so f is homogeneous of degree 1.
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(b) We ﬁnd that x(tv1 , tv2 , . . . , tvn ) = A δ1 (tv1 )− + δ2 (tv2 )− + · · · + δn (tvn )−
− = A t − (δ1 v1 + δ2 v2 + · · · + δn vn ) − − − − −μ/ −μ/ − = (t − )−μ/ A δ1 v1 + δ2 v2 + · · · + δn vn − = t μ A δ1 v1 + δ2 v2 + · · · + δn vn − − −μ/ −μ/

= t μ x(x1 , x2 , x3 )

so x is homogeneous of degree μ. 5. (a) We use deﬁnition (12.7.6). Suppose (x1 y1 )2 +1 = (x2 y2 )2 +1. Then (x1 y1 )2 = (x2 y2 )2 . If t > 0, then (tx1 ty1 )2 + 1 = (tx2 ty2 )2 + 1 ⇐⇒ t 4 (x1 y1 )2 + 1 = t 4 (x2 y2 )2 + 1 ⇐⇒ t 4 (x1 y1 )2 = t 4 (x2 y2 )2 ⇐⇒ (x1 y1 )2 = (x2 y2 )2 , so f is homothetic. 2(x1 y1 )2 2(x2 y2 )2 (b) From = , we get 2(x1 y1 )2 [(x2 y2 )2 + 1] = 2(x2 y2 )2 [(x1 y1 )2 + 1] and (x1 y1 )2 + 1 (x2 y2 )2 + 1 2(tx2 ty2 )2 2t 4 (x1 y1 )2 2(tx1 ty1 )2 = ⇐⇒ = so (x1 y1 )2 = (x2 y2 )2 . If t > 0, then (tx1 ty1 )2 + 1 (tx2 ty2 )2 + 1 t 4 (x1 y1 )2 + 1 (x1 y1 )2 (x2 y2 )2 2t 4 (x2 y2 )2 ⇐⇒ 4 = 4 ⇐⇒ (x1 y1 )2 = (x2 y2 )2 , so f is homothetic. t 4 (x2 y2 )2 + 1 t (x1 y1 )2 + 1 t (x2 y2 )2 + 1 (c) f (1, 0) = 1 = f (0, 1), but f (2, 0) = 4 = 8 = f (0, 2). This is enough to show that f is not homothetic. (d) g(x, y) = x 2 y is homogeneous of degree 3 and u → eu is strictly increasing, so f is homothetic according to (12.7.7). 7. Deﬁne = ln C(tw, y)−ln C(w, y). It sufﬁces to prove that e = t. We ﬁnd that n = ln t, because then C(tw, y)/C(w, y) = n = i=1 1 ai [ln(twi ) − ln wi ] + 2

n

aij [ln(twi ) ln(twj ) − ln wi ln wj ] + ln y i,j =1 i=1

bi [ln(twi ) − ln wi ]

Since ln(twi )−ln wi = ln t +ln wi −ln wi = ln t and ln(twi ) ln(twj )−ln wi ln wj = (ln t)2 +ln t ln wi + ln t ln wj , this reduces to n = ln t i=1 1 ai + (ln t)2 2

n

aij + i,j =1

1 ln t 2

n

n

ln wi j =1 i=1

aij +

1 ln t 2

n

n

n

ln wj i=1 j =1

aij + ln y ln t i=1 bi

Hence,

= ln t + 0 + 0 + 0 + 0 = ln t, because of the restrictions on the parameters ai , aij , and bi .

12.8
7. We use formula (12.8.3). (a) Here, ∂z/∂x = 2x and ∂z/∂y = 2y. At (1, 2, 5), we get ∂z/∂y = 2 and ∂z/∂x = 4, so the tangent plane has the equation z − 5 = 2(x − 1) + 4(y − 2) ⇐⇒ z = 2x + 4y − 5. (b) From z = (y −x 2 )(y −2x 2 ) = y 2 −3x 2 y +2x 4 we get ∂z/∂x = −6xy +8x 3 and ∂z/∂y = 2y −3x 2 . Thus, at (1, 3, 2) we have ∂z/∂x = −10 and ∂z/∂y = 3. The tangent plane is given by the equation z − 2 = −10(x − 1) + 3(y − 3) ⇐⇒ z = −10x + 3y + 3. 8. g(0) = f (x0 ), g(1) = f (x). Using formula (12.2.3), it follows that
0 0 g (t) = f1 (x0 + t (x − x0 ))(x1 − x1 ) + · · · + fn (x0 + t (x − x0 ))(xn − xn ) 0 0 Putting t = 0 gives g (0) = f1 (x0 )(x1 − x1 ) + · · · + fn (x0 )(xn − xn ), and the conclusion follows.

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12.9
1 4. T (x, y, z) = [x 2 + y 2 + z2 ]1/2 = u1/2 , where u = x 2 + y 2 + z2 . Then dT = 2 u−1/2 du = u−1/2 (x dx + y dy + z dz). For x = 2, y = 3, and z = 6, we have u = 49, T = 7 and dT = 1 1 7 (x dx + y dy + z dz) = 7 (2 dx + 3 dy + 6 dz). Thus, T (2 + 0.01, 3 − 0.01, 6 + 0.02) ≈ T (2, 3, 6) + 1 1 7 [2 · 0.01 + 3(−0.01) + 6 · 0.02] = 7 + 7 · 0.11 ≈ 7.015714. (A calculator or computer gives a better √ approximation: 49.2206 ≈ 7.015739.)

12.11
3. Since we are asked to ﬁnd the partials of y1 and y2 w.r.t. x1 only, we might as well differentiate the system partially w.r.t. x1 : (i) 3− ∂y1 2 ∂y2 − 9y2 = 0, ∂x1 ∂x1 (ii)
2 2 3x1 + 6y1

∂y1 ∂y2 − =0 ∂x1 ∂x1

Solving these two simultaneous equations for the partials gives the answers in the text. (An alternative method, in particular if one needs all the partials, is to use total differentiation: (i)
2 3 dx1 + 2x2 dx2 − dy1 − 9y2 dy2 = 0,

(ii)

2 2 3x1 dx1 − 2 dx2 + 6y1 dy1 − dy2 = 0

Letting dx2 = 0 and solving for dy1 and dy2 leads to dy1 = A dx1 and dy2 = B dx1 , where A = ∂y1 /∂x1 and B = ∂y2 /∂x1 .) 4. Differentiating with respect to M gives (i) I (r)rM = S (Y )YM , (ii) aYM + L (r)rM = 1. (Remember that Y and r are functions of the independent variables a and M.) Writing this as a linear equation system in standard form, we get −S (Y )YM + I (r)rM = 0 aYM + L (r)rM = 1 Using either ordinary elimination or formula (2.4.2) gives YM = I (r) S (Y )L (r) + aI (r) and rM = S (Y ) S (Y )L (r) + aI (r)

Review Problems for Chapter 12
4. X = Ng(u), where u = ϕ(N)/N . Then du/dN = [ϕ (N)N − ϕ(N )]/N 2 = (1/N)(ϕ (N) − u), and by the product rule and the chain rule, dX du = g(u) + Ng (u) = g(u) + g (u)(ϕ (N) − u), dN dN Differentiating g(u) + g (u)(ϕ (N) − u) w.r.t. N gives du du du d 2X + g (u) ϕ (N) − u + g (u) ϕ (N) − = g (u) dN 2 dN dN dN 1 2 = g ϕ(N)/N ϕ (N) − ϕ(N)/N + g ϕ(N)/N ϕ (N) N
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

u=

ϕ(N ) N

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11. Taking the elasticity of each side of the equation gives Elx (y 2 ex e1/y ) = Elx y 2 + Elx ex + Elx e1/y = 0. Here Elx y 2 = 2 Elx y and Elx ex = x. Moreover, Elx e1/y = Elx eu , where u = 1/y, so Elx eu = u Elx (1/y) = (1/y)(Elx 1 − Elx y) = −(1/y) Elx y. All in all, 2 Elx y + x − (1/y) Elx y = 0, so Elx y = xy/(1 − 2y). (We used the rules for elasticities in Problem 7.7.9. If you are not comfortable with these rules, you can ﬁnd y by implicit differentiation and then use Elx y = (x/y)y .) 16. (a) Differentiating and then gathering all terms in dp and dL on the left-hand side, one obtains (i) F (L) dp + pF (L) dL = dw, (ii) F (L) dp + (pF (L) − w) dL = L dw + dB

Since we know that pF (L) = w, (ii) implies that dp = (Ldw + dB)/F (L). Substituting this into (i) and solving for dL, we obtain dL = [(F (L) − LF (L))dw − F (L)dB]/pF (L)F (L). It follows that L ∂p = , ∂w F (L) ∂p 1 = , ∂B F (L) ∂L F (L) − LF (L) = , ∂w pF (L)F (L) ∂L F (L) =− ∂B pF (L)F (L)

(b) Because all variables including p are positive, whereas F (L) > 0 and F (L) < 0, it is clear that ∂p/∂w, ∂p/∂B, and ∂L/∂B are all positive. The sign of ∂L/∂w is the opposite of the sign of F (L) − LF (L). From the equations in the model, we get F (L) = w/p and F (L) = (wL+B)/p, so F (L)−LF (L) = B/p > 0. Therefore ∂L/∂w < 0. 19. (a) The ﬁrst-order necessary condition for maximum is P (t) = V (t)e−rt − rV (t)e−rt − me−rt = 0. Cancelling e−rt , we see that t ∗ can only maximize the present value provided (∗) is satisﬁed. The equation says that the marginal increase V (t ∗ ) in market value per unit of time from keeping the car a little longer must equal the sum of the interest cost rV (t ∗ ) per unit time from waiting to receive the sales revenue, plus the maintenance cost m per unit of time. (b) Differentiating P (t) again gives P (t) = V (t)e−rt −rV (t)e−rt −rV (t)e−rt +r 2 V (t)e−rt +rme−rt . Gathering terms, we have P (t) = [V (t)−rV (t)]e−rt +[V (t)−rV (t)−m](−re−rt ). At the stationary point t ∗ the last square bracket is 0, so the condition P (t ∗ ) < 0 reduces to D = V (t ∗ ) − rV (t ∗ ) < 0. (c) Taking the differential of (∗) yields V (t ∗ ) dt ∗ = dr V (t ∗ ) + r V (t ∗ ) dt ∗ = dm. Hence V (t ∗ ) V (t ∗ ) ∂t ∗ = = ∂r V (t ∗ ) − rV (t ∗ ) D and 1 ∂t ∗ 1 = = ∗ ) − rV (t ∗ ) ∂m V (t D

Assuming that V (t ∗ ) > 0 (otherwise it would be better to scrap the car immediately), both partial derivatives are negative. A small increase in either the interest rate or the maintenance cost makes the owner want to sell the car a bit sooner.

Chapter 13 Multivariable Optimization
13.2
3. Solving the budget equation to express x as a function of y and z yields x = 108 − 3y − 4z. Then utility as a function of y and z is U = (108 − 3y − 4z)yz. Necessary ﬁrst-order conditions for a maximum are Uy = 108z − 6yz − 4z2 = 0 and Uz = 108y − 3y 2 − 8yz = 0. Because y and z are assumed to be positive, these two equations reduce to 6y + 4z = 108 and 3y + 8z = 108, with solution y = 12 and z = 9. (Theorem 13.2.1 cannot be used directly to prove optimality. However, it can be applied to the equivalent problem of maximizing ln U . See Theorem 13.6.3.)
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13.3
3. (a) The ﬁrst and second order derivatives of f are f1 (x, y) = (2x − ay)ey , f2 (x, y) = x(x − ay − a)ey , f11 (x, y) = 2ey , f12 (x, y) = (2x − ay − a)ey , and f22 (x, y) = x(x − ay − 2a)ey . The stationary points are the solutions of the two-equation system (1) 2x − ay = 0; (2) x(x − ay − a) = 0. If x = 0, then (1) 1 gives y = 0 (because a = 0). If x = 0, then (2) gives x = ay + a, whereas (1) gives x = 2 ay. Hence x = −a and y = −2. Conclusion: There are two stationary points, (0, 0) and (−a, −2). To determine the nature of each stationary point (x0 , y0 ), we use the second-derivative test, with A = f11 (x0 , y0 ), B = f12 (x0 , y0 ), and C = f22 (x0 , y0 ). The test gives Point (0, 0) (−a, −2) A 2 2e−2 B −a −ae−2 C 0 a 2 e−2 AC − B 2 −a 2 a 2 e−4 Result Saddle point Local minimum

(b) (x ∗ , y ∗ ) = (−a, −2), and therefore f ∗ (a) = f (−a, −2) = −a 2 e−2 On the other hand, if fˆ(x, y, a) = (x 2 − axy)ey , then fˆ3 (x, y, a) = −xyey and
∗ fˆ3 (x ∗ , y ∗ , a) = −x ∗ y ∗ ey = −2ae−2

and

df ∗ (a)/da = −2ae−2

Thus the equation fˆ3 (x ∗ , y ∗ , a) = df ∗ (a)/da is true. (This is also what the envelope theorem tells us. See formula (13.7.2).) 4. (a) Vt (t, x) = ft (t, x)e−rt − rf (t, x)e−rt = 0, Vx (t, x) = fx (t, x)e−rt − 1 = 0, so at the optimum, ∗ ft (t ∗ , x ∗ ) = rf (t ∗ , x ∗ ) and fx (t ∗ , x ∗ ) = ert . (b) See the text. (c) V (t, x) = g(t)h(x)e−rt − x, so Vt = h(x)(g (t) − rg(t))e−rt , Vx = g(t)h (x)e−rt − 1. Moreover, Vtt = h(x)(g (t) − 2rg (t) + r 2 g(t))e−rt , Vtx = h (x)(g (t) − rg(t))e−rt , and Vxx = g(t)h (x)e−rt . Because the ﬁrst-order condition g (t ∗ ) = rg(t ∗ ) is satisﬁed at (t ∗ , x ∗ ), there one has Vtx = 0, as well ∗ as Vxx < 0 provided that h (x ∗ ) < 0, and Vtt = h(x ∗ )[g (t ∗ ) − r 2 g(t ∗ )]e−rt < 0 provided that g (t ∗ ) < r 2 g(t ∗ ). When both stated conditions are satisﬁed, one also has Vxx Vtt − (Vxt )2 > 0. These inequalities are sufﬁcient to ensure that (t ∗ , x ∗ ) is a local maximum point. √ √ √ ∗ ∗ (d) The ﬁrst-order conditions in (b) reduce to e t /2 t ∗ = re t , so t ∗ = 1/4r 2 , and 1/(x ∗ + 1) = e1/4r /e1/2r , or x ∗ = e1/4r − 1. We check that the two conditions in (c) are satisﬁed. Obviously, h (x ∗ ) = √ √ 1 ∗ −(1 + x ∗ )−2 < 0. Moreover, g (t ∗ ) = √ e t ( t ∗ − 1) = r 2 (1 − 2r)e1/2r , whereas r 2 g(t ∗ ) = ∗ t∗ 4t √ ∗ r 2 e t = r 2 e1/2r . Hence g (t ∗ ) < r 2 g(t ∗ ) provided that r 2 (1 − 2r) < r 2 , which is true for all r > 0. 6. (a) We need to have 1 + x 2 y > 0. When x = 0, f (0, y) = 0. For x = 0, 1 + x 2 y > 0 ⇐⇒ y > −1/x 2 . (The ﬁgure in the text shows a part of the graph of f . Note that f = 0 on the x-axis and on the y-axis.) 2y − 2x 2 y 2 2x −x 4 , f12 (x, y) = , and f22 (x, y) = . (b) See the text. (c) f11 (x, y) = (1 + x 2 y)2 (1 + x 2 y)2 (1 + x 2 y)2 The second-order derivatives at all points of the form (0, b) are f11 (0, b) = 2b, f12 (0, b) = 0, and f22 (0, b) = 0. Hence AC − B 2 = 0 at all the stationary points, so the second-derivative test tells us nothing about these points. (d) See the text.
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13.4
2. (a) See the text. (b) The new proﬁt function is π = −bp2 −dp2 +(a +βb)p +(c +βd)p −α −β(a +c) ˆ a + c + β(b + d) . and the price that maximizes proﬁts is easily seen to be p = ˆ 2(b + d) a c and q ∗ = , with maximized (c) In the case β = 0, the answers in part (a) simplify to p ∗ = 2b 2d c2 a2 + − α. But when price discrimination is prohibited, the answer in part (b) proﬁt π(p ∗ , q ∗ ) = 4b 4d (a + c)2 a+c , with maximized proﬁt π(p) = ˆ ˆ − α. The ﬁrm’s loss of proﬁt is becomes p = ˆ 2(b + d) 4(b + d) (ad − bc)2 ≥ 0. Note that this loss is 0 if and only if ad = bc, in which case ˆ ˆ π(p ∗ , q ∗ ) − π(p) = 4bd(b + d) p ∗ = q ∗ , so the ﬁrm wants to charge the same price in each market anyway.

4. (a) The four data points are (x0 , y0 ) = (0, 11.29), (x1 , y1 ) = (1, 11.40), (x2 , y2 ) = (2, 11.49), and (x3 , y3 ) = (3, 11.61), where x0 corresponds to 1970, etc. (The numbers yt are approximate, as are most subsequent results.) 1 Using the method of least squares set out in Example 4, we ﬁnd that μx = 4 (0 + 1 + 2 + 3) = 1.5, 1 1 μy = 4 (11.29 + 11.40 + 11.49 + 11.61) = 11.45, and σxx = 4 [(0 − 1.5)2 + (1 − 1.5)2 + (2 − 1.5)2 + (3 − 1.5)2 ] = 1.25. 1 Moreover, σxy = 4 [(−1.5)(11.29 − 11.45) + (−0.5)(11.40 − 11.45) + (0.5)(11.49 − 11.45) + (1.5)(11.61 − 11.45)], which is equal to 0.13125, so formula (∗∗) implies that a = σxy /σxx = 0.105 ˆ ˆ and b = μy − aμx ≈ 11.45 − 0.105 · 1.5 = 11.29. ˆ (b) With z0 = ln 274, z1 = ln 307, z2 = ln 436, and z3 = ln 524, the four data points are (x0 , z0 ) = (0, 5.61), (x1 , z1 ) = (1, 5.73), (x2 , z2 ) = (2, 6.08), and (x3 , z3 ) = (3, 6.26). As before, μx = 1.5 and 1 1 σxx = 1.25. Moreover, μz = 4 (5.61 + 5.73 + 6.08 + 6.26) = 5.92 and σxz ≈ 4 [(−1.5)(5.61 − 5.92) + (−0.5)(5.73 − 5.92) + (0.5)(6.08 − 5.92) + (1.5)(6.26 − 5.92)] = 0.2875. Hence c = σxz /σxx = 0.23, ˆ ˆ = μz − cμx = 5.92 − 0.23 · 1.5 = 5.575. ˆ d (c) If the time trends ln(GDP) = ax + b and ln(FA) = cx + d had continued, then FA would have grown to equal 1% of GNP by the time x that solves ln(FA/GNP) = ln 0.01 or (c − a)x + d − b = ln 0.01. Hence x = (b − d + ln 0.01)/(c − a). Inserting the numerical estimates found in parts (a) and (b) gives x ≈ (11.29 − 5.575 − 4.605)/(0.23 − 0.105) = 1.11/0.125 = 8.88. The goal would be reached in late 1978. 5. (a) The two ﬁrms’ combined proﬁt is px + qy − (5 + x) − (3 + 2y), or substituting for x and y, (p −1)(29−5p +4q)+(q −2)(16+4p −6q)−8, which simpliﬁes to 26p +24q −5p 2 −6q 2 +8pq −69. This is a concave function of p and q. The ﬁrst-order conditions are the two equations 26 − 10p + 8q = 0 and 24−12q +8p = 0. The unique solution is p = 9, q = 8, which gives a maximum. The corresponding production levels are x = 16 and y = 4. Firm A’s proﬁt is 123, whereas B’s is 21. (b) Firm A’s proﬁt is now πA (p) = (p − 1)(29 − 5p + 4q) − 5 = 34p − 5p 2 + 4pq − 4q − 34, with q ﬁxed. This quadratic polynomial is maximized at p = pA (q) = 1 (2q + 17). Likewise, ﬁrm B’s proﬁt is 5 now πB (q) = qy − 3 − 2y = 28q − 6q 2 + 4pq − 8p − 35, with p ﬁxed. This quadratic polynomial is maximized at q = qB (p) = 1 (p + 7). 3 (c) Equilibrium occurs where the price pair (p, q) satisﬁes the two equations p = pA (q) = 1 (2q + 17) 5 and q = qB (p) = 1 (p + 7) simultaneously. Substituting from the second equation into the ﬁrst yields 3
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p = 1 2 1 (p + 7) + 17 or, after clearing fractions, 15p = 2p + 14 + 51. Hence prices are p = 5 and 5 3 q = 4, whereas production levels are x = 20, y = 12, and proﬁts are 75 for A and 21 for B, respectively. (d) Starting at (9, 8), ﬁrst ﬁrm A moves to pA (8) = 33/5 = 6.6, then ﬁrm B answers by moving to qA (6.6) = 13.6/3 ≈ 4.53, then ﬁrm A responds by moving to near pA (4.53) = 26.06/5 = 5.212, and so on. After the ﬁrst horizontal move away from (9, 8), the process keeps switching between moves vertically down from the curve p = pA (q), and moves horizontally across to the curve q = qB (p), as shown in Fig. SM13.4.5. These moves never cross either curve, and in the limit the process converges to the equilibrium (5, 4) found in part (c). q 10 8 6 4 2 2 4 6 Figure SM13.4.5 8 10 12 14 p
Figure SM13.5.2

p = pA (q) (9, 8) q = qB (p) (5, 4)

y 4 (3, 3)

4

x

13.5
2. (a) The continuous function f is deﬁned on a closed, bounded set S (see Fig. SM13.5.2), so the extreme value theorem ensures that f attains both a maximum and a minimum over S. Stationary points are where (i) f1 (x, y) = 3x 2 − 9y = 0 and (ii) f2 (x, y) = 3y 2 − 9x = 0. From (i), y = 1 x 2 , which inserted 3 into (ii) yields 1 x(x 3 − 27) = 0. The only solutions are x = 0 and x = 3. Thus the only stationary 3 point in the interior of S is (x, y) = (3, 3). We proceed by examining the behaviour of f (x, y) along the boundary of S, i.e. along the four edges of S. (I) y = 0, x ∈ [0, 4]. Then f (x, 0) = x 3 + 27, which has minimum at x = 0, and maximum at x = 4. (II) x = 4, y ∈ [0, 4]. Then f (4, y) √ y 3 − 36y + 91. The function g(y) = y 3 − 36y + 91,√ ∈ [0, 4] = y has g (y) = 3y 2 − 36 = 0 at y = 12. Possible extreme points along (II) are (4, 0), (4, 12), and (4, 4). (III) y = 4, x ∈ [0, 4]. Then f (x, 4) = x 3 − 36x + 91, and as in (II) we see that possible extreme points √ are (0, 4), ( 12, 4), and (4, 4). (IV) x = 0, y ∈ [0, 4]. As in case (I) we obtain the possible extreme points (0, 0) and (0, 4). This results in six candidates, where the function values are f (3, 3) = 0, f (0, 0) = 27, f (4, 0) = √ √ √ f (0, 4) = 91, f (4, 12) = f ( 12, 4) = 91 − 24 12 ≈ 7.86, f (0, 0) = 27. The conclusion follows. (b) The constraint set S = (x, y) : x 2 + y 2 ≤ 1 consists of points that lie on or inside a circle around the origin with radius 1. This is a closed and bounded set, and f (x, y) = x 2 + 2y 2 − x is continuous. Therefore the extreme value theorem ensures that f attains both a maximum and a minimum over S. Stationary points for f occur where fx (x, y) = 2x − 1 = 0 and fy (x, y) = 4y = 0. So the only stationary point for f is (x1 , y1 ) = (1/2, 0), which is an interior point of S. An extreme point that does not lie in the interior of S must lie on the boundary of S, that is, on the circle x 2 + y 2 = 1. Along this circle we have y 2 = 1 − x 2 , and therefore f (x, y) = x 2 + 2y 2 − x = x 2 + 2(1 − x 2 ) − x = 2 − x − x 2 where x runs through the interval [−1, 1]. (It is a common error to overlook this restriction.) The function g(x) = 2 − x − x 2 has one stationary point in the interior of [−1, 1], namely x = −1/2, so any extreme
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values of g(x) must occur either for this value of x or at one the endpoints ±1 of the interval [−1, 1]. Any extreme points for f (x, y) on the boundary of S must therefore be among the points √ √ 1 1 1 1 (x2 , y2 ) = (− 2 , 2 3), (x3 , y3 ) = (− 2 , − 2 3), (x4 , y4 ) = (1, 0), (x5 , y5 ) = (−1, 0) √ 1 1 1 1 9 Now, f ( 2 , 0) = − 4 , f (− 2 , ± 2 3) = 4 , f (1, 0) = 0, and f (−1, 0) = 2. The conclusion in the text follows. 3. The set S is shown in Fig. A13.5.3 in the book. It is clearly closed and bounded, so the continuous function f has a maximum in S. The stationary points are where ∂f/∂x = 9 − 12(x + y) = 0 and ∂f/∂y = 8 − 12(x + y) = 0. But 12(x + y) = 9 and 12(x + y) = 8 give a contradiction. Hence, there are no stationary points at all. The maximum value of f must therefore occur on the boundary, which consists of ﬁve parts. Either the maximum value occurs at one of the ﬁve corners or “extreme points” of the boundary, or else at an interior point of one of ﬁve straight “edges.” The function values at the ﬁve corners are f (0, 0) = 0, f (5, 0) = −105, f (5, 3) = −315, f (4, 3) = −234, and f (0, 1) = 2. We proceed to examine the behaviour of f at interior points along each of the ﬁve edges. (I) y = 0, x ∈ (0, 5). The behaviour of f is determined by the function g1 (x) = f (x, 0) = 9x − 6x 2 for x ∈ (0, 5). If this function of one variable has a maximum in (0, 5), it must occur at a stationary point where g1 (x) = 9 − 12x = 0, and so at x = 3/4. We ﬁnd that g1 (3/4) = f (3/4, 0) = 27/8. (II) x = 5, y ∈ (0, 3). Deﬁne g2 (y) = f (5, y) = 45 + 8y − 6(5 + y)2 for y ∈ (0, 3). Here g2 (y) = −52 − 12y, which is negative throughout (0, 3), so there are no stationary points on this edge. (III) y = 3, x ∈ (4, 5). Deﬁne g3 (x) = f (x, 3) = 9x + 24 − 6(x + 3)2 for x ∈ (4, 5). Here g3 (x) = −27 − 12x, which is negative throughout (4, 5), so there are no stationary points on this edge either. (IV) −x + 2y = 2, or y = x/2 + 1, with x ∈ (0, 4). Deﬁne the function g4 (x) = f (x, x/2 + 1) = −27x 2 /2 − 5x + 2 for x ∈ (0, 4). Here g4 (x) = −27x − 5, which is negative in (0, 4), so there are no stationary points here. (V) x = 0, y ∈ (0, 1). Deﬁne g5 (y) = f (0, y) = 8y − 6y 2 . Then g5 (y) = 8 − 12y = 0 at y = 2/3, with g5 (2/3) = f (0, 2/3) = 8/3. After comparing the values of f at the ﬁve corners of the boundary and at the two points found on the edges labelled (I) and (V) rewspectively, we conclude that the maximum value of f is 27/8, which is achieved at (3/4, 0). 5. (a) First, f1 (x, y) = e−x (1 − x)(y − 4)y = 0 when x = 1, or y = 0, or y = 4. Second, f2 (x, y) = 2xe−x (y − 2) = 0 when x = 0 or y = 2. It follows that the stationary points are (1, 2), (0, 0) and (0, 4). Moreover, f11 (x, y) = e−x (x − 2)(y 2 − 4y), f12 (x, y) = e−x (1 − x)(2y − 4), and f22 = 2xe−x . Classiﬁcation of the stationary points: (x, y) (1, 2) (0, 0) (0, 4) A 4e−1 0 0 B 0 −4 4 C 2e−1 0 0 AC − B 2 8e−2 −16 −16 Type of point Local minimum Saddle point Saddle point

(b) We show that the range of f is unbounded both above and below. Indeed, there is no global maximum because f (1, y) = e−1 (y 2 −4y) → ∞ as y → ∞. Nor is there any global minimum because f (−1, y) = −e(y 2 − 4y) → −∞ as y → ∞.
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(c) The set S is obviously bounded. The boundary of S consists of the four edges of the rectangle, and all points on these line segments belong to S. Hence S is closed. Since f is continuous, the extreme value theorem tells us that f has global maximum and minimum points in S. These global extreme points must be either stationary points if f in the interior of S, or points on the boundary of S. The only stationary point of f in the interior of S is (1, 2). The function value at this point is f (1, 2) = −4e−1 ≈ −1.4715. The four edges are most easily investigated separately: (i) Along (I), y = 0 and f (x, y) = f (x, 0) is identically 0. (ii) Along (II), x = 5 and f (x, y) = 5e−5 (y 2 − 4y) for y ∈ [0, 4]. This has its least value for y = 2 and its greatest value for y = 0 and for y = 4. The function values are f (5, 2) = −20e−5 ≈ −0.1348 and f (5, 0) = f (5, 4) = 0. (iii) Along edge (III), y = 4 and f (x, y) = f (x, 4) = 0. (iv) Finally, along (IV), x = 0 and f (x, y) = f (0, y) = 0. Collecting all these results, we see that f attains its least value (on S) at the point (1, 2) and its greatest value (namely 0) at all points of the line segments (I), (III) and (IV). f (x, y) e−x (1 − x)(y − 4)y (x − 1)(y − 4)y (d) y = − 1 =− = = 0 when x = 1. −x (y − 2) f2 (x, y) 2xe 2x(y − 2)

13.6
4. To calculate fx is routine. The derivative of y2 z t2 y e

dt w.r.t. y, keeping z constant, can be found using

z 2 (9.3.7): it is −e . The derivative of y et dt w.r.t. z, keeping y constant, can be found using (9.3.6): it is 2 2 2 ez . Thus fy = 2 − ey and fz = −3 + ez . Since each of the three partials depends only on one variable

and is 0 for two different values of that variable, there are eight stationary points (as indicated in the text).

13.7
2. (a) With proﬁts π as given in the text, ﬁrst-order conditions for a maximum are πK = 2 pK −1/3 − r = 0, 3 1 πL = 2 pL−1/2 − w = 0, πT = 1 pT −2/3 − q = 0. Thus, K −1/3 = 3r/2p, L−1/2 = 2w/p, and 3 T −2/3 = 3q/p. Raising each side of K −1/3 = 3r/2p to the power of −3 yields, K = (3r/2p)−3 = (2p/3r)3 = (8/27)p3 r −3 . In a similar way we ﬁnd L and T . (b) Routine algebra: see the text. 5. (a) Differentiating pFK (K ∗ , L∗ ) = r using the product rule gives dp FK (K ∗ , L∗ ) + pd(FK (K ∗ , L∗ )) = dr. Moreover, d(FK (K ∗ , L∗ )) = FKK (K ∗ , L∗ ) dK ∗ + FKL (K ∗ , L∗ ) dL∗ . (To see why, note that dg(K ∗ , L∗ ) = gK (K ∗ , L∗ ) dK ∗ + gL (K ∗ , L∗ ) dL∗ . Then let g = FK .) This explains the ﬁrst displayed equation (replacing dK by dK ∗ and dL by dL∗ ). The second is derived in the same way. (b) Rearrange the equation system by moving the differentials of the exogenous prices p, r, and w to the right-hand side, while suppressing the notation indicating that the partials are evaluated at (K ∗ , L∗ ): pFKK dK ∗ + pFKL dL∗ = dr − FK dp pFLK dK ∗ + pFLL dL∗ = dw − FL dp Putting = FKK FLL − FKL FLK = FKK FLL − (FKL )2 , then using (2.4.2) and cancelling p, we get dK ∗ = F −FKL −FK FLL + FL FKL dp + LL dr + dw p p p

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In the same way dL∗ =

−FLK F −FL FKK + FK FLK dp + dr + KK dw p p p

We can now read off the required partials. (c) See the text. (Recall that FLL < 0 follows from (∗∗) in Example 13.3.3.)
∗ ∗ ∗ ∗ 6. (a) (i) R1 (x1 , x2 ) + s = C1 (x1 , x2 ) (marginal revenue plus subsidy equals marginal cost). ∗ ∗ ∗ ∗ (ii) R2 (x1 , x2 ) = C2 (x1 , x2 ) + t = 0 (marginal revenue equals marginal cost plus tax). (b) See the text, which also introduces the notation D = (R11 − C11 )(R22 − C22 ) − (R12 − C12 )2 . (c) Taking the total differentials of (i) and (ii) yields ∗ ∗ (R11 − C11 )dx1 + (R12 − C12 )dx2 = −ds, ∗ ∗ Solving for dx1 and dx2 yields, after rearranging, ∗ dx1 = ∗ ∗ (R21 − C21 )dx1 + (R22 − C22 )dx2 = dt

−(R22 − C22 )ds − (R12 − C12 )dt (R − C21 )ds + (R11 − C11 )dt ∗ , dx2 = 21 D D

From this we ﬁnd that the partial derivatives are
∗ ∗ ∗ ∗ −R22 + C22 ∂x1 −R12 + C12 ∂x2 R − C21 ∂x2 R − C11 ∂x1 = > 0, = > 0, = 21 < 0, = 11 0 from (b). Note that these signs accord with economic intuition. For example, if the tax on good 2 increases, then the production of good 1 increases, while the production of good 2 decreases. (d) Follows from the expressions in (c) because R12 = R21 and C12 = C21 .

Review Problems for Chapter 13
2. (a) The proﬁt function is π(Q1 , Q2 ) = 120Q1 + 90Q2 − 0.1Q2 − 0.1Q1 Q2 − 0.1Q2 . First-order 1 2 conditions for maximal proﬁt are: π1 (Q1 , Q2 ) = 120 − 0.2Q1 − 0.1Q2 = 0 and π2 (Q1 , Q2 ) = 90 − 0.1Q1 − 0.2Q2 = 0. We ﬁnd (Q1 , Q2 ) = (500, 200). Moreover, π11 (Q1 , Q2 ) = −0.2 ≤ 0, π12 (Q1 , Q2 ) = −0.1, and π22 (Q1 , Q2 ) = −0.2 ≤ 0. Since also π11 π22 − (π12 )2 = 0.03 ≥ 0, (500, 200) maximizes proﬁts. (b) The proﬁt function is now π(Q1 , Q2 ) = P1 Q1 + 90Q2 − 0.1Q2 − 0.1Q1 Q2 − 0.1Q2 . First-order ˆ 1 2 conditions for maximal proﬁt become π1 = P1 −0.2Q1 −0.1Q2 = 0 and π2 = 90−0.1Q1 −0.2Q2 = 0. ˆ ˆ In order to induce the choice Q1 = 400, the ﬁrst-order conditions imply that P1 − 80 − 0.1Q2 = 0 and 90 − 40 − 0.2Q2 = 0. It follows that Q2 = 250 and P1 = 105. 4. (a) Stationary points: (i) f1 (x, y) = 3x 2 − 2xy = x(3x − 2y) = 0, (ii) f2 (x, y) = −x 2 + 2y = 0. From (i), x = 0 or 3x = 2y. If x = 0, then (ii) gives y = 0. If 3x = 2y, then (ii) gives 3x = x 2 , and so x = 0 or x = 3. If x = 3, then (ii) gives y = x 2 /2 = 9/2. So the stationary points are (0, 0) and (3, 9/2). 2 2 2 2 (b) (i) f1 (x, y) = ye4x −5xy+y (8x 2 − 5xy + 1) = 0, (ii) f2 (x, y) = xe4x −5xy+y (2y 2 − 5xy + 1) = 0. If y = 0, then (i) is satisﬁed and (ii) holds only when x = 0. If x = 0, then (ii) is satisﬁed and (i) holds only if y = 0. Thus, in addition to (0, 0), any other stationary point must satisfy both 8x 2 − 5xy + 1 = 0 and 2y 2 − 5xy + 1 = 0. Subtracting the second of these equations from the ﬁrst yields 8x 2 = 2y 2 , or y = ±2x. Inserting y = −2x into 8x 2 − 5xy + 1 = 0 yields 18x 2 + 1 = 0, which has no solutions. √ 1 −2x + Inserting y = 2x into 8x 2 − 5xy + 1 = 0 yields√ 2 √ 1 = 0, so x = ± 2 2. We conclude that the √ √ 1 1 stationary points are: (0, 0) and ( 2 2, 2), (− 2 2, − 2).
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(c) Stationary points occur where: (i) f1 (x, y) = 24xy − 48x = 24x(y − 2) = 0; and (ii) f2 (x, y) = 12y 2 + 12x 2 − 48y = 12(x 2 + y 2 − 4y) = 0. From (i) x = 0 or y = 2. If x = 0, then (ii) gives y(y − 4) = 0, so y = 0 or y = 4. So (0, 0) and (0, 4) are stationary points. If y = 2, then (ii) gives x 2 − 4 = 0, so x = ±2. Hence (2, 2) and (−2, 2) are also stationary points. 6. (a) With π = p(K a +Lb +T c )−rK −wL−qT , the ﬁrst-order conditions for (K ∗ , L∗ , T ∗ ) to maximize π are πK = pa(K ∗ )a−1 − r = 0, πL = pb(L∗ )a−1 − w = 0, πT = pc(T ∗ )a−1 − q = 0 Hence, K ∗ = (ap/r)1/(1−a) , L∗ = (bp/w)1/(1−a) , T ∗ = (cp/q)1/(1−a) . (b) π ∗ = + terms that do not depend on r, where = p(ap/r)a/(1−a) − r(ap/r)1/(1−a) . Some algebraic manipulations yield = (a/r)a/(1−a) p1/(1−a) − (ap)1/(1−a) r −a/(1−a) = (a a/(1−a) − a 1/(1−a) )p1/(1−a) r −a/(1−a) = (1 − a)a a/(1−a) p1/(1−a) r −a/(1−a) Then ∂ ∂π ∗ = = −aa a/(1−a) p1/(1−a) r −a/(1−a)−1 = −a 1/(1−a) p1/(1−a) r −1/(1−a) = −(ap/r)1/(1−a) . ∂r ∂r (c) We apply (13.7.2) to this case, where π(K, L, T , p, r, w, q) = pQ − rK − wL − qT with Q = K a +Lb +T c , and π ∗ (p, r, w, q) = pQ∗ −rK ∗ −wL∗ −qT ∗ . With the partial derivatives of π evaluated at (K ∗ , L∗ , T ∗ , p, r, w, q) where output is Q∗ , one should have ∂π ∗ /∂p = πp = Q∗ , ∂π ∗ /∂r = πr = −K ∗ , ∂π ∗ /∂w = πw = −L∗ , and ∂π ∗ /∂w = πw = −T ∗ . From (b), we have the second property. The other three equations can be veriﬁed by rather tedious algebra in a similar way.

8. (a) f1 (x, y) = 2x − y − 3x 2 , f2 (x, y) = −2y − x, f11 (x, y) = 2 − 6x, f12 (x, y) = −1, f22 (x, y) = −2. Stationary points occur where 2x − y − 3x 2 = 0 and −2y − x = 0. The last equation yields y = −x/2, 5 which inserted into the ﬁrst equation gives 2 x − 3x 2 = 0. It follows that there are two stationary points, (x1 , y1 ) = (0, 0) and (x2 , y2 ) = (5/6, −5/12). These points are classiﬁed in the following table: (x, y) (0, 0)
5 ( 5 , − 12 ) 6

A 2 −3

B −1 −1

C −2 −2

AC − B 2 −5 5

Type of point Saddle point Local maximum

(b) f is concave in the domain where f11 ≤ 0, f22 ≤ 0, and f11 f22 − (f12 )2 ≥ 0, i.e. where 2 − 6x ≤ 0, −2 ≤ 0, and (2 − 6x)(−2) − (−1)2 ≥ 0. These conditions are equivalent to x ≥ 1/3 and x ≥ 5/12. Since 5/12 > 1/3, f is concave in the set S consisting of all (x, y) where x ≥ 5/12. (c) The stationary point (x2 , y2 ) = (5/6, −5/12) found in (a) does belong to S. Since f concave in S, 25 25 125 125 this is a (global) maximum point for f in S, and fmax = 25 − 144 + 72 − 216 = 432 . 36 9. (a) Stationary points require that x −1 = −ay and a(x −1) = y 2 −2a 2 y. Multiplying the ﬁrst equation by a gives −a 2 y = a(x − 1) = y 2 − 2a 2 y, by the second equation. Hence a 2 y = y 2 , implying that y = 0 or y = a 2 . Since x = 1 − ay, the stationary points are (1, 0) and (1 − a 3 , a 2 ). (Since we were asked only to show that (1 − a 3 , a 2 ) is a stationary point, it would sufﬁce to verify that it makes both partials equal to 0.) 1 (b) The function value at the stationary point in (a) is 2 (1 − a 3 )2 − (1 − a 3 ) + a 3 (−a 3 ) − 1 a 6 + a 2 · a 4 = 3 1 − 2 + 1 a 6 , whose derivative w.r.t. a is a 5 . On the other hand, the partial derivative of f w.r.t. a, keeping 6 x and y constant, is ∂f/∂a = y(x − 1) + 2ay 2 . Evaluated at x = 1 − a 3 , y = a 2 , this partial derivative
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is also a 5 , thus conﬁrming the envelope theorem. (c) f11 = 1, f22 = −2y + 2a 2 , f12 = a, and f11 f22 − (f12 )2 = a 2 − 2y. Thus f is convex if and only if −2y + 2a 2 ≥ 0 and −2y + a 2 ≥ 0, which is equivalent to a 2 ≥ y and a 2 ≥ 2y. It follows that f (x, y) is convex in that part of the xy-plane where 1 y ≤ 2 a2. 10. Actually, we have nothing to add to the answer that is already in the main text. Sorry!

Chapter 14 Constrained Optimization
14.1
4. (a) With L(x, y) = x 2 + y 2 − λ(x + 2y − 4), the ﬁrst-order conditions are L1 = 2x − λ = 0 and L2 = 2y − 2λ = 0. From these equations we get 2x = y, which inserted into the constraint gives x + 4x = 4. So x = 4/5 and y = 2x = 8/5, with λ = 2x = 8/5. (b) The same method as in (a) gives 2x − λ = 0 and 4y − λ = 0, so x = 2y. From the constraint we get x = 8 and y = 4, with λ = 16. (c) The ﬁrst-order conditions imply that 2x + 3y = λ = 3x + 2y, which gives x = y. So the solution is (x, y) = (50, 50) with λ = 250. 9. (a) With L = x a + y − λ(px + y − m), the ﬁrst-order conditions for (x ∗ , y ∗ ) to solve the problem are (i) Lx = a(x ∗ )a−1 − λp = 0; (ii) Ly = 1 − λ = 0. Thus λ = 1, and x ∗ = x ∗ (p, m) = kp−1/(1−a) where k = a 1/(1−a) . Then y ∗ = y ∗ (p, m) = m − kp −a/(1−a) . (b) ∂x ∗ /∂p = −x ∗ /p(1 − a) < 0, ∂x ∗ /∂m = 0, ∂y ∗ /∂p = ax ∗ /(1 − a) > 0, and ∂y ∗ /∂m = 1. (c) The optimal expenditure on good x is px ∗ (p, m) = kp−a/(1−a) , so Elp px ∗ (p, m) = −a/(1−a) < 0. In particular, the expenditure on good x will decrease as its price increases. √ (d) We see that x ∗ = (1/2p)2 , y ∗ = m − 1/4p, so U ∗ (p, m) = x ∗ + y ∗ = (1/2p) + m − 1/4p = m + 1/4p, and the required identity is obvious. 10. (a) With L(x, y) = 100 − e−x − e−y − λ(px + qy − m), the ﬁrst-order conditions Lx = Ly = 0 imply that e−x = λp and e−y = λq. Hence, x = − ln(λp) = − ln λ − ln p, y = − ln λ − ln q. Inserting these expressions for x and y into the constraint yields −p(ln λ + ln p) − q(ln λ + ln q) = m and so ln λ = −(m + p ln p + q ln q)/(p + q). Therefore x(p, q, m) = [m + q ln(q/p)]/(p + q), y(p, q, m) = [m + p ln(p/q)]/(p + q). (b) x(tp, tq, tm) = [tm + tq ln(tq/tp)]/(tp + tq) = x(p, q, m), so x is homogeneous of degree 0. In the same way we see that y(p, q, m) is homogeneous of degree 0.

14.2
√ 4. (a) With L(x, y) = √+ y − λ(x + 4y − 100), the ﬁrst-order conditions for (x ∗ , y ∗ ) to solve the problem x are: (i) ∂L/∂x = 1/2 x ∗ − λ = 0 (ii) ∂L/∂y = 1 − 4λ = 0. From (ii), λ = 1/4, which inserted into √ √ 1 (i) yields x ∗ = 2, so x ∗ = 4. Then y ∗ = 25 − 4 4 = 24, and maximal utility is U ∗ = x ∗ + y ∗ = 26. (b) Denote the new optimal values of x and y by x and y. If 100 is changed to 101, still λ = 1/4 and ˆ ˆ √ ˆ x = 4. The constraint now gives 4 + 4y = 101, so that y = 97/4 = 24.25, with U = x + y = 26.25. ˆ ˆ ˆ ˆ ˆ ˆ − U ∗ = 0.25 = λ. (In general, the increase in utility is The increase in maximum utility is therefore U approximately equal to the value of the Lagrange multiplier.)
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√ (c) The necessary conditions for optimality are now ∂L/∂x = 1/2 x ∗ − λp = 0, ∂L/∂y = 1 − λq = 0. √ Proceeding in the same way as in (a), we ﬁnd λ = 1/q, x ∗ = q/2p, and so x ∗ = q 2 /4p2 , with y ∗ = m/q − q/4p. (Note that y ∗ > 0 ⇐⇒ m > q 2 /4p.) (If we solve the constraint for y, the utility √ √ function is u(x) = x + (m − px)/q. We see that u (x) = 1/2 x − p/q = 0 for x ∗ = q 2 /4p2 and u (x) = −(1/4)x −3/2 < 0 when x > 0. So we have found the maximum.) 5. (a) The ﬁrst-order conditions given in the main text imply that px ∗ = pa + α/λ and qy ∗ = qb + β/λ. Substituting these into the budget constraint gives m = px ∗ +qy ∗ = pa+qb+(α+β)/λ = pa+qb+1/λ, so 1/λ = m − (pa + qb). The expressions given in (∗∗) are now easily established. (We can interpret a and b as minimum subsistence quantities of the two goods, in which case the assumption pa + qb < m means that the consumer can afford to buy (a, b).) (b) With U ∗ as given in the answer provided in the text, differentiating partially while remembering that ∂U ∗ α β 1 α + β = 1 gives = + = = λ > 0. Moreover, ∂m m − (pa + qb) m − (pa + qb) m − (pa + qb) ∂U ∗ −αa α −βa −a α α = − + = − = −aλ − , whereas ∂p m − (pa + qb) p m − (pa + qb) m − (pa + qb) p p ∗ ∗ ∗ α ∂U ∂U ∗ α ∂U ∗ = −aλ − , so =− x . The last equality is shown in the same − x = −λ a + λp p ∂p ∂m ∂m way. 6. f (x, T ) = x =x
T 0 0 1 1 1 [− 4 t 4 +(αT 2 +T −1) 1 t 3 +(T −αT 3 ) 2 t 2 ] = − 1 αxT 5 + 12 xT 4 + 1 xT 3 = 3 6 6 T 0 1 3 12 xT (2+T T

[−t 3 + (αT 2 + T − 1)t 2 + (T − αT 3 )t] dt −2αT 2 )

A similar but easier calculation shows that g(x, T ) =

(xtT − xt 2 ) dt = 1 xT 3 . The Lagrangian for 6

1 the producer’s problem is L = 12 xT 3 (2 +T −2αT 2 )−λ( 1 xT 3 −M). The two ﬁrst-order conditions are 6 1 3 1 1 1 2 3 2 2 2 12 T (2+T −2αT )− 6 λT = 0 and 12 xT (6+4T −10αT )− 2 λxT = 0. These equations imply that 1 λ = 2 (2+T −2αT 2 ) = 1 (6+4T −10αT 2 ). It follows that 4αT 2 = T . One solution is T = 0, but this is 6 inconsistent with the constraint 1 xT 3 = M. Hence, the solution we are interested is T = 1/4α, implying 6 that λ = 1 + 1/16α. Substituting into the constraint g(x, T ) = M determines x = 6MT −3 = 384Mα 3 . Because xT 3 = 6M, the maximum proﬁt is f ∗ (M) = M + M/8α − αM/16α 2 = M + M/16α, whose derivative w.r.t. M is indeed λ. We note that the maximum by itself is much easier to ﬁnd if one substitutes the constraint M = 1 xT 3 6 1 into the objective function f (x, T ), which then becomes the function −αMT 2 + 2 MT + M of T alone, with α and M as parameters. The ﬁrst-order condition for T to maximize this expression is 1 −2αMT + 2 M = 0, implying that T = 1/4α. However, this method does not ﬁnd λ.

14.3
1. (a) With L(x, y) = 3xy − λ(x 2 + y 2 − 8), the ﬁrst-order conditions are L1 = 3y − 2λx = 0 and L2 = 3x − 2λy = 0. These can be rewritten as (i) 3y = 2λx and (ii) 3x = 2λy. If x = 0, then (i) gives y = 0; conversely, if y = 0, then (ii) gives x = 0. But (x, y) = (0, 0) does not satisfy the constraint. Hence x = 0 and y = 0. Equating the ratio of the left-hand sides of (i) and (ii) to the ratio of their right-hand sides, one has y/x = x/y or x 2 = y 2 . Finally, using the constraint gives x 2 = y 2 = 4. The four solution candidates are therefore (2, 2) and (−2, −2) with λ = 3/2, as well as
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(2, −2) and (−2, 2) with λ = −3/2. The corresponding function values are f (2, 2) = f (−2, −2) = 12 and f (2, −2) = f (−2, 2) = −12. Because f is continuous and the constraint set is a circle, which is closed and bounded, the extreme value theorem implies that a maximum and minimum do exist. The function values tell us that (2, 2) and (−2, −2) solve the maximization problem, whereas (−2, 2) and (2, −2) solve the minimization problem. (b) With L = x + y − λ(x 2 + 3xy + 3y 2 − 3), the ﬁrst-order conditions are 1 − 2λx − 3λy = 0 and 1 − 3λx − 6λy = 0. These equations give us 1 = 2λx + 3λy = 3λx + 6λy. In particular, λ(3y + x) = 0. Here λ = 0 is impossible, so x = −3y. Inserting this into the constraint reduces it to 3y 2 = 3, with solutions y = ±1. So the ﬁrst-order conditions give two solution candidates (x, y, λ) = (3, −1, 1 ) and 3 (x, y, λ) = (−3, 1, − 1 ). Because the objective function is continuous and the constraint curve is closed 3 and bounded (actually, an ellipse — see (5.5.5)), the extreme value ensures that both a maximum and minimum exist. The function values f (3, −1) = 2 and f (−3, 1) = −2 tell us that the maximum is at (3, −1), the minimum at (−3, 1). 2. (a) With L = x 2 + y 2 − 2x + 1 − λ(x 2 + 4y 2 − 16), the ﬁrst-order conditions are (i) 2x − 2 − 2λx = 0 and (ii) 2y −8λy = 0. Equation (i) implies that x = 0 and then λ = 1−1/x, whereas equation (ii) shows that y = 0 or λ = 1/4. If y = 0, then x 2 = 16 − 4y 2 = 16, so x = ±4, which then gives λ = 1 ∓ 1/4. If y = 0, then λ = 1/4 and (i) gives 2x − 2 − x/2 = 0, so x = √ The constraint x 2 + 4y 2 = 16 4/3. √ 2 = 16 − 16/9 = 128/9, so y = ± 32/9 = ±4 2/3. Thus, there are four solution now yields 4y √ candidates: (i) (x, y, λ) = (4, 0, 3/4), (ii) (x, y, λ) = (−4, 0, 5/4), (iii) (x, y, λ) = (4/3, 4 2/3, 1/4), √ and (iv) (x, y, λ) = (4/3, −4 2/3, 1/4). Of these four, checking function values shows that (i) and (ii) both give a maximum, whereas (iii) and (iv) both give a minimum. (b) The Lagrangian is L = ln (2 + x 2 )+y 2 −λ(x 2 +2y −2). Hence, the necessary ﬁrst-order conditions for (x, y) to be a minimum point are (i) ∂L/∂x = 2x/(2 + x 2 ) − 2λx = 0 (ii) ∂L/∂y = 2y − 2λ = 0, (iii) x 2 + 2y = 2. From (i) we get x 1/(2 + x 2 ) − λ = 0, so x = 0 or λ = 1/(2 + x 2 ). (I) If x = 0, then (iii) gives y = 1, so (x1 , y1 ) = (0, 1) is a candidate. (II) If x = 0, then y = λ = 1/(2 + x 2 ) , where we used (ii). Inserting y = 1/(2 + x 2 ) into (iii) gives √ 4 + 2 x 2 + 2/(2 + x 2 ) = 2 ⇐⇒ 2x 2√ x 4 + 2 = 4 + 2x 2 ⇐⇒ x 4 =√ ⇐⇒ x = ± 2. √ √ √ 1 1 1 1 From (iii), y = 1 − 2 x 2 = 1 − 2 2. Thus, (x2 , y2 ) = ( 4 2, 1 − 2 2) and (x3 , y3 ) = (− 4 2, 1 − 2 2) are two other candidates. Comparing function values, we see that f (x1 , y1 ) = f (0, 1) = ln 2+1 ≈ 1.69, √ √ √ √ 1 3 f (x2 , y2 ) = f (x3 , y3 ) = ln (2 + 2 ) + (1 − 2 2 )2 = ln (2 + 2 ) + 2 − 2 ≈ 1.31. Hence, the minimum points for f (x, y) subject to the constraint are (x2 , y2 ) and (x3 , y3 ). 4. (a) With L = 24x − x 2 + 16y − 2y 2 − λ(x 2 + 2y 2 − 44), the ﬁrst-order conditions are (i) L1 = 24 − 2x − 2λx = 0 and (ii) L2 = 16 − 4y − 4λy = 0. From (i) x(1 + λ) = 12 and from (ii) y(1 + λ) = 4. Eliminating λ from (i) and (ii) we get x = 3y = 12/(1 + λ), with λ = −1. Inserted into the constraint, 11y 2 = 44, so y = ±2, and then x = ±6. So there are two candidates, (x, y) = (6, 2) and (−6, −2), with λ = 1. Computing the objective function at these two points, the only possible maximum is at (x, y) = (6, 2). Since the objective function is continuous and the constraint curve is closed and bounded (an ellipse), the extreme value theorem assures us that there is indeed a maximum at this point. (b) According to (14.2.3) the approximate change is λ · 1 = 1.

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14.4
4. Before trying to ﬁnd the minimum, consider the graph of the curve g(x, y) = 0, as shown in Fig. SM14.4.4. It consists of three pieces: √ (i) the continuous curve y = x(x + 1) in the positive quadrant; (ii) √ the continuous curve y = − x(x + 1), which is the reﬂection of curve (i) about the x-axis; (iii) the isolated point (−1, 0). The problem is to minimize the square of the distance d from the point (−2, 0) to a point on this graph. The minimum of f obviously occurs at the isolated point (−1, 0), with f (−1, 0) = 1. With the Lagrangian L = (x + 2)2 + y 2 − λ(y 2 − x(x + 1)2 ), we have L1 = 2(x + 2) + λ((x + 1)2 + 2x(x + 1)), L2 = 2y(1 − λ) y 6 5 4 3 2 d 1 −3−2−1 −1 −2 −3 −4 −5 −6

y 2 = x(x + 1)2

(x, y) 1 2 3 4 x

Note that L2 = 2y(1 − λ) = 0 only if λ = 1 or y = 0. For λ = 1, we Figure SM14.4.4 ﬁnd that L1 = 3(x + 1)2 + 2 > 0 for all x. For y = 0, the constraint gives x = 0 or x = −1. At x = 0, we have L1 = 4 + λ = 0 and L1 = 2 at x = −1. Thus the Lagrange multiplier method produces a unique solution candidate (x, y) = (0, 0) with λ = −4, which does correspond to a local minimum. The global minimum at (−1, 0), however, fails to satisfy the ﬁrst-order conditions L1 = L2 = 0 for any value of λ, because L1 = 2 at this point. So the Lagrange multiplier method cannot locate this minimum. Note that at (−1, 0) both g1 (−1, 0) and g2 (−1, 0) are 0.

14.5
4. With L = x a + y a − λ(px + qy − m), the ﬁrst-order conditions are L1 = ax a−1 − λp = 0 and L2 = ay a−1 − λq = 0. It follows that λ = 0 and then x = (λp/a)1/(a−1) , y = (λq/a)1/(a−1) . Inserting these values of x and y into the budget constraint gives (λ/a)1/(a−1) (pa/(a−1) + q a/(a−1) ) = m. To reduce notation, deﬁne R = pa/(a−1) + q a/(a−1) as in the text answer. Then we have (λ/a)1/(1−a) = m/R. Therefore x = mp1/(a−1) /R and y = mq 1/(a−1) /R.

14.6
7. The Lagrangian is L = x + y − λ(x 2 + 2y 2 + z2 − 1) − μ(x + y + z − 1), which is stationary when (i) Lx = 1 − 2λx − μ = 0; (ii) Ly = 1 − 4λy − μ = 0; (iii) Lz = −2λz − μ = 0. From (ii) and (iii) we get 1 = λ(4y − 2z), and in particular λ = 0. From (i) and (ii), λ(x − 2y) = 0 and so x = 2y. Substituting this value for x into the constraints gives 6y 2 + z2 = 1 and 3y + z = 1. Thus z = 1 − 3y, implying that 1 = 6y 2 + (1 − 3y)2 = 15y 2 − 6y + 1. Hence y = 0 or y = 2/5, implying that x = 0 or 4/5, and that z = 1 or −1/5. The only two solution candidates are (x, y, z) = (0, 0, 1) with λ = −1/2, μ = 1, and (x, y, z) = (4/5, 2/5, −1/5) with λ = 1/2, μ = 1/5. Because x + y is 0 at (0, 0, 1) and 6/5 at (4/5, 2/5, −1/5), these are respectively the minimum and the maximum. (The two constraints determine the curve in three dimensions which is the intersection of an ellipsoid (see Fig. 11.4.2) and a plane. Because an ellipsoid is a closed bounded set, so is this curve. By the extreme value theorem, the continuous function x + y does attain a maximum and a minimum over this closed bounded set.)
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8. (a) For the given Cobb–Douglas utility function, one has Uj (x) = αj U (x)/xj . So (14.6.6) with k = 1 implies that pj /p1 = Uj (x)/U1 (x) = αj x1 /α1 xj . Thus pj xj = (aj /a1 )p1 x1 . Inserting this into the budget constraint for j = 2, . . . , n, gives p1 x1 + (a2 /a1 )p1 x1 + · · · + (an /a1 )p1 x1 = m, which implies that p1 x1 = a1 m/(a1 + · · · + an ). Similarly, pj xj = aj m/(a1 + · · · + an ) for k = 1, . . . , n. a−1 = pj /p1 and so xj /x1 = (pj /p1 )−1/(1−a) , or (b) From (14.6.6) with k = 1, we get xja−1 /x1 pj xj /p1 x1 = (pj /p1 )1−1/(1−a) = (pj /p1 )−a/(1−a) . Inserting this into the budget constraint for j = 2, . . . , n, gives p1 x1 1 + (p2 /p1 )−a/(1−a) + · · · + (pn /p1 )−a/(1−a) = m

so p1 x1 =

−a/(1−a) mp1 −1/(1−a)

n

pi i=1 n

−a/(1−a)

Arguing similarly for each k, we get xj = mpj

pi i=1 −a/(1−a)

for k = 1, . . . , n.

14.7
2. Here L = x + 4y + 3z − λ(x 2 + 2y 2 + 1 z2 − b). So necessary ﬁrst-order conditions are: 3 (i) L1 = 1 − 2λx = 0; (ii) L2 = 4 − 4λy = 0; (iii) L3 = 3 − 2 λz = 0. It follows that λ = 0, and so 3 x = 1/2λ, y = 1/λ, z = 9/2λ. Inserting these values into the constraint yields [(1/4)+2+(27/4)]λ−2 = √ b and so λ2 = 9/b, implying that λ = ±3/ b. The value of the objective function is x +4y +3z = 18/λ, √ √ so λ = −3/ b determines the minimum point. This is (x, y, z) = (a, 2a, 9a), where a = − b/6. For the last veriﬁcation, see the answer in the book. 4. (a) With L = x 2 + y 2 + z − λ(x 2 + 2y 2 + 4z2 − 1), necessary conditions are: (i) ∂L/∂x = 2x − 2λx = 0, (ii) ∂L/∂y = 2y − 4λy = 0, (iii) ∂L/∂z = 1 − 8λz = 0. From (i), 2x(1 − λ) = 0, so there are two possibilities: x = 0 or λ = 1. (A) Suppose x = 0. From (ii), 2y(1 − 2λ) = 0, so y = 0 or λ = 1/2. (A.1) If y = 0, then the constraint gives 4z2 = 1, so z2 = 1/4, or z = ±1/2. Equation (iii) gives λ = 1/8z, so we have two solution candidates: P1 = (0, 0, 1/2) with λ = 1/4; and P2 = (0, 0, −1/2) with λ = −1/4. 3/4 (A.2) If λ = 1/2, then (iii) gives z = 1/8λ = 1/4. It follows from the constraint that 2y 2 =√ (recall √ √ that we assumed x = 0), and hence y = ± 3/8 = ± 6/4. So new candidates are: P3 = (0, 6/4, 1/4) √ with λ = 1/2; and P4 = (0, − 6/4, 1/4) with λ = 1/2. (B) Suppose λ = 1. Equation (iii) yields z = 1/8, and (ii) gives y = 0. From the constraint, x 2 = 15/16, √ √ √ so x = ± 15/4. Candidates: P5 = ( 15/4, 0, 1/8) with λ = 1; and P6 = (− 15/4, 0, 1/8) with λ = 1. For k = 1, 2, . . . , 6, let fk denote the value of the criterion function f at the candidate point Pk . Routine calculation shows that f1 = 1/2, f2 = −1/2, f3 = f4 = 5/8, and f5 = f6 = 17/16. It follows that both P5 and P6 solve the maximization problem, whereas P2 solves the minimization problem. (b) See the text. 5. The Lagrangian is L = rK + wL − λ(K 1/2 L1/4 − Q), so necessary conditions for (K ∗ , L∗ ) to solve 1 1 the problem are: (i) LK = r − 2 λ(K ∗ )−1/2 (L∗ )1/4 = 0, (ii) LL = w − 4 λ(K ∗ )1/2 (L∗ )−3/4 = 0, (iii) (K ∗ )1/2 (L∗ )1/4 = Q. Together (i) and (ii) imply that r/w = 2L∗ /K ∗ and so L∗ = rK ∗ /2w. Inserting this into (iii) gives Q = (K ∗ )1/2 (rK ∗ /2w)1/4 = (K ∗ )3/4 2−1/4 r 1/4 w−1/4 . Solving for K ∗ gives the answer in the text. The answers for L∗ and C ∗ = rK ∗ +wL∗ follow if we observe that 21/3 = 2·2−2/3 . The veriﬁcation of (∗) in Example 14.7.3 is easy.
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14.8
5. (a) With the Lagrangian L(x, y) = 2 − (x − 1)2 − ey − λ(x 2 + y 2 − a), the Kuhn–Tucker conditions 2 are: (i) −2(x − 1) − 2λx = 0; (ii) −2yey − 2λy = 0; (iii) λ ≥ 0, with λ = 0 if x 2 + y 2 < a. 2 2 From (i), x = (1 + λ)−1 . Moreover, (ii) reduces to y(ey + λ) = 0, and so y = 0 (because ey + λ is always positive). (I): Assume that λ = 0. Then equation (i) gives x = 1. In this case we must have a ≥ x 2 + y 2 = 1. √ (II): Assume that λ > 0. Then (iii) gives x 2 + y 2 = a, and so x = ± a (remember that y = 0). Because √ x = 1/(1 + λ) and λ > 0 we must have 0 < x < 1, so x = a and a = x 2 < 1. It remains to ﬁnd the √ value of λ and check that it is > 0. From equation (i) we get λ = 1/x − 1 = 1/ a − 1 > 0. Conclusion: The only point that satisﬁes the Kuhn–Tucker conditions is (x, y) = (1, 0) if a ≥ 1 and √ √ ( a, 0) if 0 < a < 1. The corresponding value of λ is 0 or 1/ a − 1, respectively. In both cases it follows from Theorem 14.8.1 that we have found the maximum point, because L is concave in (x, y), as L11 L12 −2 − 2λ 0 . we can see by studying the Hessian = 2 L21 L22 0 −ey (2 + 4y 2 ) − 2λ √ √ √ (b) If a ∈ (0, 1) we have f ∗ (a) = f ( a, 0) = 2 − ( a − 1)2 − 1 = 2 a − a, and for a ≥ 1 we get f ∗ (a) = f (1, 0) = 1. The derivative of f ∗ is as given in the book, but note that in order to ﬁnd the derivative df ∗ (a)/da when a = 1, we need to show that the right and left derivatives (see page 243 in the book) f ∗ (1 + h) − f ∗ (1) f ∗ (1 + h) − f ∗ (1) (f ∗ ) (1+ ) = lim+ and (f ∗ ) (1− ) = lim− h→0 h→0 h h exist and are equal. But the left and right derivatives are respectively equal to the derivatives of the √ differentiable functions g− (a) = 2 a − a and g+ (a) = 1 at a = 1, which are both 0. Hence (f ∗ ) (1) exists and equals 0.
2

14.9
2. The Lagrangian is L = α ln x + (1 − α) ln y − λ(px + qy − m) − μ(x − x), and the Kuhn–Tucker ¯ conditions for (x ∗ , y ∗ ) to solve the problem are α − λp − μ = 0 x∗ 1−α − λq = 0 L2 = y∗ λ ≥ 0, and λ = 0 if px ∗ + qy ∗ < m L1 = μ ≥ 0, and μ = 0 if x∗ 0 and so (iii) entails px ∗ + qy ∗ = m. Suppose μ = 0. Then from (i) and (ii) α/px ∗ = (1 − α)/qy ∗ , so qy ∗ = (1 − α)px ∗ /α. Then the budget constraint implies that px ∗ + (1 − α)px ∗ /α = m, from which it follows that x ∗ = mα/p and ¯ ¯ y ∗ = (1 − α)m/q, with λ = 1/m. This is valid as long as x ∗ ≤ x, that is m ≤ px/α. ∗ = x and y ∗ = m/q − p x/q = (m − p x)/q, with λ = (1 − α)/(m − p x) Suppose μ > 0. Then x ¯ ¯ ¯ ¯ and μ = α/x − λp = (αm − p x)/x(m − px). Note that if m > px/α, then m > px since α < 1. We ¯ ¯ ¯ ¯ ¯ ¯ conclude that if m > px/α, then λ and μ are both positive and conditions (i)–(iv) are satisﬁed. ¯ Since L11 = −α/x 2 < 0, L22 = −α/y 2 < 0, and L12 = 0, the Lagrangian L(x, y) is concave, so we have found the solution in both cases.
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3. (a) See the answer in the text. (b) With the constraints g1 (x, y) = −x − y ≤ −4, g2 (x, y) = −x ≤ 1, g3 (x, y) = −y ≤ −1, the Lagrangian is L = x+y−ex −ex+y −λ1 (−x−y+4)−λ2 (−x−1)−λ3 (−y+1). The Kuhn–Tucker conditions are that there exist nonnegative numbers λ1 , λ2 , and λ3 such that (i) Lx = 1 − ex − ex+y + λ1 + λ2 = 0; (ii) Ly = 1 − ex+y + λ1 + λ3 = 0; (iii) λ1 (−x − y + 4) = 0; (iv) λ2 (−x − 1) = 0; (v) λ3 (−y + 1) = 0. (We formulate the complementary slackness conditions as in (14.8.5).) From (ii), ex+y = 1 + λ1 + λ3 . Inserting this into (i) yields λ2 = ex + λ3 ≥ ex > 0. Because λ2 > 0, (iv) implies that x = −1. So any solution must lie on the line (II) in the ﬁgure, which shows that the third constraint y ≥ 1 must be slack. (Algebraically, because x + y ≥ 4 and x = −1, we have y ≥ 4 − x = 5 > 1.) So from (v) we get λ3 = 0, and then (ii) gives λ1 = ex+y − 1 ≥ e4 − 1 > 0. Thus from (iii), the ﬁrst constraint is active, so y = 4 − x = 5. Hence the only possible solution is (x ∗ , y ∗ ) = (−1, 5). Because L(x, y) is concave, we have found the optimal point. 4. (a) The feasible set is shown in Fig. A14.9.4 in the book. (The function to be maximized is f (x, y) = x + ay. The level curves of this function are straight lines with slope −1/a if a = 0, and vertical lines if a = 0. The dashed line in the ﬁgure is such a level curve (for a ≈ −0.25). The maximum point for f is that point in the feasible region that we shall ﬁnd if we make a parallel displacement of this line as far to the right as possible (why to the right?) without losing contact with the shaded region.) The Lagrangian is L(x, y) = x + ay − λ1 (x 2 + y 2 − 1) + λ2 (x + y) (the second constraint must be written as −x − y ≤ 0), so the Kuhn–Tucker conditions are: (i) L1 (x, y) = 1 − 2λ1 x + λ2 = 0; (ii) L2 (x, y) = a − 2λ1 y + λ2 = 0; (iii) λ1 ≥ 0, with λ1 = 0 if x 2 + y 2 < 1; (iv) λ2 ≥ 0, with λ2 = 0 if x + y > 0. (b) From (i), 2λ1 x = 1 + λ2 ≥ 1. But (iii) implies that λ1 ≥ 0, so in fact λ1 > 0 and x > 0. Because λ1 > 0, it follows from (iii) that x 2 + y 2 = 1, so any maximum point must lie on the circle. √ 1 (I) First consider the case x + y = 0. Then y = −x, and since x 2 + y 2 = 1, we get x = 2 2 (recall √ 1 that we have seen that x must be positive) and y = − 2 2. Adding equations (i) and (ii), we get 1 + a − 2λ1 (x + y) + 2λ2 = 0 and since x + y = 0, we ﬁnd that λ2 = −(1 + a)/2. Now, λ2 must be ≥ 0, and therefore a ≤ −1 in this √ case. Equation (i) gives λ1 = (1 + λ2 )/2x = (1 − a)/4x = 2(1 − a)/4. (II) Second, consider the other case x + y > 0. Then (iv) implies that λ2 = 0, so (i) and (ii) reduce to 1 − 2λ1 x = 0 and a − 2λ1 y = 0, and so x = 1/(2λ1 ) and y = a/(2λ1 ). Inserting these into x 2 + y 2 = 1 √ 1 a and y = √ . yields (1/4λ1 )2 (1 + a 2 ) = 1, and so 2λ1 = 1 + a 2 . This gives x = √ 1 + a2 1 + a2 Because x + y = (1 + a)/(2λ1 ), and because x + y is now assumed to be positive, we must have a > −1 in this case. Conclusion: The only points satisfying the Kuhn–Tucker conditions are the ones given in the text. Since the feasible set is closed and bounded and f is continuous, it follows from the extreme value theorem that extreme points exists. 5. The Lagrangian is L = y − x 2 + λy + μ(y − x + 2) − ν(y 2 − x), which is stationary when (i) −2x − μ + ν = 0; (ii) 1 + λ + μ − 2νy = 0. In addition, complementary slackness requires (iii) λ ≥ 0, with λ = 0 if y > 0; (iv) μ ≥ 0, with μ = 0 if y − x > −2; (v) ν ≥ 0, with ν = 0 if y 2 < x. From (ii), 2νy = 1 + λ + μ > 0, so y > 0. Then (iii) implies λ = 0, and 2νy = 1 + μ. From (i), 1 x = 2 (ν − μ). But x ≥ y 2 > 0, so ν > μ ≥ 0, and from (v), y 2 = x. Suppose μ > 0. Then y − x + 2 = y − y 2 + 2 = 0 with roots y = −1 and y = 2. Only y = 2 is feasible. Then x = y 2 = 4. Because λ = 0, conditions (i) and (ii) become −μ+ν = 8 and μ−4ν = −1,
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so ν = −7/3, which contradicts ν ≥ 0, so (x, y) = (4, 2) is not a candidate. Therefore μ = 0 after 1 all. Thus x = 2 ν = y 2 and, by (ii), 1 = 2νy = 4y 3 . Hence y = 4−1/3 , x = 4−2/3 . This is the only remaining candidate. It is the solution with λ = 0, μ = 0, and ν = 1/2y = 4−1/6 . 6. (a) See Fig. A14.9.6 in the text. Note that for (x, y) to be admissible, e−x ≤ y ≤ 2/3, and so ex ≥ 3/2, which implies, in particular, that x > 0. 1 1 (b) The Lagrangian is L = −(x + 2 )2 − 2 y 2 − λ1 (e−x − y) − λ2 (y − 2 ), and the ﬁrst-order conditions 3 are: (i) −(2x + 1) + λ1 e−x = 0; (ii) −y + λ1 − λ2 = 0; (iii) λ1 ≥ 0, with λ1 = 0 if e−x < y; (iv) λ2 ≥ 0, with λ2 = 0 if y < 2/3. From (i), λ1 = (2x + 1)ex ≥ 3/2, because of (a), implying that y = e−x . From (ii), λ2 = λ1 − y ≥ 3/2 − 2/3 > 0, so y = 2/3 because of (iv). This gives the solution candidate (x ∗ , y ∗ ) = (ln(3/2), 2/3), with λ1 = 3[ln(3/2) + 1/2] and λ2 = 3 ln(3/2) + 5/6. The Lagrangian is easily seen to be concave as a function of (x, y) when λ1 ≥ 0, so this is indeed the solution. Alternative argument: Suppose λ1 = 0. Then from (ii), y = −λ2 ≤ 0, contradicting y ≥ e−x . So λ1 > 0, and (iii) gives y = e−x . Suppose λ2 = 0. Then from (ii), λ1 = y = e−x and (i) gives e−2x = 2x + 1. Deﬁne g(x) = 2x + 1 − e−2x . Then g(0) = 0 and g (x) = 2 + 2e−2x > 0. So the equation e−2x = 2x + 1 has no solution except x = 0. Thus λ2 > 0, etc.

14.10
2. The Lagrangian without Lagrange multipliers for the nonnegativity constraints is L = xey−x − 2ey − λ(y − 1 − x/2), so the ﬁrst-order conditions (14.10.3) and (14.10.4) are 1 (i) Lx = ey−x − xey−x + 2 λ ≤ 0 (= 0 if x > 0); (ii) Ly = xey−x − 2e − λ ≤ 0 (= 0 if y > 0); and 1 (iii) λ ≥ 0 with λ = 0 if y < 1 + 2 x. 1 If x = 0, then (i) implies ey + 2 λ ≤ 0, which is impossible, so x > 0. Then from (i) we get 1 (iv) xey−x = ey−x + 2 λ. 1 Suppose ﬁrst that λ > 0. Then (iii) and y ≤ 1 + x/2 imply (v) y = 1 + 2 x. Thus y > 0 and from 1 (ii) we have xey−x = 2e + λ. Using (iv) and (v), we get λ = 2ey−x − 4e = 2e(e− 2 x − 2). But then 1 λ > 0 implies that e− 2 x > 2, which contradicts x ≥ 0. This leaves λ = 0 as the only possibility. Then (iv) gives x = 1. If y > 0, then (ii) yields ey−1 = 2e, 1 and so y − 1 = ln(2e) = ln 2 + 1. With x = 1 this contradicts the constraint y ≤ 1 + 2 x. Hence y = 0, so we see that (x, y) = (1, 0) is the only point satisfying all the ﬁrst-order conditions, with λ = 0. (The extreme value theorem cannot be applied because the feasible set, i.e. the set of all points satisfying the constraints, is unbounded—it includes points (x, 0) for arbitrarily large x. However, you were told to assume that the problem has a solution.)
2 2 ∗ ∗ 3. The Lagrangian is L = x1 + 3x2 − x1 − x2 − k 2 − λ(x1 − k) − μ(x2 − k). A feasible triple (x1 , x2 , k ∗ ) ∗ ∗ solves the problem if and only if there exist numbers λ and μ such that (i) 1−2x1 −λ ≤ 0 (= 0 if x1 > 0); ∗ ∗ ∗ + λ + μ ≤ 0 (= 0 if k ∗ > 0); (iv) λ ≥ 0 with λ = 0 (ii) 3 − 2x2 − μ ≤ 0 (= 0 if x2 > 0); (iii) −2k ∗ ∗ if x1 < k ∗ ; (v) μ ≥ 0 with μ = 0 if x2 < k ∗ . ∗ = 0, then feasibility requires x ∗ = 0 and x ∗ = 0, and so (i) and (ii) imply that λ ≥ 1 and If k 2 1 ∗ μ ≥ 3, which contradicts (iii). Thus, k ∗ > 0. Next, if μ = 0, then (ii) and (iii) imply that x2 ≥ 3/2 ∗ ∗ ∗ and λ = 2k ∗ > 0. So x1 = k ∗ = 1/4, contradicting x2 ≤ k ∗ . So μ > 0, which implies that x2 = k ∗ . ∗ ∗ , then λ = 0, which contradicts (i). So 0 < x ∗ = 1 (1 − λ). Next, if λ > 0, then Now, if x1 = 0 < k 1 2 1 1 1 ∗ ∗ x1 = k ∗ = x2 = 2 (1 − λ) = 2 (3 − μ) = 2 (λ + μ) by (i), (ii), and (iii) respectively. But the last two equalities are only satisﬁed when λ = −1/3 and μ = 5/3, which contradicts λ ≥ 0. So λ = 0 after all,

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1 1 ∗ ∗ with x2 = k ∗ > 0, μ > 0, x1 = 2 (1 − λ) = 2 . Now, from (iii) it follows that μ = 2k ∗ and so, from (ii), ∗ ∗ ∗ that 3 = 2x2 + μ = 4k ∗ . The only possible solution is, therefore, (x1 , x2 , k ∗ ) = (1/2, 3/4, 3/4), with λ = 0 and μ = 3/2. 3 3 2 2 Finally, with λ = 0 and μ = 2 , the Lagrangian x1 + 3x2 − x1 − x2 − k 2 − 2 (x2 − k) is a quadratic ∗ ∗ function of (x1 , x2 , k), which has a maximum at the stationary point (x1 , x2 , k ∗ ). As stated at the end of ∗ ∗ ∗ the recipe in Section 14.9, this is sufﬁcient for the same (x1 , x2 , k ) to solve the problem.

Review Problems for Chapter 14
3. (a) If sales x of the ﬁrst commodity are increased, the increase in net proﬁt per unit increase in x is the sum of three terms: (i) p(x ∗ ), which is the gain in revenue due to the extra output; (ii) −p (x ∗ )x ∗ , which is the loss in revenue from selling x ∗ units due to the reduced price; (iii) −C1 (x ∗ , y ∗ ), which is minus the marginal cost of the additional output. In fact p(x ∗ ) + p (x ∗ )x ∗ is the derivative of the revenue function R(x) = p(x)x at x = x ∗ , usually called the marginal revenue. The ﬁrst necessary condition therefore states that the marginal revenue from increasing x must equal the (partial) marginal cost. The argument when considering any variation in the sales y of the second commodity is just the same. (b) With the restriction x + y ≤ m, we have to add the condition λ ≥ 0, with λ = 0 if x + y < m. ˆ ˆ y 4 (x1 , y1 )

2 (x4 , y4 ) -4

-2 -2

2

(x3 , y3 ) 4 6

x

(x2 , y2 )

-4

Figure SM14.R.5
1 5. (a) With L = x 2 + y 2 − 2x + 1 − λ( 4 x 2 + y 2 − b), the ﬁrst-order conditions are: 1 1 (i) L1 = 2x − 2 − 2 λx = 0; (ii) L2 = 2y − 2λy = 0; (iii) 4 x 2 + y 2 = b. From (ii), (1 − λ)y = 0, and thus λ = 1 or y = 0. 1 (I) Suppose ﬁrst that λ = 1. Then (i) gives x = 4 , and from (iii) we have y 2 = b− 4 x 2 = b− 4 , which gives 3 9

y = ± b − 4 . This gives two candidates: (x1 , y1 ) = (4/3, b − 4 ) and (x2 , y2 ) = (4/3, − b − 4 ). 9 9 9 √ (II) If y = 0, then from (iii), x 2 = 4b, i.e. x = ±2 b. This gives two further candidates: (x3 , y3 ) = √ √ The (2 b, 0) and (x4 , y4 ) = (−2 b, 0).√ objective function evaluated at the candidates are: f (x1 , y1 ) = √ √ √ f (x2 , y2 ) = b−1/3, f (x3 , y3 ) = (2 b−1)2 = 4b−4 b+1, f (x4 , y4 ) = (−2 b−1)2 = 4b+4 b+1 Clearly, (x4 , y4 ) is the maximum point. To decide which of the points (x3 , y3 ), (x1 , y1 ), or (x2 , y2 ) give √ the minimum, we have to decide which of the two values 4b − 4 b + 1 and b − 1 is smaller. The 3 √ √ √ 2 difference is 4b − 4 b + 1 − b − 1 = 3 b − 4 b + 4 = 3 b − 2 > 0 since b > 4 . Thus the 3 3 9 3 9 minimum occurs at (x1 , y1 ) and (x2 , y2 ).
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The constraint x 2 /4+y 2 = b describes the ellipse indicated in Fig. SM14.R.5. The objective function f (x, y) = (x − 1)2 + y 2 is the square of the distance between (x, y) and the point (1, 0). The level curves for f are therefore circles centred at (1, 0), and in the ﬁgure we see those two that pass through the maximum and minimum points. √ √ 7. (a) With L = x 2 − 2x + 1 + y 2 − 2y − λ[(x + y) x + y + b − 2 a], the ﬁrst-order conditions are: √ √ (i) L1 = 2x − 2 − λ[ x + y + b + (x + y)/2 x + y + b ] = 0, √ √ (ii) L2 = 2y − 2 − λ[ x + y + b + (x + y)/2 x + y + b ] = 0. From these equations it follows immediately that 2x − 2 = 2y − 2, so x = y. The constraint gives √ √ 2x 2x + b = 2 a. Cancelling 2 and then squaring each side. one obtains the second equation in (∗). (b) Differentiating yields: (i) dx = dy; (ii) 6x 2 dx + x 2 db + 2bx dx = da. From these equations we easily read off the ﬁrst-order partials of x and y w.r.t. a and b. Further, ∂ ∂x ∂ 2x = ∂a 2 ∂a ∂a = ∂ 1 12x + 2b ∂x 12x + 2b 6x + b =− =− =− 3 ∂a 6x 2 + 2bx (6x 2 + 2bx)2 ∂a (6x 2 + 2bx)3 4x (3x + b)3

8. With L = 10 − (x − 2)2 − (y − 1)2 − λ(x 2 + y 2 − a), the Kuhn–Tucker conditions are: (i) Lx = −2(x − 2) − 2λx = 0; (ii) Ly = −2(y − 1) − 2λy = 0; (iii) λ ≥ 0, with λ = 0 if x 2 + y 2 < a. Since the Lagrangian is concave when λ ≥ 0, these conditions are sufﬁcient for a maximum. One case occurs when λ = 0, implying that (x, y) = (2, 1). This is valid when a ≥ x 2 + y 2 = 5. The other case is when λ > 0. Then (i) implies that x = 2/(1 + λ) and (ii) implies that y = 1/(1 + λ). √ Because (iii) implies that x 2 + y 2 = a, we have 5/(1 + λ)2 = a and so λ = 5/a − 1, which is positive √ √ when a < 5. The solution then is (x, y) = (2 a/5, a/5). 9. (a) See main text. (b) The numbers (i)–(vi) in the following refer to the answer to (a) in the main text. From (ii) and (vi) we see that λ1 = 0 is impossible. Thus λ1 > 0, and from (iii) and (v), we see that (vii) (x ∗ )2 + r(y ∗ )2 = m. (I): Assume λ2 = 0. Then from (i) and (ii), y ∗ = 2λ1 x ∗ and x ∗ = 2λ1 ry ∗ , so x ∗ = 4λ2 rx ∗ . But (vi) 1 √ √ implies that x ∗ = 0. Hence λ2 = 1/4r and thus λ1 = 1/2 r. Then y ∗ = x ∗ / r, which inserted into 1 √ √ √ m/2 ≥ (vii) and solved for x ∗ yields x ∗ = m/2 and then y ∗ = m/2r. Note that x ∗ ≥ 1 ⇐⇒ √ √ ∗ = ∗ = 1 ⇐⇒ m ≥ 2. Thus, for m ≥ 2, a solution candidate is x m/2 and y m/2r, with √ λ1 = 1/2 r and λ2 = 0. (II): Assume λ2 > 0. Then x ∗ = 1 and from (vii) we have r(y ∗ )2 = m − 1. By (ii), one has y ∗ ≥ 0, √ so y ∗ = (m − 1)/r. Inserting these values into (i) and (ii), then solving for λ1 and λ2 , one obtains √ √ λ1 = 1/2 r(m − 1) and furthermore, λ2 = (2 − m)/ r(m − 1). Note that λ2 > 0 ⇐⇒ 1 < m < 2. √ √ Thus, for 1 < m < 2, the only solution candidate is x ∗ = 1, y ∗ = (m − 1)/r, with λ1 = 1/2 r(m − 1) √ and λ2 = (2 − m)/ r(m − 1). The objective function is continuous and the constraint set is obviously closed and bounded, so by the extreme value theorem there has to be a maximum. The solution candidates we have found are therefore optimal. (Alternatively, L11 = −2λ1 ≤ 0, L22 = −2rλ1 ≤ 0, and = L11 L22 − (L12 )2 = 4rλ2 − 1. 1 In the case m ≥ 2, = 0, and in the case 1 < m < 2, = 1/(m − 1) > 0. Thus in both cases, L(x, y) is concave.) √ √ √ (c) For m ≥ 2, V (r, m) = m/2 r, so Vm = 1/2 r = λ1 , and Vr = −m/4 r 3 , whereas Lr = √ √ −λ1 (y ∗ )2 = −(1/2 r)m/2r = −m/4 r 3 . √ √ For 1 < m < 2, V (r, m) = (m − 1)/r, so Vm = 1/2 r(m − 1) = λ1 , and Vr = −(1/2) (m − 1)/r 3 , √ whereas Lr = −λ1 (y ∗ )2 = −[1/2 r(m − 1)](m − 1)/r = −(1/2) (m − 1)/r 3 .
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Chapter 15 Matrix and Vector Algebra
15.1
6. The equation system is: ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 0.712y 0.158x x− − c = −95.05 34.30 0 c= .

− s + 0.158c = y−s+

x = 93.53 Solve the ﬁrst equation for y as a function of c. Insert this expression for y and x = 93.53 into the third equation. Solve it to get s as a function of c. Insert the results into the second equation and solve for c, and then solve for y and s in turn. The answer is given in the main text.

15.3
6. (a) We know that A is an m × n matrix. Let B be a p × q matrix. The matrix product AB is deﬁned if and only if n = p, and BA is deﬁned if and only if q = m. So for both AB and BA to be deﬁned, it is necessary and sufﬁcient that B is an n × m matrix. x y (b) We know from part (a) that B must be a 2 × 2 matrix. So let B = . Then BA = z w x y 1 2 x + 2y 2x + 3y 1 2 x y x + 2z y + 2w = ; AB = = . z w 2 3 z + 2w 2z + 3w 2 3 z w 2x + 3z 2y + 3w Hence, BA = AB if and only if (i) x + 2y = x + 2z, (ii) 2x + 3y = y + 2w, (iii) z + 2w = 2x + 3z, and (iv) 2z + 3w = 2y + 3w. Equations (i) and (iv) are both true if and only if y = z; then equations (ii) and (iii) are also both true if and only if, in addition, x = w − y. To summarize, all four equations hold if and only if y = z and x = w − y. Hence, the matrices B that commute with A are precisely those of the form w−y y 1 0 −1 1 B= =w +y y w 0 1 1 0 where y and w can be any real numbers.

15.4

⎞⎛ ⎞ ⎛ ⎞ a d e x ax + dy + ez 2. We start by performing the multiplication ⎝ d b f ⎠ ⎝ y ⎠ = ⎝ dx + by + f z ⎠. Next, e f c z ex + fy + cz ⎛ ⎞ ax + dy + ez (x, y, z) ⎝ dx + by + f z ⎠ = (ax 2 + by 2 + cz2 + 2dxy + 2exz + 2fyz), which is a 1 × 1 matrix. ex + fy + cz a 2 + bc ac + cd ab + bd bc + d 2

⎛

8. (a) Direct veriﬁcation yields (i) A2 = (a + d)A − (ad − bc)I2 =

(b) For the matrix A in (a), A2 = 0 if a + d = 0 and ad = bc, so one example with A2 = 0 = A is 1 1 A= . −1 −1 (c) By part (a), A3 = (a + d)A2 − (ad − bc)A. So A3 = 0 implies that (a + d)A2 = (ad − bc)A and then, multiplying each side by A once more, that (a + d)A3 = (ad − bc)A2 . If A3 = 0, therefore, one
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has two cases: (i) A2 = 0; (ii) ad − bc = 0. But in the second case, one has (a + d)A2 = 0, which gives rise to two subcases: (a) A2 = 0; (b) ad − bc = 0 and (a + d) = 0. Now, even in case (ii)(b), the result of part (a) implies that A2 = 0, which is therefore true in every case.

15.5
6. In general, for any natural number n > 3, one has ((A1 A2 · · · An−1 )An ) = An (A1 A2 · · · An−1 ) . As the induction hypothesis, suppose the result is true for n − 1. Then the last expression becomes An An−1 · · · , A2 A1 , so the result is true for n.
1 1 8. (a) TS = S is shown in the text. A similar argument shows that T2 = 2 T + 2 S. To prove the last equality, we do not have to consider the individual elements. Instead, we premultiply the last equation by T and 1 1 1 1 1 1 1 1 1 3 then use TS = S to obtain T3 = TT2 = T( 2 T + 2 S) = 2 T2 + 2 TS = 2 ( 2 T + 2 S) + 2 S = 4 T + 4 S.

(b) We prove by induction that the appropriate formula is (∗) Tn = 21−n T + (1 − 21−n )S. This formula is correct for n = 1 (and, by part (a), also for n = 2, 3). Suppose (∗) is true for n = k. Then premultiplying by T and using the two ﬁrst equalities in (a), one obtains Tk+1 = TTk = T(21−k T + (1 − 21−k )S) = 1 1 21−k T2 +(1−21−k )TS = 21−k ( 2 T+ 2 S)+(1−21−k )S = 2−k T+2−k S+S−2·2−k S = 2−k T+(1−2−k )S, which is formula (∗) for n = k + 1.

15.6
3. By using the following elementary operations successively: (i) subtract the third equation from the ﬁrst; (ii) subtract the new ﬁrst equation from the two others; (iii) interchange the second and the third equation; (iv) multiply the second equation by −3 and add it to the third equation, we ﬁnd that w x y z ⎞ ⎞ ⎛ ⎛ 1 0 2 2 1 − c2 2 1 4 3 1 ⎠ ⎝ 1 3 2 −1 3c ⎠ ∼ ⎝ 0 1 0 −1 2c2 − 1 1 1 2 1 c2 0 0 0 0 −5c2 + 3c + 2 We can tell from the last matrix that the system has solutions if and only if −5c2 + 3c + 2 = 0, that is, if and only if c = 1 or c = −2/5. For these particular values of c we get the solutions in the text. (The ﬁnal answer can take many equivalent forms depending on how you arrange the elementary operations.) 4. (a) After moving the ﬁrst row down to row number three, then applying elementary row operations, we obtain the successive augmented matrices ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ 1 2 1 b2 1 2 1 b2 1 2 1 b2 3 1 ⎝3 4 7 b3 ⎠ ∼ ⎝ 0 −2 4 b3 − 3b2 ⎠ ∼ ⎝ 0 1 −2 2 b2 − 2 b3 ⎠ a 1 a + 1 b1 0 1 − 2a 1 b1 − ab2 0 1 − 2a 1 b1 − ab2 ⎛ ⎞ 1 2 1 b2 3 1 ⎝0 1 ⎠ ∼ −2 2 b2 − 2 b3 3 1 0 0 3 − 4a b1 + (2a − 2 )b2 + ( 2 − a)b3 Obviously, there is a unique solution if and only if a = 3/4.
1 (b) Put a = 3/4 in part (a). Then the last row in the matrix in (a) becomes (0, 0, 0, b1 − 4 b3 ). It follows 1 1 that if b1 = 4 b3 there is no solution. If b1 = 4 b3 there is an inﬁnite set of solutions. For an arbitrary real 3 1 t, there is a unique solution with z = t. Then the second equation gives y = 2 b2 − 2 b3 + 2t, and ﬁnally the ﬁrst equation gives x = −2b2 + b3 − 5t.

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15.8
2. (a) See the text. (b) According to the point–point formula, the line L through (3, 1) and (−1, 2) has the 1 7 equation x2 = − 4 x1 + 4 or x1 + 4x2 = 7. The line segment S is traced out by having x1 run through [3, −1] as x2 runs through [1, 2]. Now, λa + (1 − λ)b = (−1 + 4λ, 2 − λ). Any point (x1 , x2 ) on L 1 satisﬁes x1 + 4x2 = 7 and equals (−1 + 4λ, 2 − λ) for λ = 4 (x1 + 1) = 2 − x2 . So there is a one-to-one correspondence between points: (i) that lie on the line segment joining a = (3, 1) and b = (−1, 2); (ii) whose coordinates can be written as (−1 + 4λ, 2 − λ) for some λ in [0, 1].

15.9
3. First, note that (5, 2, 1) − (1, 0, 2) = (4, 2, −1) and (2, −1, 4) − (1, 0, 2) = (1, −1, 2) are two vectors in the plane. The normal (p1 , p2 , p3 ) to the plane must be orthogonal to both these vectors, so (4, 2, −1) · (p1 , p2 , p3 ) = 4p1 + 2p2 − p3 = 0 and (1, −1, 2) · (p1 , p2 , p3 ) = p1 − p2 + 2p3 = 0. One solution of these two equations is (p1 , p2 , p3 ) = (1, −3, −2). Then using formula (4) with (a1 , a2 , a3 ) = (2, −1, 4) yields (1, −3, −2) · (x1 − 2, x2 + 1, x3 − 4) = 0, or x1 − 3x2 − 2x3 = −3. A more pedestrian approach is to assume that the equation is ax + by + cz = d and require the three points to satisfy the equation: a + 2c = d, 5a + 2b + c = d, 2a − b + 4c = d. Solve for a, b, and c in terms of d, insert the results into the equation ax + by + cz = d and cancel d.

Review Problems for Chapter 15
2 ⎝1 8. (b) 3 ⎛ 1 ⎝0 ∼ 0 ⎛ 1 ⎝0 ∼ ⎛0 1 ⎝0 ∼ 0 (c) ∼ 1 5 ⎛

⎞ ⎛ ⎞ ⎛ 2 −1 2 ← 1 −3 1 0 −2 −3 1 −3 1 ⎠ ← ∼ ⎝2 ⎠← ⎝0 ∼ −3 1 0 2 −1 2 8 −3 4 −1 1 3 4 −1 1 ← 0 13 −4 ⎛ ⎞ ⎞ 1 −3 1 0 −3 1 0 1 −3/8 1/4 ⎠ 1 −3/8 1/4 ⎠ −13 ∼ ⎝ 0 0 0 7/8 −9/4 8/7 13 −4 1 ← ⎛ ⎞ ⎞ 1 0 −1/8 3/4 ← −3 1 0 ← ⎠← ⎠ 3 ∼ ⎝ 0 1 −3/8 1/4 1 −3/8 1/4 3/8 1/8 0 0 1 −18/7 0 1 −18/7 ⎞ 0 0 3/7 1 0 −5/7 ⎠. The solution is x1 = 3/7, x2 = −5/7, x3 = −18/7. 0 1 −18/7 3 1 4 1 0 0 −5 ∼ ← 1 0 3 −14 4 0 −19 0 −1/14

⎞ 0 2 ⎠ 1/8 1

1 0 −1/14 0 1 3 4 0 ← ∼ −3 0 1 19/14 0 0 1 19/14 0 The solution is x1 = (1/14)x3 , x2 = −(19/14)x3 , where x3 is arbitrary. (One degree of freedom.) 11. (a) See main text. (b) In (a) we saw that a can be produced even without throwing away outputs. For b to be possible if we are allowed to throw away output, there must exist a λ in [0, 1] such that 6λ + 2 ≥ 7, −2λ + 6 ≥ 5, and −6λ + 10 ≥ 5. These inequalities reduce to λ ≥ 5/6, λ ≤ 1/2, λ ≤ 5/6, which are incompatible. (c) Revenue = R(λ) = p1 x1 + p2 x2 + p3 x3 = (6p1 − 2p2 − 6p3 )λ + 2p1 + 6p2 + 10p3 . If the constant slope 6p1 − 2p2 − 6p3 is > 0, then R(λ) is maximized at λ = 1; if 6p1 − 2p2 − 6p3 is < 0, then R(λ)
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is maximized at λ = 0. Only in the special case where 6p1 − 2p2 − 6p3 = 0 can the two plants both remain in use. 12. If PQ − QP = P, then PQ = QP + P, and so P2 Q = P(PQ) = P(QP + P) = (PQ)P + P2 = (QP+P)P+P2 = QP2 +2P2 . Thus, P2 Q−QP2 = 2P2 . Moreover, P3 Q = P(P2 Q) = P(QP2 +2P2 ) = (PQ)P2 + 2P3 = (QP + P)P2 + 2P3 = QP3 + 3P3 Hence, P3 Q − QP3 = 3P3 . To prove the result for general k, suppose as the induction hypothesis that Pn Q − QPn = nPn for n = k. Then, for n = k + 1, one has Pn Q = P(Pk Q) = P(QPk + kPk ) = (PQ)Pk + kPk+1 = (QP + P)Pk + kPk+1 = QPk+1 + (k + 1)Pk+1 , so the induction hypothesis is also true for n = k + 1.

Chapter 16 Determinants and Inverse Matrices
16.1
9. (a) See answer in main text. (b) The suggested substitutions produce the two equations Y2 = c2 Y2 + A2 + m1 Y1 − m2 Y2 −m1 Y1 + (1 − c2 − m2 )Y2 = A2 Y1 = c1 Y1 + A1 + m2 Y2 − m1 Y1 ; or (1 − c1 − m1 )Y1 − m2 Y2 = A1 ; which can be rewritten in the matrix form 1 − c 1 − m1 −m1 −m2 1 − c 2 − m2 Y1 Y2 = A1 A2

For these to soluble, we need to assume that D = (1 − c1 − m1 )(1 − c2 − m2 ) − m1 m2 = 0. When D = 0, the answers in the text can be derived using Cramer’s rule. (c) Y2 depends linearly on A1 . Economists usually assume that D given in part (b) is positive, as it will be provided that the parameters c1 , c2 , m1 , m2 are all sufﬁciently small. Then increasing A1 by one unit changes Y2 by the factor m1 /D ≥ 0, so Y2 increases when A1 increases. Here is an economic explanation: An increase in A1 increases nation 1’s income, Y1 . This in turn increases nation 1’s imports, M1 . However, nation 1’s imports are nation 2’s exports, so this causes nation 2’s income, Y2 , to increase, and so on.

16.2
1 −1 0 1. (a) Sarrus’s rule yields: 1 3 2 = 0 − 2 + 0 − 0 − 0 − 0 = −2. 1 0 0 1 −1 0 (b) By Sarrus’s rule, 1 3 2 = 3 − 2 − 0 − 0 − 4 − (−1) = −2. 1 2 1 (c) Because a21 = a31 = a32 = 0, the only non-zero term in the expansion (3) is the product of the terms on the main diagonal. The determinant is therefore adf . Alternatively, Sarrus’s rule gives the same answer. a 0 b (d) By Sarrus’s rule, 0 e 0 = aed + 0 + 0 − bec − 0 − 0 = e(ad − bc). c 0 d
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3. (a) The determinant of the coefﬁcient matrix is |A| = The numerators in (16.2.4) are (verify!) 2 0 −6 −1 1 1 −1 = −4 , −1 −1 1 1 −1 2 0 −6

1 1 −1

−1 1 1 −1 = −4. −1 −1 1 1 −1 −1 1 −1 2 0 = −12 −6

1 −1 = −8 , −1

Hence, (4) yields the solution x1 = 1, x2 = 2, and x3 = 3. Inserting this into the original system of equations conﬁrms that this is a correct answer. (b) The determinant of the coefﬁcient matrix is equal to −2, and the numerators in (16.2.4) are all 0, so the unique solution is x1 = x2 = x3 = 0. (c). Follow the pattern in (a) to get the answer in the main text. 8. (a) Substituting T = d + tY into the expression for C gives C = a − bd + b(1 − t)Y . Substituting for C in the expression for Y then yields Y = a + b(Y − d − tY ) + A0 . Then solve for Y , T , and C in turn to derive the answers given in (b) below. ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ 1 −1 0 1 −1 0 Y A0 (b) We write the system as ⎝ −b 1 b = 1+bt −b, 1 b ⎠ ⎝ C ⎠ = ⎝ a ⎠. With D = −b d −t 0 1 −t 0 1 T Cramer’s rule yields 1 A0 −1 0 1 b −b a a − bd + A0 0 1 −t d = , C= D 1 − b(1 − t) D 1 −1 A0 −b 1 a t (a + A0 ) + (1 − b)d −t 0 d = D 1 − b(1 − t) A0 a d 0 b 1

Y =

=

a − bd + A0 b(1 − t) 1 − b(1 − t)

T =

(This problem is meant to train you in using Cramer’s rule. It is also a warning against its overuse, since solving the equations by systematic elimination is much more efﬁcient.)

16.3
1. Each of the three determinants is a sum of 4! = 24 terms. In (a) there is only one nonzero term. In fact, according to (16.3.4), the value of the determinant is 24. (b) Only two terms in the sum are nonzero: the product of the elements on the main diagonal, which is 1 · 1 · 1 · d, with a plus sign; and the term 1 0 0 a 0 1 0 b 0 0 1 c 1 0 0 d

Since there are 5 rising lines between the pairs, the sign of the product 1 · 1 · 1 · a must be minus. So the value of the determinant is d − a. (c) 4 terms are nonzero. See the text.
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16.4
14. The description in the answer in the text amounts to the following steps: a+b a Dn = . . . a a a+b . . . a ··· a ← 1 ··· a . .. . . . ··· a ··· ··· .. . 1 a . . . ··· ← na + b ··· a = . .. . . . a 1 −a ← na + b a+b . . . a · · · na + b ··· a . .. . . . ··· a + b 1 0 . . . 0 1 b . . . 0 ··· ··· .. . 1 0 . . .

1 a = (na + b) . . . a

1 a+b . . . a

· · · −a ··· = (na + b) .. .

··· a + b ←

··· b

According to (16.3.4), the last determinant is bn−1 . Thus Dn = (na + b)bn−1 .

16.5
1. (a) See the text. (b) One possibility is to expand by the second row or the third column (because they both have two zero entries). But it is easier ﬁrst to use elementary operations to get a row or a column with at most one non-zero element. For example: 1 0 2 −2 2 −1 −1 0 3 0 0 −1 0 −6 5 −2 4 11 3 ← 3 ← 2 1 0 = 0 0 2 −1 −5 4 0 −6 5 3 0 −6 5 4 11 −5 11 −60 = −(−330 + 300) = 30 55

−1 = −5 4

11 −5 4 −1 = −5 ← 0 11 ← 0

11 −6 −60 = −1 5 55

(c) See text for the simple answer. When computing determinants one can use elementary column as well as row operations, but column operations become meaningless when solving linear equation systems using Gaussian elimination.

16.6
8. B2 +B = 3/2 −5 −1/2 5 1 0 + = = I. One can either verify by direct matrix −1/4 3/2 1/4 −1/2 0 1 multiplication that B3 − 2B + I = 0, or somewhat more easily, use the relation B2 + B = I to argue that B2 = I − B and so B3 − 2B + I = B(I − B) − 2B + I = B − B2 − 2B + I = −B2 − B + I = 0. Furthermore B2 + B = I implies that B(B + I) = I. It follows from (16.6.4) that B−1 = B + I.

16.7
1. (a) |A| = 10 − 12 = −2, and the adjoint is
1 A−1 = − 2

C11 C12 −3 2

C21 C22 =

=

5 −4

−3 , so the inverse is 2 3/2 −1

5 −4

−5/2 2

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(b) The adjoint of B is

C11 adj B = ⎝ C12 C13

⎛

C21 C22 C23

⎞ ⎛ 1 C31 C32 ⎠ = ⎝ 2 C33 4

4 −1 −2

⎞ 2 4⎠ −1

and |B| = b11 C11 + b21 C21 + b31 C31 = 1 · 1 + 2 · 4 + 0 · 2 = 9 by expansion along the ﬁrst column. Hence, ⎛ ⎞ 1 4 2 B−1 = 1 (adj B) = 1 ⎝ 2 −1 4⎠ 9 9 4 −2 −1 (c) Since the second column of C equals −2 times its third column, the determinant of C is zero, so there is no inverse. ⎛ ⎞ 0.72 0.64 0.40 3. The determinant of I − A is |I − A| = 0.496, and the adjoint is adj(I − A) = ⎝ 0.08 0.76 0.32 ⎠. 0.16 0.28 0.64 ⎛ ⎞ 1.45161 1.29032 0.80645 1 · adj(I − A) ≈ ⎝ 0.16129 1.53226 0.64516 ⎠, rounded to ﬁve decimal Hence (I − A)−1 = 0.496 0.32258 0.56452 1.29032 ⎛ ⎞ 0.72 0.64 0.40 125 1000 = and adj(I − A) = ⎝ 0.08 0.76 0.32 ⎠ = places. If you want an exact answer, note that 496 62 0.16 0.28 0.64 ⎛ ⎞ ⎛ ⎞ 18 16 10 18 16 10 5 ⎝ 1 ⎝ 2 19 8 ⎠. This gives (I − A)−1 = 2 19 8 ⎠. 25 62 4 7 16 4 7 16 4. Let B denote the n × p matrix whose kth column has the elements b1k , b2k , . . . , bnk . The p systems of n equations in n unknowns can be expressed as AX = B, where A is n × n and X is n × p. Following the method illustrated in Example 2, exactly the same row operations that transform the n × 2n matrix (A : I) into (I : A−1 ) will also transform the n × (n + p) matrix (A : B) into (I : B∗ ), where ∗ B∗ is the matrix with elements bij . (In fact, because these row operations are together equivalent to −1 , it must be true that B∗ = A−1 B.) When k = r, the solution to the system is premultiplication by A ∗ ∗ ∗ x1 = b1r , x2 = b2r , . . . , xn = bnr . 5. (a) 1 2 3 4 1 0 ∼ 0 1 ⎛ 1 ⎝2 3 ⎛ 1 ∼⎝0 0 ⎛ 1 ∼⎝0 0 −3 1 0 ∼ 0 1 ← −2 1 3 1 −2 2 2 4 5 3 5 6 1 0 0 0 1 0 1 0 2 −2 1 −3 0 1
1 −2

∼

1 0

2 1

1
3 2

0 ← 1 −2 −2

(b)

2 3 −1 −3 0 −1 0 −3 1 3 0 1

⎞ 0 0⎠ 1 1 0 −3 0 −2 1 −5 0 3 0 2 −1

⎛ −2 −3 1 2 ∼ ⎝0 ← 0 ← 0 −1 ⎞ ⎛ 1 2 0 1 ⎠ −1 ∼ ⎝ 0 1 0 −1 0 0 ⎞ ⎛ ← 2 1 ∼ ⎝0 −1 ⎠ ← 0 −3 3 0

3 −1 −3 3 3 1 0 1 0

1 −2 −3 1 0 3 0 2 −1 0 1 0 −3 2 1

0 1 0

⎞ 0 0⎠ ← 1 ← ⎞ 0 ← −1 ⎠ −2 0 ⎞ −3 2 3 −1 ⎠ −1 0

(c) We see that the third row equals the ﬁrst row multiplied by −3, so the matrix has no inverse.
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16.8
1 1. (a) The determinant |A| of the coefﬁcient matrix is |A| = 2 1 The determinants in (16.8.2) are −5 6 −3 2 −1 −1 −1 1 = 19 , −3 1 2 1 2 −1 −1 −1 1 = 19. −3 1 2 1 2 −1 −1 −5 6 = 38 −3

−5 −1 6 1 = −38 , −3 −3

According to (16.8.4) the solution is x = 19/19 = 1, y = −38/19 = −2, and z = 38/19 = 2. Inserting this into the original system of equations conﬁrms that this is the correct answer. 1 1 0 0 1 0 1 0 (b) The determinant |A| of the coefﬁcient matrix is . Subtracting the fourth column from 0 1 1 1 0 1 0 1 the second leaves only one non-zero element in the second column, and so reduces the determinant to 1 1 0 − 0 1 1 = −1. We ask you to check that the other determinants in (16.8.2) are 0 0 1 3 2 6 1 1 0 1 1 0 0 1 0 = 3, 1 1 0 1 1 1 0 0 3 2 6 1 0 1 1 0 0 0 = −6 , 1 1 1 1 0 0 1 0 1 1 3 2 6 1 0 0 = −5 , 1 1 1 1 0 0 1 0 1 1 0 1 1 0 2 3 =5 6 1

According to (16.8.4) the solution is x = −3, y = 6, z = 5, and u = −5. Inserting this into the original system of equations conﬁrms that this is the correct answer. (Of course, there is a much quicker way to solve these four equations. Subtracting the fourth from the third yields z = 5 immediately. Then the second equation gives x = −3; the ﬁrst gives y = 6; and the last gives u = −5.) 3. According to Theorem 16.8.2, the system has nontrivial solutions if and only if the determinant of the coefﬁcient equal to 0. Expansion along the ﬁrst row gives a b c b c a c c a =a a b a b −b b c a b +c b c c a

= a(bc − a 2 ) − b(b2 − ac) + c(ab − c2 ) = 3abc − a 3 − b3 − c3 . Thus the system has nontrivial solutions if and only if 3abc − a 3 − b3 − c3 = 0.

Review Problems for Chapter 16 q −1 q − 2 1 −p q 5. Expanding along column 3 gives |A| = 1 −p 2 − p = (q − 2) − (2 − p) 2 −1 2 2 −1 0 (q − 2)(−1 + 2p) − (2 − p)(−q + 2) = (q − 2)(p + 1), but there are many other ways.
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|A + E| =

q +1 0 2 1−p 3 0

q −1 q +1 q −1 = 2(p − 1)(q − 2) 3 − p = (1 − p) 3 1 1

For the rest, see the answer in the text. 8. (a) This becomes easy after noting that ⎛ ⎞⎛ ⎞ ⎛ ⎞ n n ... n 1 1 ... 1 1 1 ... 1 ⎜1 1 ... 1⎟⎜1 1 ... 1⎟ ⎜n n ... n⎟ U2 = ⎜ . . . . . ⎟ ⎜ . . .. . ⎟ = ⎜ . . .. . ⎟ = nU ⎝. . . . ⎠⎝ . . . .⎠ ⎝. . . .⎠ . . . . . . . . . 1 1 ... 1 1 1 ... 1 n n ... n (b) The trick is to note that ⎛ 4 3 ⎝3 4 A= 3 3 ⎞ ⎛ 3 1 ⎠ = ⎝0 3 4 0 0 1 0 ⎞ ⎛ ⎞ 0 3 3 3 0 ⎠ + ⎝ 3 3 3 ⎠ = I3 + 3U 1 3 3 3

From (a), (I3 + 3U)(I3 + bU) = I3 + (3 + b + 3 · 3bU) = I3 + (3 + 10b)U. This can be made equal to I3 by choosing b = −3/10. It follows that ⎛ ⎞ ⎛ ⎞ ⎛ 3 3 3 ⎞ 7 −3 −3 1 0 0 10 10 10 1 3 3 3 A−1 = (I3 + 3U)−1 = I3 − (3/10)U = ⎝ 0 1 0 ⎠ − ⎝ 10 10 10 ⎠ = 10 ⎝ −3 7 −3 ⎠ 3 3 3 −3 −3 7 0 0 1 10 10 10 10. (a) Gaussian elimination with the indicated elementary row operations yields ⎛ ⎞ ⎞ ⎛ a 1 4 2 ← −2 −a 1 0 −3 a ⎝ 2 1 a2 2 ⎠ ⎝ 2 1 a2 2 ⎠ ← ∼ 1 0 −3 a ← a 1 4 2 ← ⎞ ⎛ 1 0 −3 a ∼ ⎝ 0 1 a 2 + 6 −2a + 2 ⎠ −1 0 1 3a + 4 −a 2 + 2 ← ⎞ ⎛ 1 0 −3 a ∼ ⎝0 1 −2a + 2 ⎠ a2 + 6 2 + 3a − 2 −a 2 + 2a 0 0 −a It follows that the system has a unique solution if and only if −a 2 + 3a − 2 = 0, i.e. if and only if a = 1 and a = 2. If a = 2, the last row consists only of 0’s so there are inﬁnitely many solutions, whereas if a = 1, there are no solutions. (b) If we perform the same elementary operations on the associated extended matrix as in (a), then the ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ b3 b3 b1 fourth column is transformed from ⎝ b2 ⎠ ﬁrst to ⎝ b2 ⎠, then second to ⎝ b2 − 2b3 ⎠, and ﬁnally third b3 b1 b1 − ab3 ⎛ ⎞ b3 ⎠. Thus, the ﬁnal extended matrix is to ⎝ b2 − 2b3 b1 − b2 + (2 − a)b3 ⎛ ⎞ 1 0 −3 b3 ⎝0 1 ⎠ a2 + 6 b2 − 2b3 0 0 −a 2 + 3a − 2 b1 − b2 + (2 − a)b3

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We see that there are inﬁnitely many solutions if and only if all elements in the last row are 0, which is true if and only if either (i) a = 1 and b1 − b2 + b3 = 0, or (ii) a = 2 and b1 = b2 . 15. (a) See the text. (b) The trick is to note that the cofactor expansions of |A|, |B| and |C| along the rth n n n row take the respective forms j =1 arj Crj , j =1 brj Crj and j =1 (arj + brj )Crj for exactly the same collection of cofactors Crj (j = 1, 2, . . . , n). Then, of course, n n n

|C| = j =1

(arj + brj )Crj = j =1

arj Crj + j =1

brj Crj = |A| + |B|

16. It is a bad idea to use “brute force” here. Note instead that rows 1 and 3 and rows 2 and 4 in the determinant have “much in common”. So begin by subtracting row 3 from row 1, and row 4 from row 2. According to Theorem 16.4.1(F), this does not change the value of the determinant. This gives, if we thereafter use Theorem 16.4.1(C), 0 b−a x a a−b 0 b x 0 a−b x b b−a 0 0 −1 = (a − b)2 a x x a 1 0 b x 0 1 x b −1 0 0 −1 = (a − b)2 a x x a 1 0 b x 0 1 x b 0 0 a+b 2x

The last equality is obtained by adding column 2 to column 4 in the middle determinant. If we expand the last determinant by the row 1, we obtain successively −(a − b)2 −1 x a 1 x b 0 a + b = −(a − b)2 [−2x 2 + b(a + b) − 2x 2 + a(a + b)] 2x = (a − b)2 [4x 2 − (a + b)2 ] = (a − b)2 [2x − (a + b)][2x + (a + b)] The conclusion follows.

Chapter 17 Linear Programming
17.1
3. The set A corresponds to the shaded polygon in Fig. SM17.1. The following arguments explain the answers given in the text. (a) The solution is obviously at the uppermost point P in the polygon because it has the largest x2 coordinate among all points in A. Point P is where the two lines −2x1 + x2 = 2 and x1 + 2x2 = 8 intersect, and the solution of these two equations is (x1 , x2 ) = (4/5, 18/5). (b) The point in A with the largest x1 coordinate is obviously Q = (8, 0). (c) The line 3x1 + 2x2 = c for one typical value of c is the dashed line in Fig. SM17.1.3. As c increases, the line moves out farther and farther to the north-east. The line that has the largest value of c, and still has a point in common with A, is the one that passes through the point Q in the ﬁgure. (d) The line 2x1 − 2x2 = c (or x2 = x1 − c/2) makes a 45◦ angle with the x1 axis, and intersects the x1 axis at c/2. As c decreases, the line moves up and to the left. The line in this family that has the smallest value of c, and still has a point in common with A, is the one that passes through the point P in the ﬁgure.
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(e) The line 2x1 + 4x2 = c is parallel to the line x1 + 2x2 = 8 in the ﬁgure. As c increases, the line moves out farther and farther to the north-east. The line with points in common with A that has the largest value of c is obviously the one that coincides with the line x1 + 2x2 = 8. So all points on the line segment between P and Q are solutions. (f) The line −3x1 − 2x2 = c is parallel to the dashed line in the ﬁgure, and intersects the x1 axis at −c/3. As c decreases, the line moves out farther and farther to the north-east, so the solution is at Q = (8, 0). (We could also argue like this: Minimizing −3x1 − 2x2 subject to (x1 , x2 ) ∈ A is obviously equivalent to maximizing 3x1 + 2x2 subject to (x1 , x2 ) ∈ A, so the solution is the same as the one in part (c).) x2 −2x1 + x2 = 2

4

P

3x1 + 2x2 = c

A Q 8 x1 x1 + 2x2 = 8
Figure SM17.1.3

17.2
1. (a) See Fig. A17.1.1a in the text. When 3x1 + 2x2 ≤ 6 is replaced by 3x1 + 2x2 ≤ 7 in Problem 17.1.1, the feasible set expands because the steeper line through P is moved out to the right. The new optimal point is at the intersection of the lines 3x1 + 2x2 = 7 and x1 + 4x2 = 4, and it follows that the solution is (x1 , x2 ) = (2, 1/2). The old maximum value of the objective function was 36/5. The new optimal value 1 is 3 · 2 + 4 · 2 = 8 = 40/5, and the difference in optimal value is u∗ = 4/5. 1 (b) When x1 +4x2 ≤ 4 is replaced by x1 +4x2 ≤ 5, the feasible set expands because the line x1 +4x2 = 4 is moved up. The new optimal point is at the intersection of the lines 3x1 + 2x2 = 6 and x1 + 4x2 = 5, and it follows that the solution is (x1 , x2 ) = (7/5, 9/10). The old maximum value of the objective function was 36/5. The new optimal value is 39/5, and the difference in optimal value is u∗ = 3/5. 2 (c) See answer in the main text.

17.3
1. (a) From Fig. A17.3.1a in the text it is clear as c increases, the dashed line moves out farther and farther to the north-east. The line that has the largest value of c and still has a point in common with the feasible set, is the one that passes through the point P , which has coordinates (x, y) = (0, 3), where the associated maximum value is 2 · 0 + 3 · 7 = 21.
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(b) In Fig. A17.3.1b, as c decreases, the dashed line moves farther and farther to the south-west. The line that has the smallest value of c and still has a point in common with the feasible set, is the one that passes through the point P , which has coordinates (u1 , u2 ) = (0, 1). The associated minimum value is 20u1 + 21u2 = 21. (c) Yes, because the maximum of the primal in (a) and the minimum of the dual in (b) both equal 21. 3. (a) See the text. (b) The problem is illustrated graphically in Fig. SM17.3.3, which makes clear the answer given in the main text. (c) Relaxing the ﬁrst constraint to 2x1 + x2 ≤ 17 allows the solution to move out to the intersection of the two lines 2x1 + x2 = 17 and x1 + 2x2 = 11. So the new solution is x1 = 23/3, x2 = 5/3, where the proﬁt is 3900. Relaxing the second constraint to x1 + 4x2 ≤ 17 makes no difference, because some capacity in division 2 remained unused anyway at (7, 2). Relaxing the third constraint to x1 +2x2 ≤ 12, the solution is no longer where the lines 2x1 +x2 = 16 and x1 + 2x2 = 12 intersect, namely at (x1 , x2 ) = (20/3, 8/3), because this would violate the second constraint x1 + 4x2 ≤ 16. Instead, as a carefully drawn graph shows, the solution occurs where the ﬁrst and second constraints both bind, at the intersection of the two lines 2x1 + x2 = 16 and x1 + 4x2 = 16, namely at (x1 , x2 ) = (48/7, 16/7). The resulting proﬁt is 27200/7 = 3885 5 < 3900. So it is division 7 1 that should have its capacity increased. Indeed, if the capacity of division 3 is increased by 1 hour per day, some of that increase has to go to waste because of the limited capacities in divisions 1 and 2. x2 2x1 + x2 = 16 x1 + 2x2 = 11 5
∗ ∗ (x1 , x2 ) = (7, 2)

x1 + 4x2 = 16 x1

5 400x1 + 500x2 = constant

10

Figure SM17.3.3

17.4
2. (a) The problem is similar to Problem 17.3.3. The linear program is set out in the main text. As a carefully drawn graph will show, the solution occurs at the intersection of the two lines 6x1 + 3x2 = 54 and 5x1 + 5x2 = 50, where (x1 , x2 ) = (8, 2). Note that the maximum proﬁt is 300 · 8 + 200 · 2 = 2800. (b) The dual problem is ⎧ ⎪ 6u1 + 4u2 + 5u3 ≥ 300 ⎨ min (54u1 + 48u2 + 50u3 ) subject to ⎪ ⎩ 3u1 + 6u2 + 5u3 ≥ 200 u1 , u2 , u3 ≥ 0

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

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∗ ∗ The optimal solution of the primal is x1 = 8, x2 = 2. Since the optimal solution of the primal has ∗ ∗ both x1 and x2 positive, the ﬁrst two constraints in the dual are satisﬁed with equality at the optimal triple (u∗ , u∗ , u∗ ). But the second constraint of the primal problem is satisﬁed with inequality, because 1 2 3 ∗ ∗ 4x1 + 6x2 = 44 < 48. Hence u∗ = 0, with 6u∗ + 5u∗ = 300 and 3u∗ + 5u∗ = 200. It follows that 2 1 3 1 3 u∗ = 100/3, u∗ = 0, and u∗ = 20. Moreover, the optimum value 54u∗ + 48u∗ + 50u∗ = 2800 of the 1 2 3 1 2 3 dual equals the optimum value of the primal. (c) See the text.

17.5
3. (a) See the text. (b) The dual is given in the text. It is illustrated graphically in Fig. SM17.5.3, where (1) labels the ﬁrst, (2) the second, and (3) the third constraint. The parallel dashed lines are level curves of the objective 300x1 + 500x2 . We see from the ﬁgure that optimum occurs at the point where the ﬁrst and ∗ ∗ ∗ ∗ the third constraint are satisﬁed with equality—i.e., where 10x1 + 20x2 = 10 000 and 20x1 + 20x2 = ∗ ∗ 11 000. The solution is x1 = 100 and x2 = 450. The maximum value of the objective function is 300 · 100 + 500 · 450 = 255 000. y 1000 900 800 700 600 500 400 300 200 100 (3) x (1) (2)

100 200 300 400 500 600 700 800 900 1000
Figure SM17.5.3

By complementary slackness, those constraints in the dual problem that correspond to variables which are positive at the optimum of the primal must be satisﬁed with equality. Since the second constraint in the ∗ primal in the optimum is satisﬁed with strict inequality, (20·100+10·450 < 8000), we have y2 = 0. Hence ∗ ∗ ∗ ∗ ∗ 10y1 + 20y3 = 300, 20y1 + 20y3 = 500. It follows that the solution of the primal problem is y1 = 20, ∗ ∗ y2 = 0, y3 = 5. The maximum value of the objective function is 10 000 · 20 + 8 000 · 0 + 11 000 · 5 = 255 000. (c) If the cost per hour in factory 1 increases by 100, this has no effect on the constraints in the primal, but does increase the right-hand side of the ﬁrst constraint of the dual by 100. An approximate answer
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

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CHAPTER 17

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of 100 · 20 = 2000 is the increase in cost that would result from choosing the same feasible point ∗ ∗ ∗ (y1 , y2 , y3 ) = (20, 0, 5) in the primal. This may over-estimate the increased minimum cost, however, because it may be better to switch some production away from factory 1, now it has become more expensive. To ﬁnd the exact answer, we check whether any production is switched by considering the dual linear program again. It is unchanged except that constraint (1) becomes 10x1 + 20x2 ≤ 10100. Looking again at Fig. SM17.5.3, the solution to the dual still occurs at the intersection of the lines (1) and (3), even after (1) has been shifted out. In particular, therefore, the solution to the altered primal satisﬁes exactly the same constraints (including nonnegativity constraints) with equality, and is therefore the same point. So the earlier estimate of 2000 for the increased cost is indeed accurate.

Review Problems for Chapter 17
2. (a) Regard the given LP problem as the primal and denote it by (P). Its dual is shown in answer section and is denoted by (D). If you draw the feasible set for (D) and a line −x1 + x2 = c, you see that as c increases, the line moves to the northwest. The line with the largest value of c which intersects the feasible set does so at the point (0, 8), at the intersection of three lines −x1 + 2x2 = 16, −2x1 − x2 = −8, and x1 = 0 that deﬁne three of the constraints, but the constraint x1 ≥ 0 is redundant. (b) We see that when x1 = 0 and x2 = 8, the second and fourth constraints in (D) are satisﬁed with strict inequality, so y2 = y4 = 0 at the optimum of (P). Also, since x2 = 8 > 0, the second constraint in (P) must be satisﬁed with equality at the optimum — i.e., 2y1 − y3 = 1. But then we see that the objective function in (P) can be reduced from 16y1 + 6y2 − 8y3 − 15y4 to 16y1 − 8y3 = 8(2y1 − y3 ) = 8. 1 We conclude that any (y1 , y2 , y3 , y4 ) of the form (y1 , y2 , y3 , y4 ) = ( 2 (1 + b), 0, b, 0) must solve (P) provided its components are nonnegative and the ﬁrst constraint in (P) is satisﬁed. (The second constraint 1 we already know is satisﬁed with equality.) The ﬁrst constraint reduces to − 2 (1 + b) − 2b ≥ −1, or 1 b ≤ 1 . We conclude that ( 2 (1 + b), 0, b, 0) is optimal provided 0 ≤ b ≤ 1 . 5 5 (c) The objective function in (D) changes to kx1 + x2 , but the constraints remain the same. The solution (0, 8) found in part (a) also remains unchanged provided that the slope −k of the level curve x2 = 8 − kx1 through the point (0, 8) remains positive and no less than the slope 1/2 of the line −x1 + 2x2 = 16. Hence 1 k ≤ −2. 4. (a) If a = 0, it is a linear programming problem, whose answer appears in the main text. (b) When a ≥ 0 we follow the techniques in Section 14.10 and consider the Lagrangian L = (500 − ax1 )x1 + 250x2 − λ1 (0.04x1 + 0.03x2 − 100) − λ2 (0.025x1 + 0.05x2 − 100) − λ3 (0.05x1 − 100) − λ4 (0.08x2 − 100) Then the Kuhn–Tucker conditions (with nonnegativity constraints) are: there exist numbers λ1 , λ2 , λ3 , and λ4 , such that ∂L/∂x1 = 500 − 2ax1 − 0.04λ1 − 0.025λ2 − 0.05λ3 ≤ 0 (= 0 if x1 > 0) ∂L/∂x2 = 250 − 0.03λ1 − 0.05λ2 − 0.08λ4 ≤ 0 (= 0 if x2 > 0) λ1 ≥ 0, λ2 ≥ 0, and λ1 = 0 and λ2 = 0 λ3 ≥ 0, λ4 ≥ 0, and if 0.04x1 + 0.03x2 < 100 λ3 = 0 if 0.05x1 < 100 if 0.08x2 < 100 if 0.025x1 + 0.05x2 < 100 (i) (ii) (iii) (iv) (v) (vi)

and λ4 = 0

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

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(c) The Kuhn–Tucker conditions are sufﬁcient for optimality since the Lagrangian is easily seen to be concave in (x1 , x2 ) for a ≥ 0. To ﬁnd when (x1 , x2 ) = (2000, 2000/3) remains optimal, we must ﬁnd Lagrange multipliers such that all the Kuhn–Tucker conditions are still satisﬁed. Note that (i) and (ii) must still be satisﬁed with equality. Moreover, the inequalities in (iv) and (vi) are strict, so λ2 = λ4 = 0. Then (ii) gives λ1 = 25000/3. It remains to check for which values of a one can satisfy (i) for a λ3 that also satisﬁes (v). From (i), 0.05λ3 = 500 − 4000a − 0.04(25000/3) = 500/3 − 4000a ≥ 0 if and only if a ≤ 1/24. 5. (a) See the answer in the main text. (b) The admissible set is shown as the shaded inﬁnite polygon in Fig. SM17.R.5. Lines I, II, and III show where the respective constraints are satisﬁed with equality. The dotted lines are level curves for the objective function 100y1 + 100y2 . We see that the objective function ∗ ∗ has its smallest value at P , the intersection of the two lines I and II. The coordinates (y1 , y2 ) for P are ∗ ∗ ∗ ∗ ∗ ∗ given by the two equations 3y1 + 2y2 = 6 and y1 + 2y2 = 3. Thus P = (y1 , y2 ) = (3/2, 3/4). The ∗ ∗ optimal value of the objective function is 100(y1 + y2 ) = 225. (c) Since the dual has a solution, the duality theorem tells us that (a) also has an optimal solution, which ∗ ∗ ∗ we denote by (x1 , x2 , x3 ). Since the third constraint in the dual is satisﬁed with inequality, we must have ∗ x3 = 0. Moreover, both constraints in the primal must be satisﬁed with equality at the optimum because ∗ ∗ ∗ ∗ both dual variables are positive in optimum. Hence 3x1 + x2 = 100 and 2x1 + 2x2 = 100, which gives ∗ ∗ ∗ ∗ ∗ x1 = x2 = 25. The maximal proﬁt is 6x1 + 3x2 + 4x3 = 225, equal to the value of the primal, as expected.
∗ (d) To a ﬁrst-order approximation, proﬁt increases by y1 b1 = 1.5, so the new maximal proﬁt is 226.5. For this approximation to be exact, the optimal point in the dual must not change when b1 is increased from 100 to 101. This is obviously true, as one sees from Fig. SM17.R.5.

(e) The maximum value in the primal is equal to the minimum value in the dual. Given the same approxi∗ ∗ mation as in part (d), this equals b1 y1 + b2 y2 , which is obviously homogeneous of degree 1 in b1 and b2 . More generally, let F (b1 , b2 ) denote the minimum value of the dual set out in part (b). Given any α > 0, ∗ ∗ note that minimizing αb1 y1 + αb2 y2 over the constraint set of the dual gives the same solution (y1 , y2 ) as minimizing b1 y1 + b2 y2 over the same constraint set. Hence F (αb1 , αb2 ) = αF (b1 , b2 ) for all α > 0, so F is homogeneous of degree 1. y2 5 4 3 2 1 P y1 100y1 + 100y2 = constant

1 III
Figure SM17.R.5

2 I

3 II

4

5

6

7

© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012

Essential Mathematics for Economics

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