Free Essay: Student’s Manual Essential Mathematics for Economic Analysis

Student’s Manual Essential Mathematics for Economic Analysis

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

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Preface
This student’s solutions manual accompanies Essential Mathematics for Economic Analysis (3rd edition, FT SM ⊃ Prentice Hall, 2008). Its main purpose is to provide more detailed solutions to the problems marked ⊂ in the text. This Manual should be used in conjunction with the answers in the text. In some few cases only a part of the problem is done in detail, because the rest follows the same pattern. We are grateful to Carren Pindiriri for help with the proofreading. We would appreciate suggestions for improvements from our readers as well as help in weeding out inaccuracies and errors. Oslo and Coventry, July 2008 Knut Sydsæter (knutsy@econ.uio.no) Arne Strøm (arne.strom@econ.uio.no) Peter Hammond (hammond@stanford.edu) Version: 9 July 2008

Contents
1 2 3 4 5 6 7 8 9 10 11 13 14 15 16 17 Introductory Topics I: Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Introductory Topics II: Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Introductory Topics III: Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Functions of One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Properties of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Derivatives in Use . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Single-Variable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Interest Rates and Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Functions of Many Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Multivariable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Constrained Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Matrix and Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Determinants and Inverse Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

12 Tools for Comparative Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

CHAPTER 1

INTRODUCTORY TOPICS I: ALGEBRA

1

Chapter 1 Introductory Topics I: Algebra
1.1
1. (a) True (b) False. −5 is smaller than −3, so on the number line it is to the left of −3. (See Fig. 1.1.1 in the book.) (c) False. −13 is an integer, but not a natural number. (d) True. Any natural number is rational. For example 5 = 5/1. (e) √ False, since 3.1415 = 31415/10000, the quotient of two integers. √ (f) False. Counterexample: 2 + (− 2) = 0. (g) True. (h) True.

1.3
9. (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 (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) (x − y − z)2 = (x−y−z)(x−y−z) = x 2 −xy−xz−xy+y 2 −yz−xz−yz+z2 , so (x+y+z)2 −(x−y−z)2 = 4xy+4xz 13. (a) a 2 +4ab+4b2 = (a+2b)2 using the ﬁrst quadratic identity. (d) 9z2 −16w 2 = (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) 4 x−2 x+2−x+2 1 x+2 1 = 2 − = − = (x − 2)(x + 2) x −4 (x − 2)(x + 2) (x + 2)(x − 2) x−2 x+2

(b) Since 4x + 2 = 2(2x + 1) and 4x 2 − 1 = (2x + 1)(2x − 1), the LCD is 2(2x + 1)(2x − 1). Then (6x + 25)(2x − 1) − 2(6x 2 + x − 2) 21(2x − 1) 21 6x + 25 6x 2 + x − 2 = = = − 2−1 4x 2(2x + 1)(2x − 1) 2(2x + 1)(2x − 1) 2(2x + 1) 4x + 2 (c) (d) (e) (f) a(a + 3b) a 18b2 − a(a − 3b) + 2(a 2 − 9b2 ) a 18b2 = = +2= − 2 − 9b2 (a + 3b)(a − 3b) (a + 3b)(a − 3b) a − 3b a a + 3b 5a − 2 a 2 − 4 − a(a − 2) + 8a 2(5a − 2) 1 1 1 = = = − + 2 − 4) 2 − 4) 2 − 4) 8ab(a 4ab(a 2 − 4) 8ab(a 8ab 8b(a + 2) b(a 2t − t 2 · t +2 2t 5t − t −2 t −2
1 2

=

t (2 − t) 3t −t (t − 2) 3t −3t 2 · = · = t +2 t −2 t +2 t −2 t +2
1 a 1 − 2a = 2 − (4a − 2) = 4 − 4a = 4(1 − a) 0.25

1 a 1 − 2a a− = 1 0.25 4

= 4a − 2, so 2 −

6. (a)

1 2(x + 1) + x − 3x(x + 1) 2 − 3x 2 2 + −3= = x x+1 x(x + 1) x(x + 1) t t (2t − 1) − t (2t + 1) −2t t − = = 2 (b) 2t + 1 2t − 1 (2t + 1)(2t − 1) 4t − 1 (c) 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

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

2

CHAPTER 1

INTRODUCTORY TOPICS I: ALGEBRA

1 1 + x y = (d) 1 xy

1 1 + xy y+x x y = =x+y 1 1 · xy xy

1 1 − 2 2 x y (e) = 1 1 + 2 x2 y

1 1 − 2 2 x y 1 1 + 2 2 x y

· x2y2 = · x2y2

y2 − x2 x2 + y2

(f) Multiply the numerator and the denominator by xy. Then the fraction reduces to 8. (a)
1 4

y−x a(y − x) = . a(y + x) y+x

−

1 5

= n

5 20

−

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 + = + + = If u = x p−q , then p−q q−p 1+x 1+x 1 + u 1 + 1/u 1+u 1+u 1 1 (x 2 − 1) + 2 x+1+1 x+2 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 −2x − h 1 1 x 2 − (x + h)2 −2xh − h2 (x + h)2 x = 2 − 2 = = 2 , so 2 2 (x + h)2 2 (x + h) x x x (x + h) h x (x + h)2 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. (Needs some hints.) Multiply the denominator and the numerator by: (a) √ √ √ √ √ √ (c) 3 + 2 (d) x y − y x (e) x + h + x (f) 1 − x + 1 7. The answers will depend on the calculator you use.

7−

5

(b)

5−

3

12. (a) For x = 1 the left-hand side is 4 and the right-hand side is 2. (In fact, (2x )2 = 22x .) (b) Correct because a p−q = a p /a q (c) Correct because a −p = 1/a p (d) For x = 1 it says 5 = 1/5, which is absurd. (e) For √ = y = 1,√ says that a 2 = 2a, which is usually wrong. (In fact, a x+y = a x a y .) x √ it √ √ x · 2 y = 2 x+ y , not 2 xy . (f) 2

1.6
4. (a) 2 < −x − 7 3x + 1 − 2(2x + 4) 3x + 1 3x + 1 > 0, or >0 − 2 > 0, or has the same solutions as 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 assuming that 2x + 4 > 0. When multiplying with 2x + 4 when this number is negative, the inequality sign must be reversed. (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?) 3 3(160 − n) 120 − ≤ 0, i.e. ≤ 0. A sign diagram reveals that the (b) The inequality is equivalent to n 4 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.)

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

CHAPTER 1

INTRODUCTORY TOPICS I: ALGEBRA

3

(c) Easy: g(g − 2) ≤ 0 etc. (d) Note that p2 − 4p + 4 = (p − 2)2 , and the inequality reduces to p+1 ≥ 0. The fraction makes no sense if p = 2. The conclusion follows. (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 2 . If you do, x = 0 appears as a false solution.) (f) See the text and use a sign diagram. (Don’t cancel x 5. (a) Use a sign diagram. (b) The inequality is not satisﬁed for x = 1. If x = 1, it is obviously satisﬁed for 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. ((5x − 1)11 < 0 if x < 1/5, > 0 if x > 1/5.) −(1 + x 2 ) 3x − 1 3x − 1 − (x + 3) > 0, > 0, so x < 0. (1 + x 2 is always positive.) > x + 3, (f) x x x x−3 −2x(x + 2) x−3 > 2x − 1 or − (2x − 1) < 0 or < 0. Then use a sign diagram. (g) x+3 x+3 x+3 (h) Use the hint and a sign diagram. (Actually, this problem and the next could be postponed to Section 2.3 if you have forgotten your high school algebra.) (i) Use the hint and a sign diagram. Review Problems for Chapter 1 4. (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 (c) Cancel the common factor 4x 2 yz2 . (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 (f) x 2
3

·

8 x −2

−3

=

√ √ √ √ √ √ √ √ 8. 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 ) a.s.o. 11. (a) and (b) are easy. (c) ax +ay +2x +2y = ax +2x +ay +2y = (a +2)x +(a +2)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) p 2 − q 2 + p − q = (p − q)(p + q) + (p − q) = (p − q)(p + q + 1) (f) See the answer in the book. 15. (a) 2s s s(2s + 1) − s(2s − 1) s = 2 − = 4s − 1 (2s − 1)(2s + 1) 2s − 1 2s + 1 −7(x + 3) −7 x 24 −x(x + 3) − (1 − x)(x − 3) − 24 1−x = = (b) − 2 = − x−3 (x − 3)(x + 3) (x − 3)(x + 3) x+3 x −9 3−x 1 y−x y−x (c) Multiplying numerator and denominator by x 2 y 2 yields, 2 = = y − x2 (y − x)(y + x) x+y

x3 · 8 8 · x −2

−3

= (x 5 )−3 = x −15

16. (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 (2a − 3b)(2a + 3b) 2a + 3b 4 − 4x + x 2 (2 − x)2 2−x

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

4

CHAPTER 2

INTRODUCTORY TOPICS II: EQUATIONS

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), 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 the numerator and the denominator in the ﬁrst fraction by 1−z, leads to (1 − z)(1 + z) 6 . Multiplying 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 8 3 3 4 3 denominator 24 gives an equation with the solution p = 15/16.

2.2
2. (a) Multiply both sides by abx to obtain b+a = 2abx. Hence, x = a b 1 1 1 a+b + . = + = 2 a b 2ab 2ab 2ab (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, √ so, by squaring each side, x = p2 /4w2 . (d) Multiply each side by 1 + x to obtain 1 + x + ax = 0, so x = −1/(1 + a). (e) x 2 = b2 /a 2 , so x = ±b/a. (f) We see immediately that x = 0.

4. (a) αx − a = βx − b ⇐⇒ (α − β)x = a − b, so x = (a − b)/(α − β). √ (b) Squaring each side of pq = 3q + 5 yields pq = (3q + 5)2 , so p = (3q + 5)2 /q. (c) Y = 94 + 0.2(Y − (20 + 0.5Y )) = 94 + 0.2Y − 4 − 0.1Y , so 0.9Y = 90, and then Y = 100. 1/3 1 r . (d) Raise each side to the 4th power: K 2 2 K = Q4 , so K 3 = 2wQ4 /r, and hence K = 2wQ4 /r w 1/2 L3/4 , leads to 2L/K = r/w, (e) Multiplying numerator and denominator in the left-hand fraction by 4K 1 r 1 = r 4 . It follows from which we get L = rK/2w. (f) Raise each side to the 4th power: 16 p4 K −1 2 w 1 that K −1 = 32r 3 w/p4 , so K = 32 p4 r −3 w −1 . 5. (a) √ 1 T −t tT 1 1 = , so s = . (b) KLM = B + αL, so KLM = (B + αL)2 , and = − t T tT T −t s 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 answer in the book. (b) If the ﬁrst natural number is n, then the next is n + 1, so the requirement is that n2 + (n + 1)2 = 13, which reduces to 2n2 + 2n − 12 = 0, i.e. n2 + n − 6 = 0. This second-order equation has the solutions n = −3 and n = 2, so the two numbers are 2 and 3. (If we asked for integer solutions, we would have −3 and −2 in addition.)
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

CHAPTER 2

INTRODUCTORY TOPICS II: EQUATIONS

5

(c) If the shortest side is x, the other is x + 14, so according to Pythagoras’ Theorem (see page 633 and draw a picture), x 2 + (x + 14)2 = (34)2 , or x 2 + 14x − 480 = 0. The solutions are x = 16 and x = −30, so the shortest side is 16 cm and the longest is 30 cm. (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 chair be $x and the cost of one table $y. Then 5x + 20y = 1800 and 2x + 3y = 420. Solving this system yields x = 120, y = 60. 1 3 3 (c) Units produced of B: x. Then x + 2 x = 2 x units are produced of A, and 300 · 2 x + 200x = 13 000, or 650x = 13 000, so x = 20. Thus, 30 of quality A and 20 of quality B should be produced. (d) If she invests $x at 15 % and $y at 20 %, then x + y = 1500 and 0.15x + 0.2y = 275. The solution is x = 8000 and y = 2000.

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. Multiply to x2 + 1 x +1 the equation by (x + 1)2/3 . Then the denominator becomes x + 1 − 1 x, which is 0 for x = −3/2. 3 (d) Multiplying by x − 1 and rearranging yields x(2x − 1) = 0, and so x = 0 or x = 1/2. 3. (a) z = 0 satisﬁes the equation. If z = 0, canceling z2 yields z − a = za + zb, or z(1 − (a + b)) = a. If a + b = 1 we have a contradiction. If a + b = 1, 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. 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. Review Problems for Chapter 2 2. See Problem 2.1.3. 3. (a) x = 2 (y − 3) + y = 2 y − 2 + y = 5 y − 2, or 5 y = x + 2, so y = 3 (x + 2). 3 3 5 3 3 (b) √ − cx = b + d, or (a − c)x = b + d, so x = (b + d)/(a − c). ax (c) L = Y0 /AK, so squaring each side yields L = (Y0 /AK)2 . (d) qy = m − px, so y = (m − px)/q. (e) and (f): See the answers in the text. 5. (a) Multiply the equation by 5K 1/2 to obtain K 1/2 = 15L1/3 . 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. 9. (a) See the answer in the text. (b) Let u = 1/x and v = 1/y. Then the system reduces to 3u + 2v = 2, 2u − 3v = 1/4, with solution u = 1/2, v = 1/4. It follows that x = 1/u = 2 and y = 1/v = 4. (c) See the answer in the text.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

6

CHAPTER 3

INTRODUCTORY TOPICS II: MISCELLANEOUS

Chapter 3 Introductory Topics II: Miscellaneous
3.1
3. (a)–(d): Look at the last term and replace n by k. Sum over k from 1 to n. (e) The coefﬁcients are the powers 3n for n = 1, 2, 3, 4, 5, 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: The left-hand side is a1 + a2 /2, but the right-hand side is (1/k)(a1 + a2 ).

3.2
5. One does not have to use summation signs. The sum is a + (a + d) + (a + 2d) + · · · + (a + (n − 1)d. There are n terms. The sum of all the a’s is na. The rest is d(1 + 2 + · · · + n − 1). Then use formula (4).

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

(b)

2 s=0

= m i· (c) i=1 where we used (4) and (5).

2 2 + 6 + 8 5 6 m n 2 i=1 j =1 i ·j

=5+

4 r=2 3113 3600

rs 2 r+s

=

2 s=0

2s 2 2+s

+

3s 2 3+s

+

4s 2 4+s

=

2 2 3

+

3 2 4

+

4 2 5

+

n 2 j =1 j

1 = 2 m(m+1)· 1 n(n+1)(2n+1) = 6 n m n

1 12 m(m+1)n(n+1)(2n+1),

4. a is the mean of the as ’s because a = ¯ ¯ ¯

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

(arj − a)(asj − a) = ¯ ¯ r=1 s=1 r=1

(arj − a) ¯ s=1 (asj − a) ¯

(∗∗)

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

√ √ √ 2, 6. (a) If (i) x − 4 = x + 5 − 9, then also (ii) x − 4 = ( x + 5 − 9)√ which we get by squaring both sides in (i). Calculating the square on the right-hand side of (ii) gives x + 5 = 5, and so x + 5 = 25,

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

CHAPTER 3

INTRODUCTORY TOPICS II: MISCELLANEOUS

7

i.e. 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 the following ﬁgure. y y =9− 5 y= 5 10 √ x−4 15 20 25 x √ x+5

-5 y=
Figure SM3.4.6

√

x+5−9

√ 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.) √ 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. 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.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

8

CHAPTER 4

FUNCTIONS OF ONE VARIABLE

3.7
3. For n = 1, both sides are 1/2. Suppose (∗) 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 + 1)2 k+1 k , which is (∗) for n = k + 1. Thus, by induction, + = = k+2 k + 1 (k + 1)(k + 2) (k + 1)(k + 2) (∗) 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, ⇐ false because with y > −2 and x = 3, (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, f (x) = x 3 is strictly increasing. See Problem 6.3.3 and see the graph Fig. 7, page 88.) 9. Consider Fig. A3.6.8, page 643 in the book, and let nk denote the number of students in the set marked Sk , for k = 1, 2, . . . , 8. 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 to the problems are: (a): n1 = 100 (b): n3 + n4 = 648 + 22 = 670 (c) 1000 − 8 ni = 1000 − 905 = 95 i=1

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 = √ 2 + 1 and 2f (x) = 2x 2 + 2. Now, 4x 2 + 1 = 2x 2 + 2 ⇔ x 2 = 1/2 ⇔ x = ± 1/2 = ± 1 2. 4x 2 10. (a) No: f (2 + 1) = f (3) = 18, whereas f (2) + f (1) = 10. (b) Yes: f (2 + 1) = f (2) + f (1) = −9. √ √ (c) No: f (2 + 1) = f (3) = 3 ≈ 1.73, whereas f (2) + f (1) = 2 + 1 ≈ 2.41. 13. (a) We must 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 must 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.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

CHAPTER 4

FUNCTIONS OF ONE VARIABLE

9

15. Since g is obviously deﬁned for x ≥ −2, Dg = [−2, ∞). Note that g(−2) = 1, and g(x) ≤ 1 for all x ∈ Df . As x increases from −2 to ∞, g(x) decreases from 1 to −∞, so Rg = (−∞, 1].

4.4
3. If D = a + bP , then 200 = a + 10b, and 150 = a + 15b. Solving for a and b yields a = 300 and b = −10, so D = 300 − 10P . 4. L1 : The slope is obviously 1, and the point-slope formula with (x1 , y1 ) = (0, 2) and a = 1 give y = x +2. 0−3 L2 : Using the point-point formula with (x1 , y1 ) = (0, 3) and (x2 , y2 ) = (5, 0) yields: y − 3 = x, 5−0 or y = − 3 x + 3. L3 : Has slope 0 and equation y = 1. For L4 and L5 see the text. 5 10. The set of points that satisfy the inequality 3x + 4y ≤ 12 are those on or below the straight line 3x + 4y = 12 as explained in Example 6 for a similar inequality. Those 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, those 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 set shown in Fig. A.4.4.10.

4.5
3. The point–point formula gives C − 200 = 3 275 − 200 (x − 100), or C = x + 50. 150 − 100 2

4.6
1 2. Complementing the answers in the text. (c) Formula (4) with a = − 2 and b = −1 gives x = −1 as the 1 1 2 maximum point. (Alternatively, completing the square, f (x) = − 2 (x +2x−3) = − 2 (x 2 +2x+1−4) = 1 − 2 (x + 1)2 + 2, from which we see immediately that f (x) has maximum 2 at x = −1.) (e) Use (2.3.5), or expand, to verify the formula for f (x). Use a sign diagram to study the sign variation of f (x).

6. Expanding we get U (x) = −(1 + r 2 )x 2 + 8(r − 1)x. Then apply (4.6.4) with a = −(1 + r 2 ) and b = 8(r − 1). 9. (b) If B 2 − 4AC > 0, then according to formula (2.3.4), the equation f (x) = Ax 2 + Bx + C = 0 would have two distinct solutions, which is impossible when f (x) ≥ 0 for all x. We ﬁnd that A = 2 2 2 2 2 2 a1 + a2 + · · · + an , B = 2(a1 b1 + a2 b2 + · · · + an bn ), and C = b1 + b2 + · · · + bn , so the conclusion follows easily.

4.7
1. (a) Integer roots must divide 6. Thus ±1, ±2, ±3, and ±6 are the only possible integer solutions. We ﬁnd that −2, −1, 1, 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. (b) The same possible integer solutions. 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.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

10

CHAPTER 4

FUNCTIONS OF ONE VARIABLE

3. (a) The answer is 2x 2 + 2x + 4 + 3/(x − 1), because (2x 3 + 2x − 1) ÷ (x − 1) = 2x 2 + 2x + 4 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

+1 3x 3 + 4x 2 3x 3 + 3x 2 + 3x x 2 − 3x + 1 x2 + x + 1 − 4x remainder

(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 + − 12x 4 + 6x 3 6x 6 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 − 24x 2 + 12x 12x 4 − 12x 3 + 28x 2 − 12x + 1 + 24x − 12 − 12x 3 28x 2 − 36x + 13
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

remainder

CHAPTER 4

FUNCTIONS OF ONE VARIABLE

11

1 4. (a) y = 2 (x + 1)(x − 3). (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). Then f (1) = −4a, and since the graph passes through the point (1, −2),. f (1) = −2 = −4a. But the a = 1/2.) (b) Because the equation f (x) = 0 has 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). 1 (c) y = 2 (x + 3)(x − 2)2 . (We try a polynomial of the form f (x) = c(x − 2)2 (x + 3), with x = 2 as a double root. Then f (0) = 12c. From the graph we see that f (0) = 6, and so a = 1/2.)

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

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

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, y is close to 2. 5. (a) See the answer in the text. (b) 9t = (32 )t = 32t and (27)1/5 /3 = (33 )1/5 /3 = 33/5 /3 = 3−2/5 , and then 2t = −2/5, so t = −1/5.

4.9
10. Suppose y = Abx , with b > 0. Then in (a), since the graph passes through the points (x, y) = (0, 2) and (x, y) = (2, 8), we get 2 = Ab0 , or A = 2, and 8 = 2b2 , so b = 2. Hence, y = 2 · 2x . In (b), 2 = Ab−1 and 6 = Ab. It follows that A = 2 and b = 3, and so y = 2 · 3x . 3 1 In (c), 4 = Ab0 and 1/4 = Ab4 . It follows that A = 4 and b4 = 1/16, and so b = 1/2. Thus, y = 4( 2 )x .

4.10
3. (a) and (c) see the text. (b) Since ln x 2 = 2 ln x, 7 ln x = 6, so ln x = 6/7, and thus x = e6/7 . B 1 ln . r −s A 4. (a) ln(Aert ) = ln(Best ), so ln A + rt = ln B + st, or (r − s)t = ln(B/A), and so t = (b) t =

14 1 5.6 · 1012 1 ln ≈ 22. = ln 12 0.07 3 0.09 − 0.02 1.2 · 10 According to this, the two countries would have the same GNP in approximately 22 years, so in 2012.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

12

CHAPTER 5

PROPERTIES OF FUNCTIONS

Review Problems for Chapter 4 4. (a) We must have x 2 ≥ 1, i.e. x ≥ 1 or x ≤ −1. (Look at Fig. 4.3.6, page 88.) (b) The square root is deﬁned if x ≥ 4, but x = 4 makes the denominator 0, so we must require x > 4. (c) We must have (x − 3)(5 − x) ≥ 0, i.e. 3 ≤ x ≤ 5 (using a sign diagram). 7. (a) The point-slope formula gives y − 3 = −3(x + 2), or y = −3x − 3. 7−5 (b) The two-point formula gives: y − 5 = (x − (−3)), or y = 2x/5 + 31/5. 2 − (−3) 3b − b (x − a), or y = (2b/a)x − b. (c) y − b = 2a − a 10. (1, −3) belongs to the graph if −3 = a + b + c, (0, −6) belongs to the graph if −6 = c, and (3, 15) belongs to the graph if 15 = 9a + 3b + c. It follows that a = 2, b = 1, and c = −6. 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). 16. We use (4.7.5) and denote each polynomial by p(x). (a) p(2) = 8 − 2k = 0 for k = 4. (b) p(−2) = 4k 2 + 2k − 6 = 0 for k = −3/2 and k = 1. (c) p(−2) = −26 + k = 0 for k = 26. (d) p(1) = k 2 − 3k − 4 = 0 for k = −1 and k = 4.
1 17. Since p(2) = 0, x − 2 is a factor in p(x). We ﬁnd that p(x) ÷ (x − 2) = 4 (x 2 − 2x − 15) = 1 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.)

21. (a) ln(x/e2 ) = ln x − ln e2 = ln x − 2 (b) ln(xz/y) = ln(xz) − ln y = ln x + ln z − ln y (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|.) (d) See the text.

Chapter 5 Properties of Functions
5.1
3. The equilibrium condition is 106 − P = 10 + 2P , and thus P = 32. The corresponding quantity is Q = 106 − 32 = 74. See the graph in the answer section of the text. 6. f (y ∗ − d) = f (y ∗ ) − c gives A(y ∗ − d) + B(y ∗ − d)2 = Ay ∗ + B(y ∗ )2 − c, or Ay ∗ − Ad + B(y ∗ )2 − 2Bdy ∗ + Bd 2 = Ay ∗ + B(y ∗ )2 − c. It follows that y ∗ = [Bd 2 − Ad + c]/2Bd.

5.2
4. If f (x) = 3x + 7, then f (f (x)) = f (3x + 7) = 3(3x + 7) + 7 = 9x + 28. f (f (x ∗ )) = 100 requires 9x ∗ + 28 = 100, and so x ∗ = 8.

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 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. (b) Since f (0) = 4 and f (x) increases by 2 for each unit 1 1 increase in x, f (x) = 2x + 4. Solving y = 2x + 4 for x yields x = 2 y − 2, and thus f −1 (x) = 2 x − 2.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

CHAPTER 5

PROPERTIES OF FUNCTIONS

13

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 of f is all x = 2 Hence (with x = 2), = 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 = ln y = x + 4, so x = ln y − 4, y > 0. (b) The range is , which is the domain of the inverse. y = ln x − 4, ln x = y + 4, and then x = ey+4 . (c) The domain is . y is increasing and x → −∞, y → ln 2. Moreover, y → ∞ as x → ∞. So the range of the function is (ln 2, ∞). y = ln 2 + ex−3 , 2 + ex−3 = ey , so ex−3 = ey − 3, and thus x = 3 + ln(ey − 3), y > ln 2. ex+4 , From when From

1 1 1 11. We must solve x = 2 (ey −e−y ) for y. Multiply the equation by ey to get√ e2y − 2 = xey or e2y −2xey −1 = 2 y = z yields z2 − 2xz − 1 = 0, with solution z = x ± 2 + 1. The minus sign makes z 0. Letting e x √ √ y = x + x 2 + 1. This gives y = ln x + x 2 + 1 as the inverse function. negative, so z = e

5.4
1. (a) It is natural ﬁrst to see if the curve √ intersects the axes, by putting x = 0, and then y = 0. This gives 4 √ points. Then choose some values of − 6 < x < 6, and compute the corresponding values of y. Argue why the graph is symmetric about the x-axis and the y-axis. (The curve is called an ellipse. See the next section.) (b) This graph is also symmetric about the x-axis and the y-axis. (If (a, b) lies on the graph, so does (a, −b), (−a, b), and (−a, −b). (The graph is a hyperbola. See the next section.) 2. We see that we must have x ≥ 0 and y ≥ 0. If (a, b) lies on the graph, so does (b, a), so the graph is symmetric about the line y = x. See the answer in the text.

5.5
4. (a) See the text. (b) Since the circle has centre at (2, 5), its equation is (x − 2)2 + (x − 5)2 = r 2 . Since (−1, 3) lies on the circle, (−1 − 2)2 + (3 − 5)2 = r 2 , so r 2 = 13. 8. x 2 +y 2 +Ax+By+C = 0 ⇐⇒ x 2 +Ax+y 2 +By+C = 0 ⇐⇒ x 2 +Ax+
1 2 4 (A 2 1 1 + − 4C) ⇐⇒ x + + y + 2 B = 4 (A2 + B 2 − 4C). The last is the equation of a √ 1 1 1 circle centred at − 2 A, − 2 B with radius 2 A2 + B 2 − 4C. If A2 + B 2 = 4C, the graph consists only 1 1 of the point − 2 A, − 2 B . For A2 + B 2 < 4C, the solution set is empty.

B2

2 1 2A

2 2 1 1 2 2 A +y +By+ 2 B

=

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 surface area of a sphere is given, its volume is uniquely determined: From S = 4πr 2 , r = (S/4π)1/2 , and then V = 4 πr 3 = 4 π(S/4π )3/2 3 3 (for the formulas for the surface area and the volume of a sphere of radius r, see page 632.)
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

14

CHAPTER 6

DIFFERENTIATION

Review Problems for Chapter 5
1 3. (a) Equilibrium condition: 150 − 2 P ∗ = 20 + 2P ∗ , which gives P ∗ = 52 and Q∗ = 20 + 2P ∗ = 124. For (b) and (c) se answers in the text.

7. (a) f is deﬁned and strictly increasing for ex > 2, i.e. x > ln 2. Its range is . (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. (a) We start by using the recipe in (6.2.3) to ﬁnd the slope of the tangent. (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. (d) [f (h) − f (0)]/ h = (h3 − 2h)/ h = h2 − 2 → −2 as h → 0, so f (0) = −2. 1 + h + 1/(−1 + h) + 2 h f (−1 + h) − f (−1) =− = → 0 as h → 0, so f (0) = 0. (e) h h −1 + h (1 + h)4 − 1 h4 + 4h3 + 6h2 + 4h + 1 − 1 f (1 + h) − f (1) = = = h3 + 4h2 + 6h + 4 → 4 as (f) h h h h → 0, so f (1) = 4. √ √ √ √ √ √ √ √ √ √ 8. (a) x + h− x x + h+ x = ( x + h )2 + x + h x − x x + h−( x)2 = x +h−x = h. √ √ √ √ ( x + h − x)( x + h + x) f (x + h) − f (x) h 1 = (b) = √ √ √ √ =√ √ h h( x + h + x) h( x + h + x) x+h+ x (c) This follows from (b).

6.5
5. (a) 3h 1/3 − 2/3h 1 h−2 1 1/3 − 2/3h → as h → 2 = = = 6 3h 3h(h − 2) 3h(h − 2) h−2 (b) When x → 0, x 2 − 1 → −1 and x 2 → ∞, so the fraction has no limit, but tends to −∞. √ 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 − 2t − 3√ (t − 3)(t + 1) √ t + 1 √ √ √ √ ( h + 3 − 3)( h + 3 + 3) h+3− 3 h+3−3 1 1 (d) = = √ √ √ = √ √ → √ √ h h( h + 3 + 3) h( h + 3 + 3) h+3+ 3 2 3 as h → 0.

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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DIFFERENTIATION

15

(e)

(t + 2)(t − 2) t −2 2 t2 − 4 = = → − as t → −2. 2 + 10t + 16 t (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

6. (a)

x 2 + 2x − 3 (x − 1)(x + 3) f (x) − f (1) = x + 3 → 4 as x → 1. = = x−1 x−1 x−1 f (x) − f (1) (b) = x + 3 → 5 as x → 2. x−1 h2 + 6h (2 + h)2 + 2(2 + h) − 8 f (2 + h) − f (2) = = h + 6 → 6 as h → 0. = (c) h h h (a + h)2 + 2(a + h) − a 2 − 2a f (a + h) − f (a) = 2a + 2 + h → 2a + 2 as h → 0. = (d) h h (e) Same answer as in (d) putting x − a = h. (a + h)2 + 2a + 2h − (a − h)2 − 2a + 2h f (a + h) − f (a − h) = = 4a + 4 → 4a + 4 as h → 0. (f) h h 7. (a) x 3 − 8 = 0 has the solution x = 2, and polynomial division yields x 3 − 8 = (x − 2)(x 2 + 2x + 4). (b) and (c): see the text.

6.6
7. (a) With f (x) = x 2 , lim f (a + h) − f (a) (5 + h)2 − 52 = lim = f (5). On the other hand, f (x) = h→0 h→0 h h 2x, so f (5) = 10, and the limit is therefore 10. (b) and (c): see the text.

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 4 5 x 1 (e) y = 5 + 5 = x −4 + x −5 ⇒ y = − 5 − 6 x x x x 3(2x + 8) − 2(3x − 5) 34 3x − 5 (g) y = 3x −11 ⇒ y = −33x −12 ⇒ = (f) y = 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

6. (a) f (x) = 6x − 12 = 6(x − 2) ≥ 0 ⇐⇒ x ≥ 2, so f is increasing in [2, ∞). √ f (x) = x 3 − 3x = (b) √ √ √ 2 − 3) = x(x − 3)(x + 3), so (using a sign diagram) f is increasing in − 3, 0 and in 3, ∞ . x(x √ √ √ √ 2(2 − 2)(2 + 2) 2(2 − x 2 ) , so f is increasing in [− 2, 2]. (d) See the text. = (c) f (x) = 2 (x 2 + 2)2 (x + 2)2 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 = 4x/(x 2 + 1)2 = 1 and y = 0 when x = 1, so y = x − 1. (c) y = x 2 − x −2 , so y = 2x + 2x −3 = 17/4 and y = 15/4 when 4x 3 (x 3 + 3x 2 + x + 3) − (x 4 + 1)(3x 2 + 6x + 1) 1 =− x = 2, so y = (17/4)x − 19/4. (d) y = 2 + 1)(x + 3)]2 [(x 19 and y = 1/3 when x = 0, so y = −(x − 3)/9.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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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 √ n a t + b = at n+1/2 + bt n ⇒ y = (n + 1/2)at n−1/2 + nbt n−1 . (b) y = t 0 · (at 2 + bt + c) − 1 · (2at + b) −2at − b 1 ⇒ y = = (c) y = 2 2 + bt + c)2 (at (at 2 + bt + c)2 at + bt + c √ 12. This is rather tricky because the denominator is 0 at x1,2 = 2 ± 2. A sign diagram shows that f (x) > 0 only in (−∞, 0) and in (x1 , x2 ).

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

√ a 1 dx √ . =− √ u =− √ ap − c = b − u, with u = ap − c. Then 2 ap − c dp 2 u

12. (a), (e), and (g) are easy. For the others, you need the chain rule. In (d) you need the differentiation rules for sum, product as well as the chain rule. See the text.

6.9
4. g (t) = 2 , so g (2) = 2. (t − 1)3 2t (t − 1) − t 2 t 2 − 2t (2t − 2)(t − 1)2 − (t 2 − 2t)2(t − 1) 2(t − 1) = , g (t) = = = 2 2 4 (t − 1) (t − 1) (t − 1) (t − 1)4

5. With simpliﬁed notation: y = f g + fg , y = f g + f g + f g + fg = f g + 2f g + f g , y = f g + f g + 2f g + 2f g + f g + fg = f g + 3f g + 3f g + fg

6.10
2. (a) dx/dt = (b + 2ct)et + (a + bt + ct 2 )et = (a + b + (b + 2c)t + ct 2 )et 3qt 2 tet − (p + qt 3 )(et + tet ) dx (−qt 4 + 2qt 3 − pt − p)et (c) See the text. = (b) = dt t 2 e2t t 2 e2t 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]. (The answer in the text is wrong.)

6.11
3. For these problems we need the chain rule. That is an important rule! In particular, we need the fact that d 1 f (x) ln f (x) = f (x) = when f is a differentiable function (with f (x) > 0). dx f (x) f (x) 1 1 1 1 (a) y = ln(ln x) = ln u ⇒ y = u = = . u ln x x x ln x −2x 1 1 −x = . (b) y = ln 1 − x 2 = ln u ⇒ 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.)
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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17

1 1 = ex ln x + . x x 2 3 3 3 1 3 . (d) y = ex ln x 2 ⇒ y = 3x 2 ex ln x 2 + ex 2 2x = ex 3x 2 ln x 2 + x x x e 2x + 3 (e) y = ln(ex + 1) ⇒ y = x . (f) y = ln(x 2 + 3x − 1) ⇒ y = 2 . e +1 x + 3x − 1 4. (a) ln u is deﬁned for u > 0, so we must require x + 1 > 0, i.e. x > 0. 3x − 1 > 0 for the logarithm to be deﬁned. (b) We must have 1−x = 0 for the fraction to be deﬁned, and 1−x 3x − 1 A sign diagram (see below) shows that is deﬁned and positive if and only if 1/3 < x < 1. 1−x (c) ln |x| is deﬁned ⇐⇒ |x| > 0 ⇐⇒ x = 0. (c) y = ex ln x ⇒ y = ex ln x + ex
0 1/3 1 x

3x − 1 1−x

3x − 1 1−x

5. (a) One must have x 2 > 1, i.e. x > 1 or x < −1. (See Fig. 4.3.6 in the text.) (b) ln(ln x) is deﬁned 1 is deﬁned when ln(ln x) when ln x is deﬁned and positive, that is, for x > 1. (c) The fraction ln(ln x) − 1 is deﬁned and different from 1. From (b), ln(ln x) is deﬁned when x > 1. Further, ln(ln x) = 1 ⇐⇒ 1 is deﬁned ⇐⇒ x > 1 and x = ee . ln x = e ⇐⇒ x = ee . Conclusion: ln(ln x) − 1 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. Hence, f (x) 2x f (x) = f (x)(ln(2x) + 1) = (2x)x (ln x + ln 2 + 1). √ √ √ √ √ √ √ x ln x x x = eln x x ln x , so f (x) = e x ln x · d ( x ln x) = x x (b) f (x) = x = e √ + 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). 10. ln y = v ln u, so y /y = v ln u + (v/u)u , and so y = uv v ln u + vu u

.

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 − x)/x x = √ ( x − 1)/x, which is positive for x > 1. The conclusion follows.

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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Review Problems for Chapter 6 5. (a) y = −3 and y = −6x = −6 at x = 1, so y − (−3) = (−6)(x − 1), or y = −6x + 3. √ (b) y = −14 and y = 1/2 x − 2x = −31/4 at x = 4, so y = −(31/4)x + 17. (c) y = 0 and y = (−2x 3 − 8x 2 + 6x)/(x + 3)2 = −1/4 at x = 1, so y = (−1/4)(x − 1). 7. (a) f (x) = x 3 + x, etc. 15. (a) y = (b) Easy. (c) h(y) = y(y 2 − 1) = y 3 − y, etc. (d)–(f): use the quotient rule. ex − e−x 2 ≥ 0 ⇐⇒ ex ≥ e−x ⇐⇒ e2x ≥ 0 ⇐⇒ x ≥ 0 ln x ≥ 0 if x ≥ 1. (b) y = x x e + e−x (x − 1)(x − 2) 3x (c) y = 1 − 2 = ≥ 0 ⇐⇒ x ≤ 1 or x ≥ 2. (Use a sign diagram.) x2 + 2 x +2

Chapter 7 Derivatives in Use
7.1
2. Implicit differentiation yields (∗) 2xy + x 2 (dy/dx) = 0, and so dy/dx = −2y/x. Differentiating (∗) implicitly w.r.t. x gives 2y + 2x(dy/dx) + 2x(dy/dx) + x 2 (d 2 /dx 2 ) = 0. Inserting the result for dy/dx, and simplifying yields d 2 /dx 2 = 6y/x 2 . (Alternatively, we can differentiate −2y/x as a fraction.) These results follows more easily by differentiating y = x −2 twice. 3. (a) Implicit differentiation w.r.t. x yields (∗) 1 − y + 3y + 3xy = 0. Solving for y yields y = (1 + 3y)/(1 − 3x) = −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 (c) 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 − (144/25)x 10 y −9 = (6/25)x −4/5 . √ √ 6. (a) 2x + 2yy = 0, and solve for y . (b) 1/2 x + y /2 y = 0, and solve for y . (c) 4x 3 − 4y 3 y = 2xy 3 + x 2 3y 2 y , and solve for y . 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 ) = f (x 2 y)(2xy + x 2 y ), and solve for y . (How did we differentiate f (x 2 y) w.r.t. x? Well, if z = f (u) and u = x 2 y, then z = f (u)u where u is a product of two functions that both depend on x. So u = 2xy + x 2 y .) 10. (a) 2(x 2 +y 2 )(2x +2yy ) = a 2 (2x −2yy ), and solve for y . (b) Note that y = 0 when x 2 +y 2 = a 2 /2, √ 1 1 or y 2 = 2 a 2 − x 2 . Inserting this into the given equation yields x = ± 4 a 6. This yields the four points on the graph at which the tangent is horizontal.

7.2 dQ 1 · P 1/2 + Q 2 P −1/2 = 0. 1. Implicit differentiation w.r.t. P , with Q as a function of P , yields dP 19 dQ 1 Thus = − 2 QP −1 = − 3/2 . P dP 5. Differentiating (∗) w.r.t. P yields f (P + t) dP d 2P dP 2 d 2P 2 + 1 + f (P + t) 2 = g (P ) + g (P ) 2 . dt dt dt dt With simpliﬁed notation f (P + 1)2 + f P = g (P )2 + g P . Substituting P = f /(g − f ) and solving for P , we get P = [f (g )2 − g (f )2 ]/(g − f )3 .

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

CHAPTER 7

DERIVATIVES IN USE

19

7.3
−2x 1 4x 2 (3 − x 2 ) 2. (a) f (x) = x 2 4 − x 2 + x 3 √ = √ . For the rest, see the answer in the text. 3 2 4 − x2 3 4 − x2 5. (a) See the text. (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 + x3y 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. 4. F (1) = A and F (K) = αAK α−1 , so F (1) = αA. Then F (K) ≈ F (1) + F (1)(K − 1)= A + αA(K − 1) = A(1 + αA(K − 1)). 9. 3exy + 3xexy (y 2 + x2yy ) − 2y = 6x + 2yy . For x = 1, y = 0 this reduces to 3 − 2y = 6, so 3 y = −3/2. (b) y(x) ≈ y(1) + y (1)(x − 1) = − 2 (x − 1)
2 2

7.5
2. f (x) = (1 + x)−1 , f (x) = −(1 + x)−2 , f (x) = 2(1 + x)−3 , f (iv) (x) = −6(1 + x)−4 , f (v) (x) = 24(1 + x)−5 . Then f (0) = 0, f (0) = 1, f (0) = −1, f (0) = 2, f (iv) (0) = −6, f (v) (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 (iv) (0)x 4 + 5! f (v) (0)x 5 = x − 2 x 2 + 1 x 3 − 3 1 4 x + 1 x5 5 4 √ 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 , 5 5 f (x) = −5(1 + x)−2 + 4 (1 + x)−3/2 , and so f (0) = −5, f (0) = 2 , f (0) = − 15 . and the Taylor 4 5 1 polynomial of degree 2 about x = 0 is f (0) + f (0)x + 2 f (0)x 2 = −5 + 2 x − 15 x 2 . 8 9. h (x) = 2(p − q)x p+q−1 (px p−1 − qx q−1 )(x p + x q ) − (x p − x q )(px p−1 + qx q−1 ) = , so h (1) = (x p + x q )2 (x p + x q )2 1 1 2 (p − q). Since h(1) = 0, we get h(x) ≈ h(1) + h (1)(x − 1) = 2 (p − q)(x − 1).

7.6
1. From Problem 7.5.2, f (0) = 0, f (0) = 1, f (0) = −1, and f (c) = 2(1 + c)−3 . Then (3) gives 1 1 1 1 f (x) = f (0) + 1! f (0)x + 2! f (0)x + 3! f (c)x 3 = x − 2 x 2 + 1 (1 + c)−3 x 3 . 3 4. (a) With g(x) = (1 + x)1/3 , g (x) = 1 (1 + x)−2/3 , g (x) = − 2 (1 + x)−5/3 , and g (x) = 3 9 so g(0) = 1, g (0) = 1 , g (0) = − 2 , g (c) = 10 (1 + c)−8/3 , so 27 9 3 g(x) = 1 + 1 x − 1 x 2 + R3 (x), 3 9 where R3 (x) =
1 10 6! 27 (1 10 −8/3 , 27 (1 + x)

+ c)−8/3 x 3 =

5 81 (1

+ c)−8/3 x 3

(b) √∈ (0, x) and x ≥ 0, so (1 + c)−8/3 ≤ 1, and the inequality follows. c (a) (c) 3 1003 = 10(1+3·10−3 )1/3 ≈ 10.0099900, using part √ to approximate (1+3·10−3 )1/3 . The error 5 −3 )3 = 5 10−9 . So the error in 3 1003 is ≤ 10|R (x)| = 50 10−9 < 2 · 10−8 , in (b) is |R3 (x)| ≤ 81 (3 · 10 3 3 3 and the answer is correct to 7 decimal places.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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7.7
4. (a) Elx eax = (x/eax )aeax = ax (b) Elx ln x = (x/ ln x)(1/x) = 1/ ln x x (c) Elx (x p eax ) = p ax (px p−1 eax + x p aeax ) = p + ax x e x (px p−1 ln x + x p (1/x)) = p + 1/ ln x (d) Elx (x p ln x) = p x ln x xg x x xf x dA 9. (a) Elx A = = Elx f + Elx g (fg) = (f g + fg ) = + = 0 (b) Elx (fg) = g fg fg f A dx xg xf xg gf − fg x f f − = Elx f − Elx g = = (c) Elx = 2 f g f g g (f/g) 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 = u z du dx z dx

7.8
3. Using (7.8.4), the functions are continuous wherever they are deﬁned. So (a) and (d) are deﬁned everywhere.√ (b) we must exclude x = 1, in (c) the function is deﬁned for x < 2, in (e) we must exclude In x = ± 3 − 1, because the denominator is 0 for these values of x. Finally, in (f), the ﬁrst fraction requires x > 0, and then the other fraction is also deﬁned.

7.9
1. (b) |x| = −x for x < 0. Hence, lim− x−x x + |x| = lim− 0 = 0. = lim− x→0 x→0 x→0 x x x+x x + |x| (c) |x| = x for x > 0. Hence, lim+ = lim = lim+ 2 = 2. x→0 x→0 x √ x→0+ x √ (d) When x → 0+ , x → 0, so −1/ x → −∞. (e) When x → 3+ , x − 3 → 0+ , and so x/(x − 3) → ∞. (f) When x → 3− , 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 → ±∞.

7.10
4. Recall from Note 4.7.2 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. We are told in the problem that 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 new approximations
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

CHAPTER 7

DERIVATIVES IN USE

21

x1 = −1.9 − y1 = z1 =

f (−1.9) = −1.9 − f (−1.9) f (0.4) = 0.4 − 0.4 − f (0.4) f (1.5) = 1.5 − 1.5 − f (1.5)

−0.1749 ≈ −1.9 + 0.021 = −1.879 8.454 −0.4624 ≈ 0.4 − 0.053 = 0.347 −8.704 −0.5625 ≈ 1.5 + 0.034 = 1.534 16.75

It can be shown by more precise calculations that the actual roots, rounded to six decimals, are −1.879385, 0.347296, and 1.532089.

7.11
2. (a) When n → ∞, 2/n → 0 and so 5 − 2/n → 5. (b) When n → ∞, 3n 3n = = (c) When n → ∞, √ 2−1 2n n 2 − 1/n2 n2 − 1 = n − 1/n → ∞. n √ 3 3 3 2 . →√ = 2 3 2 2 − 1/n

7.12
2. L’Hôpital’s rule yields lim 2x x 2 − a2 “0” = lim = 2a. But note that we don’t really need = x→a 1 x→a x − a 0 x 2 − a2 = lim (x + a) = 2a. l’Hôpital’s rule here, because x 2 − a 2 = (x + a)(x − a), and therefore lim x→a x − a x→a 1/2 − 2 − x −1/2 − 1 2(1 + x) (1 + x) “0” “0” = = lim = = (b) lim x→0 2(1 + x + x 2 )1/2 − 2 − x x→0 (1 + 2x)(1 + x + x 2 )−1/2 − 1 0 0 1 − 2 (1 + x)−3/2 1 =− lim 2 )−1/2 + (1 + 2x)2 (− 1 )(1 + x + x 2 )−3/2 x→0 2(1 + x + x 3 2 f (x) 1/g(x) “0” −1/(g(x))2 g (x) (f (x))2 g (x) = lim = = lim = lim = · · x→a 1/f (x) x→a −1/(f (x))2 f (x) x→a (g(x))2 f (x) x→a g(x) 0 1 g (x) L2 lim . The conclusion follows. (Here, we have ignored problems with = L2 lim x→a f (x)/g (x) x→a f (x) “division by 0”, when either f (x) or g (x) tends to 0 as x tends to a.)

7. L = lim

Review Problems for Chapter 7 2. 5y 4 y − y 2 − 2xyy = 0, so y = meaningless, y is never 0. 6. y = 0 when 1 + 1 ln x = 0, i.e. ln x = −5, and then x = e−5 . 5 7. (a) We must have 1+x > 0, i.e −1 < x < 1. When x → 1− , f (x) → ∞. When x → −1− , 1−x f (x) → −∞. Since f (x) = 1/(1 − x 2 ) > 0 when −1 < x < 1, f is strictly increasing and the range 1+x 1+x 1 1+x , ln = 2y, so = e2y . Then solve for x. of f is . (b) From y = ln 1−x 1−x 2 1−x 9. (a) f (0) = ln 4 and f (x) = 2/(2x + 4), so f (0) = 1/2. Then f (x) ≈ f (0) + f (0)x = ln 4 + x/2. (b) g(0) = 1 and g (x) = −(1/2)(1 + x)−3/2 , so g (0) = −1/2. Then g(x) ≈ g(0) + g (0)x = 1 − x/2. (c) h(0) = 0 and h (x) = e2x + 2xe2x , so h (0) = 1. Then h(x) ≈ h(0) + h (0)x = x. y y2 = 3 . 4 − 2xy 5y 5y − 2x Because y = 0 makes the given equation

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

22

CHAPTER 8
1 2

SINGLE-VARIABLE OPTIMIZATION
1 1 1 ( 2 )2 2! 1 ( 2 )6 6! 2

12. With x =

2 and n = 5, formula (7.6.6) yields, e 2 = 1 + 1! +

+ = +

1 1 2 is some number between 0 and 2 . Now, R6 ( 2 ) = 6! ec < 1 1 the fact that since c < 2 , ec < e 2 < 2. Thus it follows that 1 1 2 2 2 2 2 e 2 ≈ 1 + 1! + 2! + 3! + 4! + 5! = 1 + 2 + 8 + 1 than 0.000043, and e 2 ≈ 1.649 correct to 3 decimals.
1 1

( 1 )6

1 1 1 1 ( 2 )3 ( 2 )4 ( 2 )5 ( 2 )6 c 3! + 4! + 5! + 6! e , where c 1 23040 ≈ 0.0004340, where we used

( 1 )2

( 1 )3

( 1 )4

( 1 )5

1 48

+

1 384

1 3840

≈ 1.6486979. The error is less

14. y + (1/y)y = 1, or (∗) yy + y = y. When y = 1, y = 1/2. Differentiating (∗) w.r.t, x yields 1 1 (y )2 + yy + y = y . With y = 1 and y = 1/2, we ﬁnd y = 1/8, so y(x) ≈ 1 + 2 x + 16 x 2 . 21. (a) lim −ex − ex + (2 − x)ex −ex + (2 − x)ex − 1 “0” (2 − x)ex − x − 2 “0” = lim = lim = = = x→0 x→0 x→0 0 3x 2 0 6x x3 x x 1 −xe −e = lim = − . (By canceling x, we needed to use l’Hôpital’s rule only twice.) lim x→0 6 x→0 6x 6 2x − 6 5 x 2 − 6x + 9 1 “0” “0” = = lim 3 = = lim 2 = − 2 (b) lim 2 − 3x + 18 x→3 3x − 8x − 3 x→3 x − 3 x→3 x − 4x 0 x −x−6 0 1 x−4 1 1 2 “0” lim = = lim = . (Can you ﬁnd another way?) = (c) lim 2 x→4 2x − 32 x→4 4x x→3 6x − 8 0 16 5 1 (1/x) − 1 “0” (−1/x 2 ) ln x − x + 1 “0” = lim = = lim =− = x→1 2(x − 1) x→1 x→1 (x − 1)2 0 0 2 2

23. (a) lim

4 7 − 3 ln(7x + 1) − ln(4x + 4) “0” 7x + 1 1 = = lim 7x + 1 4x + 4 = = lim (b) lim ln x→1 x→1 x→1 x − 1 x−1 0 1 8 4x + 4 x x (ln x + 1) − 1 x x (ln x + 1)2 + x x (1/x) xx − x “0” “0” = lim = = lim = −2 = (c) lim x→1 x→1 x→1 1 − x + ln x 0 −1 + 1/x 0 −1/x 2 (using Example 6.11.4 to differentiate x x ). √ √ 24. When x → 0, the denominator tends to b − d and the numerator to 0, so the limit does not exist when b = d. If b = d, see the text.

Chapter 8 Single-Variable Optimization
8.1
1. (a) f (0) = 2 and f (x) ≤ 2 for all x (we divide 8 by a number larger than 2), so x = 0 maximizes f (x). (b) g(−2) = −3 and g(x) ≥ −3 for all x, so x = −2 minimizes g(x). g(x) → ∞ as x → ∞, so there is no maximum. (c) h(x) has its largest value 1 when 1 + x 2 is the smallest, namely for x = 0, and h(x) has its smallest value 1/2 when 1 + x 2 is the largest, namely for x = ±1.

8.2
2. See the text, but note that the sign variation alone is not sufﬁcient to conclude that the two stationary points are extreme points. It is important to point out that h(x) → 0 as x → ±∞. Sketch the graph. (For example, f (x) = 3x − x 3 is decreasing in (−∞, −1], increasing in [−1, 1], and decreasing in [1, ∞) but has no maximum or minimum, since f (x) → ∞ as x → −∞, and f (x) → −∞ as x → ∞. In fact, x = −1 is a local minimum point and x = 1 is a local maximum point.) √ √ √ 1 3. h (t) = 1/2 t − 2 = (1 − t)/2 t. We see that h (t) ≥ 0 in [0, 1] and h (t) ≤ 0 in [1, ∞). According to Theorem 8.2.1(a), t = 1 maximizes h(t).
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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SINGLE-VARIABLE OPTIMIZATION

23

5. f (x) = 3x 2 ln x + x 3 /x = 3x 2 (ln x + 1 ). f (x) = 0 when ln x = − 1 , i.e. x = e−1/3 . We see that 3 3 f (x) ≤ 0 in (0, e−1/3 ] and f (x) ≥ 0 in [e−1/3 , ∞), so x = e−1/3 minimizes f (x). Since f (x) → ∞ as x → ∞, there is no maximum. 8. (a) y = ex − 2e−2x , y = ex + 4e−2x . We see that y = 0 when ex = 2e−2x , or e3x = 2, i.e. x = 1 ln 2. Since y > 0 everywhere, this is minimum point. (b) y = −2(x − a) − 4(x − b) = 0 3 when x = 1 (a + 2b). This is a maximum point since y = −6 for all x. 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 x0 = (1/α) ln(Aα/k). Note that x0 > 0 iff Aα > k. Moreover, f (x) < 0 if x < x0 and f (x) > 0 if x > x0 , 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 costs) increases. The signs of these responses are obviously what an economist would expect. (Not only an economist, actually.)

8.3
1 2. (a) π(Q) = Q(a −Q)−kQ = −Q2 +(a −k)Q, so π (Q) = −2Q+(a −k) = 0 for 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. (c) The new proﬁt function is 4 1 ˆ π(Q) = π(Q) + sQ = −Q2 + (a − k)Q + sQ. π (Q) = −2Q + a − k + s = 0 when Q = 2 (a − k + s). ˆ ˆ 1 ˆ Now 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. (The answer in the text has the wrong sign.)

5. T (W ) = a

pbW − bW − c pb(bW + c)p−1 W − (bW + c)p , e.t.c. = a(bW + c)p−1 2 W W2

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) is strictly decreasing so maximum is at x = 0, minimum at x = 3. (b) f (−1) = f (2) = 10 and f (x) = 3x 2 − 3 = 0 at x = ±1. f (1) = 6. (c) f (x) = x + 1/x. f (1/2) = f (2) = 5/2 and f (x) = 1 − 1/x 2 = 0 at x = ±1. f (1) = 2. √ √ √ (d) √(−1) = 4, f ( 5) = 0, and f (x) = 5x 2 (x 2 − 3) = 0 at x = 0 and x = 3. f (0) = 0, f ( 3) = f −6 3. (e) f (0) = 0, f (3000) = 4.5 · 109 , f (x) = 3(x 2 − 3000x + 2 · 106 = 3(x − 500)(x − 2000). f (1000) = 2.5 · 109 , f (2000) = 2 · 109 . 4. (a) When there are 60 + x passengers, the charter company earns 800 − 10x from each, so they earn $(60 + x)(800 − 10x). The sports club earns 1/10 of that amount. (b) See the text. 6. (a) (f (2) − f (1))/(2 − 1) = (4 − 1)/1 = 3 and f (x) = 2x, so 2x ∗ = 3, and thus x ∗ = 3/2. √ √ (b) (f (1) − f (0))/1 = −1 and f (x) = −2x/ 1 − x 2 , so 2x ∗ / 1 − (x∗)2 = 1, so x ∗ = 5/5. (From√ ∗ / 1 − (x∗)2 = 1, 1 − (x∗)2 = 2x ∗ , and then 1 − (x ∗ )2 = 4(x ∗ )2 . The positive solution is 2x √ so and x ∗ = 5/5. (c) (f (6) − f (2))/4 = −1/6√ f (x) = −2/x 2 , so −2/(x ∗ )2 = −1/6,√ x ∗ = 12. (d) (f (4) − f (0))/4 = 1/4 and f (x) = x/ 9 + x 2 , so x/ 9 + (x ∗ )2 = 1/4, so x ∗ = 3.

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CHAPTER 8

SINGLE-VARIABLE OPTIMIZATION

8.5
1 1 1. π(Q) = 10Q − 1000 Q2 − (5000 + 2Q) = 8Q − 1000 Q2 − 5000. Since π (Q) = 8 − 1 Q = 4000, and π (Q) = − 500 < 0, Q = 4000 maximizes proﬁts. 1 500 Q

= 0 for

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, Q = 450 maximizes proﬁts. (ii) π(Q) = 2200Q − 2Q2 − 5000. Since π (Q) = 2200 − 4Q = 0 for Q = 550, and π (Q) = −4 < 0, Q = 550 maximizes proﬁts.. (iii) π(Q) = −2Q2 − 100Q − 5000 is negative for all Q ≥ 0, so Q = 0 obviously maximizes proﬁts. 6. π (Q) = P − abQb−1 = 0 when Qb−1 = P /ab, i.e. Q = (P /ab)1/(b−1) . Moreover, π (Q) = −ab(b − 1)Qb−2 < 0 for all Q > 0, so this is a maximum point.

8.6
2. (a) Strictly decreasing, so no extreme points. (b) f (x) = 3x 2 − 3 = 0 for x = ±1. With f (x) = 6x, f (−1) = −6 and f (1) = 6, so x = −1 is a loc. maximum point, and x = 1 is a loc. minimum point. (c) f (x) = 1 − 1/x 2 = 0 for x = ±1. With f (x) = 2/x 3 , 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): see the text. 3. (a) 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 2 − 2x − 24 (x + 4)(x − 6) x+6+ = = = √ √ √ √ 2 x x 2 x+6 2x 2 x + 6 2x 2 x + 6 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. It follows that f is strictly increasing in [−6, −4], decreasing in [−4, 0), decreasing in (0, 6], and increasing [6, ∞). It follows from the ﬁrst-derivative test (Thm. 8.6.1) that f has two local minimum points, x1 = −6 and x2 = 6, and one local maximum point, √ √ √ 1 x3 = −4, with f (−6) = 0, f (6) = 4 8 = 8 2/3, and f (−4) = 2 2. 3 √ (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 1 ·√ = ·0=0 2 2x 2 x+6 y 8 6 4 2 −6 −4 −2 −2 −4
Figure SM8.6.3

x→∞

x→∞

2

4

6

8 10

x

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

CHAPTER 8

SINGLE-VARIABLE OPTIMIZATION

25

4. Look at the point a. Since f (x) is graphed, f (x) < 0 to the left of a, f (a) = 0, and f (x) > 0 to the right of a, so a is a local minimum point. At the points b and e, f (x) > 0 on both sides of the points, so they cannot be extreme points. 6. (a) f (x) = x 2 ex (3 + x). Use a sign diagram. (x = 0 is not an extreme point, but an inﬂection point.) (b) See the text and use a sign diagram for g (x), or check the sign of g (x) = 2x (2 + 4x ln x + x 2 (ln 2)2 ) at the stationary points. 7. f (x) = x 3 + ax + b → ∞ as x → ∞, and f (x) → −∞ as x → −∞. Thus f (x) has at least one real root. We have f (x) = 3x 2 + a. Thus, for a ≥ 0, f (x) > 0 for all x = 0, so f is strictly increasing, and there is only one real root. Note that for a ≥ 0, 4a 3 + 27b2 ≥ 0. Assume next that a < 0. Then f (x) = 0 √ √ √ √ for x = ± −a/3 = ± p, where p = −a/3 > 0. Then f has a local maximum at (− p, b + 2p p) √ √ and a local minimum at ( p, b − 2p p). If one of the local extreme values is 0, the equation has a double root, and this is the case iff 4p3 = b2 , that is, iff 4a 3 + 27b2 = 0. The equation has three real roots iff the local maximum value is positive and the local minimum value is negative. This occurs iff √ |b| < 2p p or iff b2 < 4p3 or iff 4a 3 + 27b2 < 0.

8.7
1. (a) f (x) = 3x 2 +3x−6 = 3(x−1)(x+2). Use a sign diagram and see the text. (b) f (x) = 6x+3 = 0 for x = −1/2 and f (x) changes sign around x = −1/2, so this is an inﬂection point. 3. Straightforward by using these derivatives: (a) y = −e−x (1 + x), y = xe−x 2−x x−1 ,y = (c) y = x 2 e−x (3 − x), y = xe−x (x 2 − 6x + 6) (b) y = 2 x3 x 1 − 2 ln x 6 ln x − 5 ,y = (e) y = 2ex (ex − 1), y = ex (2ex − 1) (d) y = 3 x x4 (f) y = 2e−x (2 − x 2 ), y = e−x (x 2 − 2x − 2) Review Problems for Chapter 8
1 3 2. (a) Q (L) = 24L − 20 L2 = 3L(8 − 20 L) = 0 for L∗ = 160, and Q(L) is increasing in [0, 160], ∗ = 160 maximizes Q(L). Output per worker is Q(L)/L = 12L − 1 L2 , decreasing in [160, 200], so Q 20 and this quadratic function has maximum at L∗∗ = 120. (b) See the text.

3. π = −0.0016Q2 + 44Q − 0.0004Q2 − 8Q − 64 000 = −0.002Q2 + 36Q − 64 000, and Q = 9000 maximizes this quadratic function. 0.0008Q2 + 8Q Q ≈ 0.12 when Q = 1000. C (Q) = (b) ElQ C(Q) = 0.0004Q2 + 8Q + 64 000 C(Q) 4. (a) See Problem 8.7.3(c). (b) lim f (x) = 0 according to (7.12.3), page 253 in the text. x→∞ lim f (x) = −∞ because x 3 → −∞ and e−x → ∞. (See Fig. A8.R.4, page 671 in the text.) x→−∞ 5. (a) See the text. (b) A sign diagram shows that f (x) ≥ 0 in (−1, 1] and f (x) ≤ 0 in [1, ∞). Hence x = 1 √ √ −x(x 2 + x − 1) 1 1 = 0 for x = 0 and x = 2 ( 5 − 1). (x = 2 (− 5 − 1) is a maximum point. f (x) = (x + 1)2 is outside the domain.) Since f (x) changes sign around these points, they are both inﬂection points.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

26

9

INTEGRATION

6. (a) h (x) =

ex (2 + e2x ) − ex 2e2x ex (2 − e2x ) = . See the text. (2 + e2x )2 (2 + e2x )2 (b) h is strictly increasing in (−∞, 0], lim h(x) = 0, and h(0) = 1/3. Thus, h deﬁned on (−∞, 0] has x→−∞ ex = y ⇐⇒ 2 + e2x 2 ]/2y. Since y = 1/3 y(ex )2 −ex +2y = 0. This quadratic equation in ex has the roots ex = [1 ± 1 − 8y√ for x = 0, see that we must have ex = [1 − 1 − 8y 2 ]/2y, and thus x = ln(1 − 1 − 8x 2 ) − ln(2x). √ 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.6. an inverse deﬁned on (0, 1/3] with values in (−∞, 0]. To ﬁnd the inverse, note that y 0.5 h −1.5 −1.0 −0.5 −0.5 h−1 −1.0 −1.5
Figure SM8.R.6

0.5

x

8. f (x) =

√ √ −6x 2 (x 2 − 3)(x 2 + 2) , so f is stationary when x = 0 and when x = ± 3. x = 3 is a local 4 + x 2 + 2)2 (x √ (and global) maximum point, x = − 3 is a local (and global) minimum point, and x = 0 is neither. (It is an inﬂection point.) The graph of f is shown in Fig. A8.R.8 in the text, page 671.

9 Integration
9.1
√ 1. This should be straightforward, since all the integrands are powers of x. Note that x x = x ·x 1/2 = x 3/2 , √ √ √ √ √ 1/ x = x −1/2 , and x x x = x x 3/2 = x · x 3/4 = x 7/4 = x 7/8 . 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

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

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

9
1 (x + 2)3 = 4 (x + 2)4 + C1 .

INTEGRATION

27

or: (f)

(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 − 4y + 4 (y − 2)2 = 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 , so = x 2 −x +1− x+1 x+1 x3 1 dx = 1 x 3 − 2 x 2 +x −ln |x +1|+C. 3 x+1

(b) Polynomial division: (c) Since

d 1 (1 + x 2 )16 = 16(1 + x 2 )15 · 2x = 32x(1 + x 2 )15 , x(1 + x 2 )15 dx = 32 (1 + x 2 )16 + C. dx √ 1 10. (a) Easy. (b) (ii) x + 2 = (x + 2)1/2 , and use (a). (iii) √ = (4 − x)−1/2 , and use (a). 4−x
1 1 1 11. F (x) = ( 2 ex − 2x) dx = 2 ex − x 2 + C. F (0) = 2 implies C = 0. 3 ) dx = 1 x 2 − 1 x 4 + C. F (1) = 5 implies C = 1 . (b) F (x) = (x − x 2 4 12 6

1 1 13. f (x) = (x −2 +x 3 +2) dx = −x −1 + 4 x 4 +2x +C. With f (1) = 1/4 we have 1/4 = −1+ 4 +2+C, 1 1 so C = −1. New integration yields f (x) = (−x −1 + 4 x 4 + 2x − 1) dx = − ln x + 20 x 5 + x 2 − x + D. 1 With f (1) = 0 we have 0 = − ln 1 + 20 + 1 − 1 + D, so D = −1/20.

9.2
3

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

(f)
2

1 + t dt = t −1

−2 3 2

1 2 2x

− 1 x 3 dx = 3

3 −2

1 ( 6 x3 − 9 2

1 4 12 x )

=

3 −2

1 27 ( 12 x 3 (2 − x) = − 12 + 5 2

32 12

=

5 12 .

1 ln(t − 1) + 2 t 2 = ln 2 +

−

4 2

= ln 2 +

6. (a) A sign diagram shows that f (x) > 0 when 0 √ x < 1 and when x > 2. (b) f (x) = x 3 − 3x 2 + 2x, < √ 2 − 6x + 2 = 0 for x = 1 − 3/3 and x = 1 + 3/3. We see that f (x) > 0 ⇐⇒ hence f (x) = 3x 0 1 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 ]. Hence x0 is a local maximum point and x1 is a local minimum 1 4 1 1 1 1 point. (c) See the graph in the text. 0 f (x) dx = 0 (x 3 −3x 2 +2x) dx = 0 x4 −x 3 +x 2 = 4 −0 = 4 . √ √ 3000 000 7. (a) f (x) = −1 + = 0 for x = 3000 000 = 1000 3. (Recall x > 0.) For the rest, see the x2 text.

9.3
2. (a) x p+r+1 1 1 x p+q+1 + = + p+r +1 p+q +1 p+r +1 0 p+q +1 0 (b) f (1) = 6 implies a + b = 6. Since f (x) = 2ax + b, f (1) = 18 implies 2a + b = 18. It follows that a = 12 and b = 6, so f (x) = 12x 2 − 6x. But then f (x) = (12x 2 − 6x) dx = 4x 3 − 3x 2 + C, 2 2 and since we want 0 (4x 3 − 3x 2 + C) = 18, we must have 0 (x 4 − x 3 + Cx) = 18, i.e. C = 5. (x p+q + x p+r ) dx =
1 1 1

3. (a) See text.
1

(b)

(c)
0

x2 + x + x + 1 dx = x+1

√0

(x 2 + 2)2 dx =
1 0 0

1

(x 4 + 4x 2 + 4) dx =

1 0

( 1 x 5 + 4 x 3 + 4x) = 83/15 5 3
1 0

x(x + 1) + (x + 1)1/2 dx = x+1

(x + (x + 1)−1/2 ) dx =

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

28
1 0

9

INTEGRATION

√ 1 ( 2 x 2 + 2(x + 1)1/2 ) dx = 2 2 −

3 2

(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

6. From y 2 = 3x we get x = 1 y 2 , which inserted into the other equation gives y + 1 = ( 1 y 2 − 1)2 , or 3 3 y(y 3 − 6y − 9) = 0. Here y 3 − 6y − 9 = (y − 3)(y 2 + 3y + 3), with y 2 + 3y + 3 never 0. So (0, 0) and (3, 3) are the only points of intersection. See the text. 7. W (T ) = K(1 − e− T )/ T . Here W (T ) → 0 as T → ∞, and using 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). 8. (a) f (x) = √ 2 > 0 for x > 0 and f has range (−∞, ∞), so f has an inverse √ x + 4 ( x + 4 − 2) √ deﬁned on (−∞, ∞). We ﬁnd that the inverse is g(x) = ex/2 + 4ex/4 . (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 .) (b) See Fig. A9.3.8. (c) In Fig. A9.3.8 the graphs of f and g are symmetric about the line y = x, so area A = area B. But area B is the area of a rectangle with base a and height 10 minus the area below the a graph of g over the interval [0, a]. Therefore, B = 10a− 0 (ex/2 +4ex/4 ) dx = 10a+18−2ea/2 −16ea/4 . √ √ Because a = f (10) = 4 ln( 14 − 2), this simpliﬁes to 10a + 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 . Their total income is M = n is then N = n −Br −1 = Br −2 r dr = Br −2 dr = n 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 =

nABp

γ b

r δ−1 2δ−1 − 1 = nABpγ bδ−1 . δ−1 δ−1 t 5. (a) See Fig. A9.4.5. all t. (c)
10 0 0 10

(b)
0 10 0 10 0

g(τ ) − f (τ ) dτ =
0

t

1 2τ 3 − 30τ 2 + 100τ ) dτ = 2 t 2 (t − 10)2 ≥ 0 for

p(t)f (t) =

−t 3 + 9t 2 + 11t − 11 + 11/(t + 1) dt = 940 + 11 ln 11 ≈ 966.38,

p(t)g(t)dt =

t 3 − 19t 2 + 79t + 121 − 121/(t + 1) dt = 3980/3 − 121 ln 11 ≈ 1036.52.

Proﬁle g should be chosen.

9.5
1. (a) x e−x dx = x (−e−x ) −
↑ ↑ f g ↑ f ↑ g 1 3xe4x dx = 3x · 4 e4x − ↑ f

1 · (−e−x ) dx = −xe−x +
↑ g 3 4x 16 e

e−x dx = −xe−x − e−x + C +C

(b)

1 3 3 · 4 e4x dx = 4 xe4x −

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

9

INTEGRATION

29

(c)

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

2x(−e−x ) dx = −(1 + x 2 )e−x + 2

xe−x dx

= −(1 + x 2 )e−x − 2xe−x − 2e−x + C = −(x 2 + 2x + 3)e−x + C 1 1 1 1 1 1 1 (d) xln x dx = 2 x 2 ln x − 2 x 2 dx = 2 x 2 ln x − 2 x dx = 2 x 2 ln x − 4 x 2 + C x 2. (a) See the text. d (b) Recall that dx 2x = 2x ln 2, and therefore 2x / ln 2 is the indeﬁnite integral of 2x . If follows that 2 2 2 2 3 8 4 2x 8 1 8 2x 2x − = − = − − dx = − x2x dx = x ln 2 (ln 2)2 (ln 2)2 ln 2 (ln 2)2 (ln 2)2 ln 2 ln 2 ln 2 0 0 0 0 (c) First use integration by parts on the indeﬁnite integral: 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). Inserted into
1 0

more: with f (x) = 2x and g(x) = ex , (∗) this gives
1

x 2 ex dx = x 2 ex − 2xex + 2ex + C, and hence,
1 0 T

x 2 ex dx =

1 0

(x 2 ex − 2xex + 2ex ) =
1

(e − 2e + 2e) − (0 − 0 + 2) = e − 2. Alternatively, more compactly using formula (9.5.2): x 2 ex dx = x 2 ex − 2
0 1

xex dx = e − 2
T

1 0

xex −
0

1

0

ex dx = e − 2[e − 0 ex ] = e − 2 −1 −rt −T −rT 1 e dt = e + r r r
T 0

5. (a) By formula (9.5.2), −T −rT 1 = e + r r
T T 0 −rt 2 0

te−rt dt =

t
0

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

−1 −rt e − r

T 0

e−rt dt

(b)
0 T

(a + bt)e

e−rt dt + b
T 0

T

te−rt dt, a.s.o. using (a).
T 0

(c)
0

(a − bt + ct )e t e
2 −rt

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

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. x 2 ex
3 +2

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

dx =

= 1 eu + C = 1 ex 3 3

3 +2

+ C.

ln u ln(x + 2) du. dx = 2u 2x + 4 This does not look very much simpler than the original integral. A better idea is to substitute u = ln(x +2). dx ln(x + 2) 2 2 1 1 1 Then 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 √ = 2 u5/2 − 2 u3/2 + C = 2 (1 + x)5/2 − 2 (1 + x)3/2 + C. Second attempt: u = 1 + x. Then u2 = 1 + x 3 5 3√ 5 and 2udu = dx. Then the integral is x 1 + x dx = (u2 − 1)u2u du = (2u4 − 2u3 ) du e.t.c. Check that you get the same answer. Actually, even integration by parts works in this case. Put f (x) = x and √ g (x) = 1 + x, and choose g(x) = 2 (1 + x)3/2 . (The answer looks different, but is not.) 3 (c) First attempt: u = x + 2, which gives du = dx and
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

30

9

INTEGRATION

(e) With u = 1 + x 2 , x 2 = u − 1, and du = 2xdx, so
1 2

u−1 1 du = 2 u3 √ (f) With u = 4 − x 3 , u2 = 4 − x 3 , and 2udu = −3x 2 dx, so = (4−u2 ) u (− 2 )u du = 3
1 0

x3 x2 · x dx = dx = (1 + x 2 )3 (1 + x 2 )3 1 −1 1 1 + + C. (u−2 − u−3 ) du = − 2 u−1 + 4 u−2 + C = 2(1 + x 2 ) 4(1 + x 2 )2 x 5 4 − x 3 dx = x 3 4 − x 3 x 2 dx

2 2 (− 8 u2 + 2 u4 ) du = − 8 u3 + 15 u5 +C = − 8 (4−x 3 )3/2 + 15 (4−x 3 )5/2 +C 9 3 9 3 1 0

6. (a) I =

(x 4 − x 9 )(x 5 − 1)12 dx =

−x 4 (x 5 − 1)13 dx. Introduce u = x 5 − 1. Then du = 5x 4 dx,
0 0

1 13 1 14 1 and when x = 0, u = −1, when x = 1, u = 0, and thus I = − 70 u = 70 . 5 u du = − −1 −1 √ (b) With u = x, u2 = x, and 2udu = dx, and so √ √ √ √ ln x √ dx = 2 ln u2 du = 4 ln u du = 4(u ln u − u) + C = 4 x ln x − 4 x + C = 2 x ln x − √ 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, u − 1 = x, or (u − 1)2 = x, so 2(u − 1)du = dx. When x = 0, u = 1, when x = 4, u = 3. Hence, 3 3 3 4 2(u − 1) dx = (u1/2 − u−1/2 ) du = 2 ( 2 (u3/2 − 2u1/2 ) du = 8 du = 2 √ √ 3 3 u 1 1 1 0 1+ x √ (The substitution u = 1 + x also works.)

7. (a) With u = 1 + e

√ 1 √ · e x dx. When x = 1, u = 1 + e and x = 4 gives 2 x u = 1 + e2 . Thus we get (note how we carry over the limits of integration): √ 1+e2 1+e2 4 e x 2 du √ =2 ln u = 2 ln(1 + e2 ) − 2 ln(1 + e) dx = √ x) u x (1 + e 1 1+e 1+e x +1, du = ex dx, and dx = du/ex = du/(u−1). When x = 0, u = 2, (b) A natural substitution is u = e e1/3 +1 e1/3 +1 1/3 1 1 1 dx 1/3 +1. Thus, − du = = du = when x = 1/3, u = e x +1 u−1 u u(u − 1) e 2 2 0 √ x,

u > 0, and du =

e1/3 +1 2

ln |u − 1| − ln |u| =

1 3

− ln(e1/3 + 1) + ln 2 = ln 2 − ln(e−1/3 + 1). (Verify the last equality.)

e−x , the suggested substitution t = e−x (or even better u = 1 + e−x ), 1 + e−x dt = −e−x dx works well. Verify that you get the same answer. Rewriting the integrand as

9.7
3. (a) See answer in the text. Using a simpliﬁed notation and the result in Example 1(a), we have: ∞ ∞ ∞ (b) 0 (x − 1/λ)2 λe−λx dx = − 0 (x − 1/λ)2 e−λx + 0 2 (x − 1/λ) e−λx dx ∞ ∞ −λx dx = 1/λ2 + 2/λ2 − 2/λ2 = 1/λ2 = 1/λ2 + 2 0 xe−λx dx − (2/λ) 0 e ∞ ∞ 3 −λx dx = − ∞ (c) 0 (x − 1/λ) λe (x − 1/λ)3 e−λx + 0 3 (x − 1/λ)2 e−λx dx 0 ∞ = −1/λ3 + (3/λ) 0 (x − 1/λ)2 λe−λx dx = −1/λ3 + (3/λ)(1/λ2 ) = 2/λ3 5. (a) f (x) = (1 − 3 ln x)/x 4 = 0 at x = e1/3 , and 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.)
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

9
1 (b) a x −3 ln x dx = − a 2 x −2 ln x + a ∞ −3 and a → 0. But 1 x ln x dx = 1/4. b b b 1 −3 dx 2x

INTEGRATION

31

=

b 1 (− 2 x −2 a

1 ln x − 4 x −2 ). This diverges when b = 1

√ 3 7. If both limits exist, the integral is the sum of the following two limits: I1 = limε→0+ −2+ε 1/ x + 2 dx √ √ √ √ 3 3−ε and I2 = limε→0+ −2 1/ 3 − x dx. Here I1 = lim+ −2+ε 2 x + 2 = lim+ 2 5 − 2 ε = ε→0 ε→0 √ √ √ √ √ 3−ε 2 5, and I2 = lim+ −2 −2 3 − x = lim+ −2 ε + 2 5 = 2 5. ε→0 ε→0

12. (a) The suggested substitution gives
+∞

(b) Here

−∞

1 xf (x) dx = √ π
+∞

−∞

+∞ 1 2 f (x) dx = √ e−u du = 1, by (9.7.8). π −∞ −∞ +∞ √ 2 (μ + 2σ u)e−u du = μ, using part (a) and Example 3.

+∞

+∞ √ 1 2 (2σ 2 u2 + 2 2σ μu + μ2 )e−u du x 2 f (x) dx = √ π −∞ −∞ √ +∞ μ2 2 2σ μ +∞ −u2 2σ 2 +∞ 2 −u2 2 ue du + √ e−u du = σ 2 + 0 + μ2 . (Note how u e du + √ = √ π −∞ π π −∞ −∞ +∞ √ 2 2 2 2 1 1 1 u2 e−u du = 2 π.) integration by parts gives u2 e−u du = − 2 ue−u + 2 e−u du, so

(c) Here I =

−∞

9.8
5. P (10) = 705 gives 641e10k = 705, or e10k = 705/641. Taking the natural logarithm of both sides yields 10k = ln(705/641), so k = 0.1 ln(705/641). 7. Straightforward. Note that in (9.8.10), if b = 0, there are always two constant solutions, x ≡ 0 and x ≡ a/b. The latter is obtained by letting A = 0 in (9.8.10). So in addition to the answer in the text, in (e) add x ≡ 0 as a solution, in (f) add K ≡ 0. 9. (a) Use (9.8.7). (Using (9.8.10), and then using N(0) = 1 to determine the constant, is less efﬁcient.) 1 1000 ∗ ∗ ⇐⇒ 999e−0.39t = , so e−0.39t = 1/3996, and so 0.39t ∗ = ln 3996, etc. (b) 800 = −0.39t ∗ 4 1 + 999e

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. (You cannot wait to put the

1 3 3 constant “in the end”. Wrong: 1 x 3 = 2 e2t , x 3 = 2 e2t , x = 3 2 e2t + C. This is not a solution!) 3 (b) 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). 1 (e) By (9.9.5), x = Ce2t + e2t (−t)e−2t dt = Ce2t − e2t te−2t dt. Here te−2t dt = t (− 2 )e−2t + 1 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 . 2 2 −3t 2 1 2 dt = Ce−3t + e−3t tet dt = Ce−3t + 2 et −3t (f) Formula (9.9.5): x = Ce−3t + e−3t e3t tet

3. The equation is separable: dk/k = sαeβt dt, so ln k = k(0) = k0 , we have k0 = Ce β , and thus k = k0 e
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008 sα sα βt β e sα βt (e −1) β

+ C1 , or k = e β e eC1 = Ce β e . With

sα βt

sα βt

.

32

9

INTEGRATION

5. (a) See the text. (b) Hence, K 1−α = K 1−α 6.

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 . For t = 0, K0 = β γ L0 eβt dt, so

dx dx dt t dx = a = a is separable: = a , so t x x dt x x = ea ln t+C1 = (eln t )a eC1 = Ct a , with C = eC1 .

dt . Integrating yields, ln x = a ln t + C1 , so t

Review Problems for Chapter 9
12

3. (a)
0 3

50 dx =
2

12 0

50x = 600
3 −3 1 3 (u

2

(b)
0

1 (x − 2 x 2 ) dx = 5

2 0

1 1 ( 2 x2 − 6 x3) = 5 1

2 3

(c) (e)

4 4 12 3 dt 1 v v 2 + 9 dv = 3 (v 2 + 9)3/2 = 98/3 3 ln(t + 4) = 3 ln(8/3) (f) dt = t +4 0 0 2 2 √ 5. (a) With u = 9 + x, x = (u − 9)2 and dx = 2(u − 9) du. When x = 0, u = 9 and x = 25 gives 14 14 25 14 18 1 2(u − 9) du = du = 10 − 18 ln . 2− u = 14. Thus, √ dx = u u 9 9 9 √0 9 + x (b) With u = t + 2, t = u2 − 2, and dt = 2u du. When t = 2, u = 2, and t = 7 gives u = 3. Hence, 3 3 3 7 √ 1 5 2 3 = 886/15 (u2 − 2)u · 2u du = 2 (u4 − 2u2 ) du = 2 t t + 2 dt = 5u − 3u 2 2 2 2 √ 3 (c) With u = 19x 3 + 8, u3 = 19x 3 + 8, so 3u2 du = 57x 2 dx. When x = 0, u = 2 and x = 1 gives

−3 12

(u + 1) du =

+

1)3 du

= 24

(d)
1

2 dz = z

2 ln z = 2 ln 5

u = 3. Then
0

1

57x 2 19x 3 + 8 dx =
3

3

3u3 du =

3 2

2

3 4 4u

= 195/4.

10. Equilibrium when 50/(Q∗ + 5) = 10 + Q∗ , i.e. (Q∗ )2 + 50Q∗ − 275 = 0. The only positive solution 5 5 50 − 5 dQ = [50 ln(Q + 5) − 5Q] = 50 ln 2 − 25, is Q∗ = 5, and then P ∗ = 5. CS = Q+5 0 0 PS =
0 5

(5 − 4.5 − 0.1Q) dQ = 1.25.

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 − (2 ln t − (ln t)2 ) 2t (2 · (1/t) − 2 ln t · (1/t)) t (ln t)2 − 3 ln t + 1 =8 f (t) = 4 t3 t4 (b) Stationary points: f (t) = 0 ⇐⇒ ln t (2 − ln t) = 0 ⇐⇒ ln t = 2 or ln t = 0 ⇐⇒ t = e2 or t = 1 Since f (1) = 8 > 0 and f (e2 ) = −8e−6 , 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. 4 d 4 4 1 (ln t)3 = 3(ln t)2 = f (t), so f (t) dt = (ln t)3 + C. Since f (t) ≥ 0 for all t > 0, (c) 3 t 3 dt 3 e2 e2 4 4 32 (ln t)3 = 23 − 0 = . f (t) dt = the area is 3 3 3 1 1

12. Straightforward by using (9.8.8)–(9.8.10).
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

10

INTEREST RATES AND PRESENT VALUES

33

1 1 13. (a) Separable. x −2 dx = t dt, and so −1/x = 2 t 2 + C1 , or x = 1/(C − 2 t 2 ), (with C = −C1 ). (b) and (c): Direct use of (9.9.3). (d) Using (9.9.5), x = Ce−5t + 10e−5t te5t dt. Here te5t dt = 1 1 t 1 e5t − 1 e5t dt = 1 te5t − 25 e5t . Thus x = Ce−5t + 10e−5t ( 1 te5t − 25 e5t ) = Ce−5t + 2t − 2 . 5 5 5 5 5 −t/2 + e−t/2 et/2 et dt = Ce−t/2 + e−t/2 e3t/2 dt = Ce−t/2 + e−t/2 2 e3t/2 = Ce−t/2 + 2 et . (e) x = Ce 3 3 (e) x = Ce−3t + e−3t t 2 e3t dt = Ce−3t + e−3t ( 1 t 2 e3t − 2 te3t dt) 3 3 2 1 = Ce−3t + 1 t 2 − 2 e−3t ( 1 te3t − 1 e3t dt) = Ce−3t + 3 t 2 − 2 t + 27 . 9 3 3 3 3

16. (a) and (b), see the text. (c) F (x) = f (x) = 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. (d) −∞ f (x) dx =
0 lim a→−∞ a

f (x)dx + lim

b b→∞ 0

f (x) dx = lim [F (0) − F (a)] + lim [F (b) − F (0)] = 1, by (a). a→−∞ b→∞

10 Interest Rates and Present Values
10.1
3. We solve (1 + p/100)100 = 100 for p. Raising each side to 1/100, 1 + p/100 = √ 100( 100 100 − 1) ≈ 100(1.047 − 1) = 4.7.
100

√

100, so p =

5. Use formula (10.1.2). (i) R = (1 + 0.17/2)2 − 1 = (1 + 0.085)2 − 1 = 0.177225 or 17.72% For (ii) ansd (iii) see the answers in the text.

10.2
4. If it looses 90% of its value, then e−0.1t = 1/10, so −0.1t ∗ = − ln 10, hence t ∗ = (ln 10)/0.1 ≈ 23. 6. With g(x) = (1 + r/x)x for all x > 0, then ln g(x) = x ln(1 + r/x). Differentiation gives g (x)/g(x) = ln(1 + r/x) + x(−x/r 2 )/(1 + r/x) = ln(1 + r/x) − (x/r)/(1 + r/x), as claimed in the problem. Then see the answer in the text.
∗

10.3
3. (a) We ﬁnd f (t) = 0.05(t + 5)(35 − t)e−t . Obviously, f (t) > 0 for t < 35 and f (t) < 0 for t > 35, so t = 35 maximizes f (with f (35) ≈ 278). (b) f (t) → 0 as t → ∞. See the graph in Fig. A10.3.3.

10.4
2. We use formula (10.4.5): (a) 0.1 0.1 1 = = 1 − 0.1 0.9 9 1− 5 35 a = 1 + a (e) = (d) 1 − 1/(1 + a) 1 − 3/7 4
1 5 1 5

=

1 4

(b)

(c)

517 517 · 1.1 = = 5687 1 − 1/1.1 0.1

6. Let x denote the number of years beyond 1971 that the extractable resources of iron will last. Then 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 ≈ 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.
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8. (a) The quotient of this inﬁnite series is e−rt , so the sum is f (t) = (b) f (t) =

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 ∗ = r→0 1 − e−rt r→0 t e 0 t

P (t)e−rt P (t) = rt −rt 1−e e −1

10.5
4. Offer (a) is better, because the second offer has present value 4600 1 − (1.06)−5 ≈ 20 540. 1 − (1.06)−1

7. This is a geometric series with ﬁrst term a = D/(1 + r) and quotient k = (1 + g)/(1 + r). It converges a D/(1 + r) D iff k < 1, i.e. iff 1 + g < 1 + r, or g < r. The sum is = = . 1−k 1 − (1 + g)/(1 + r) r −g

10.6
4. Schedule (b) has present value 12 000 · 1.115 [1 − (1.115)−8 ] ≈ 67 644.42. 0.115 7000 Schedule (c) has present value 22 000 + [1 − (1.115)−12 ] ≈ 66 384.08. 0.115 Thus schedule (c) is cheapest. When the interest rate becomes 12.5 %, schedules (b) and (c) have present values equal to 65907.61 and 64374.33, respectively.

10.7
5. After dividing all the amounts by $ 10 000, the equation is f (s) = s 20 + s 19 + · · · + s 2 + s − 10 = 0. Then f (0) = −10 and f (1) = 10, so by the intermediate value theorem (Theorem 7.10.1), there exists a number s ∗ such that f (s ∗ ) = 0. Here s ∗ is unique because f (s) > 0. In fact f (s) = −10+(s −s 21 )/(1−s), and f (s ∗ ) = 0 ⇐⇒ (s ∗ )10 − 11s ∗ + 10 = 0. Then s ∗ = 0.928 is an approximate root, which corresponds to an internal rate of return of about 7.8 %. (See Problem 7.R.26.)

10.8
3. Equilibrium requires αPt − β = γ − δPt+1 , or Pt+1 = −(α/δ)Pt + (β + γ )/δ. Using (10.8.4) we obtain the answer in the text. Review Problems for Chapter 10 3. If you borrow $ a at the annual interest rate of 11% with interest paid yearly, then the dept after 1 year is a(1 + 11/100) = a(1.11); if you borrow at annual interest rate 10% with interest paid monthly, your dept after 1 year will be a(1 + 10/12 · 100)12 ≈ 1.1047a, so schedule (ii) is preferable. 6. We use formula (10.4.5): (a) sum is 44 = 100 (b) The ﬁrst term is 20 and the quotient is 1/1.2, so the 1 − 0.56 3 = 5 (d) The ﬁrst term is (1/20)−2 = 400 and the quotient is (c) 1 − 2/5

20 = 120 1 − 1/1.2 400 1/20, so the sum is = 8000/19 1 − 1/20)

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35

8. (a) See the text. (c) See the text. 11. (a) f (t) = 100e

(b) We use formula (10.5.3) on the future value of an annuity. See the text.
√ t/2 −rt

1 √ − r . We see that f (t) = 0 for t ∗ = 1/16r 2 . Since f (t) > 0 for t < t ∗ 4 t and f (t) < 0 for t > t ∗ , t ∗ maximizes f (t). √ 1 (b) f (t) = 200e−1/t e−rt 2 − r . We see that f (t) = 0 for t ∗ = 1/ r. Since f (t) > 0 for t < t ∗ t and f (t) < 0 for t > t ∗ , t ∗ maximizes f (t). e
10 0

12. (a) F (10) − F (0) = not F (10).)

(1 + 0.4t) dt =

10 0

(t + 0.2t 2 ) = 30. (Note: the total revenue is F (10) − F (0),

(b) See Example 9.5.3.

11 Functions of Many Variables
11.1
6. (a) The denominator must not be 0, so the function is deﬁned for those (x, y) where y = x − 2. (b) Must require 2 − (x 2 + y 2 ) ≥ 0, i.e. x 2 + y 2 ≤ 2. (c) Put a = x 2 + y 2 . We must require (4 − a)(a − 1) ≥ 0, i.e. 1 ≤ a ≤ 4. (Use a sign diagram.) 7. For (a) and (c) see the text. (b) Since (x − a)2 ≥ 0 and (y − b)2 ≥ 0, it sufﬁces to assume that x = a and b = b, because then we take ln of positive numbers.

11.2
3. (a) and (b) are straightforward. (c) f (x, y) = (x 2 − 2y 2 )5 = u5 , where u = x 2 − 2y 2 . Then f1 (x, y) = 5u4 u1 = 5(x 2 − 2y 2 )4 2x = 10x(x 2 − 2y 2 )4 . In the same way, f2 (x, y) = 5u4 u2 = 5(x 2 − 2y 2 )4 (−4y) = −20y(x 2 − 2y 2 )4 . Finally, f12 (x, y) is the derivative of f1 (x, y) w.r.t. y, keeping x constant, so f12 (x, y) = (∂/∂y)(10x(x 2 − 2y 2 )4 ) = 10x4(x 2 − 2y 2 )3 (−4y) = −160xy(x 2 − 2y 2 )3 . 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 . In the same way, 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, y is a constant, so the rule dx a /dx = ax a−1 applies.) zyy = (∂/∂y)(x y ln x) = x y (ln x)2 . Finally, zxy = (∂/∂y)(yx y−1 ) = x y−1 + yx y−1 ln x. (Note that if w = x y−1 = x v , with v = y − 1, then wy = x v ln x · 1 = x y−1 ln x. Or: w = x y−1 = (1/x)x y etc.)

11.3
8. (a) The point (2, 3) lies on the level curve z = 8, so f (2, 3) = 8. The points (x, 3) are those on the line y = 3 parallel to the x-axis. This line intersects the level curve z = 8 when x = 2 and x = 5. (b) See the text. (c) If at A you look in the direction of the positive x-axis, you meet level sets whose values increase, so f1 (x, y) > 0. A rough estimate of f1 (x, y) at A is 2, because if you go one unit in the positive x-axis direction from A, then f (x, y) increases from 8 to 10. 9. (a) It might help to regard the ﬁgure as a map of a mountain. At P the terrain is raising in the direction of the positive x-axis, so fx (P ) > 0. The terrain is sloping downwards in the direction of the positive y-axis so fy (P ) < 0. (b) (i) The line x = 1 has no point in common with any of the given level curves.
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(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, you ﬁrst meet the level curve 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, where the straight just touches the level curve. 10. F (1, 0) = F (0, 0) + F (0, 1) = F (0, 0) + F (1, 1) = F (1, 0) +
1 1 2 0 F1 (x, 0) dx ≥ 0 2 dx = 2, F (2, 0) = F (1, 0) + 1 F1 (x, 0) dx ≥ F (1, 0) + 2, 1 1 0 F2 (0, y) dy ≤ 1, F (1, 1) = F (0, 1) + 0 F1 (x, 1) dx ≥ F (0, 1) + 2, 1 0 F2 (1, y) dy ≤ F (1, 0) + 1.

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
1 5. When r and w are constants, so is (1/r + 1/w), and thus ∂π/∂p = 2 p(1/r + 1/w).

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 .

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 2bK 5 L − aK 2 L4 2aKL5 − bK 4 L2 , so and YL = (c) YK = (aL3 + bK 3 )2 (aL3 + bK 3 )2 2aK 2 L5 − bK 5 L2 + 2bK 5 L2 − aK 2 L5 K 2 L2 (aL3 + bK 3 ) K 2 L2 = = = Y. KYK + LYL = (aL3 + bK 3 )2 (aL3 + bK 3 )2 aL3 + bK 3 (According to Section 12.6 these functions 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 = (−m/ρ)a(−ρ)K −ρ−1 Aeλt aK −ρ + bL−ρ YL = (−m/ρ)b(−ρ)L−ρ−1 Aeλt m(aK −ρ aK −ρ + bL−ρ )Aeλt
−(m/ρ)−1 −(m/ρ)−1 bL−ρ

= maK −ρ−1 Aeλt aK −ρ + bL−ρ = mbL−ρ−1 Aeλt aK −ρ +

−(m/ρ)−1

,

−(m/ρ)−1 bL−ρ .

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

aK −ρ

−(m/ρ)−1 bL−ρ

11.8
4. mDm − D p pD ∂ pD =p = 2 (mDm − D) = 2 [Elm D − 1] > 0 iff 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.)

Review Problems for Chapter 11 ∂u ∂z 7. (a) z = (x 2 y 4 +2)5 = u5 , with u = x 2 y 4 +2, so = 5u4 = 5(x 2 y 4 +2)4 2xy 4 = 10xy 4 (x 2 y 4 +2)4 ∂x ∂x For the rest see the text.
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37

11. (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 means that x = 0, or x 2 = 2y. For x = 0, (ii) yields y = −1, so (x, y) = (0, −1) is one √ solution. With x 2 = 2y, (ii) reduces to 4y − 8y + 4 = 0, or y = 1. But then √ 2 = 2, such that x = ± 2. So the two additional solutions are (x, y) = (± 2, 1). x

12 Tools for Comparative Statics
12.1
5. We look at (c) and (d). (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) = g(t)f (t) + f (t)g (t). x dx 1 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) g(t)f (t) − f (t)g (t) . = y2 (g(t))2 6. Let U (x) = u(x, h(x)) = ln[x α + (ax 4 + b)α/3 ] − α ln(ax 4 + b). 3 √ αx α−1 (3b − ax 4 ) . So U (x ∗ ) = 0 at x ∗ = 4 3b/a, whereas U (x) > 0 for Then U (x) = α + (ax 4 + b)α/3 ](ax 4 + b) 3[x 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 . Assuming 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 + F2 (x, y) = = F1 (x, y) et+s + e , etc. (b) 2 ∂t ∂t (x + y) (x + y)2 ∂t 3. It is important that you can do these problems, because in economic applications, functions are frequently not completely speciﬁed. (a) zr = Fu ur + Fv vr + Fw wr = Fu 2r + Fv · 0 + Fw (1/r) = 2rFu + (1/r)Fw . Check that you get the same answers as in the text. 7. Use the formulas in (12.2.1) and see the text. Only the notation is different. 8. (a) Let u = ln v, where v = x 3 + y 3 + z3 − 3xyz. 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 1 1 1 3v ∂u +y +z = (3x 3 − 3xyz) + (3y 3 − 3xyz) + (3z3 − 3xyz) = =3 ∂x ∂y ∂z v v v v

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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, F22 = 0. In particular, y = −F1 /F2 = −(1 + 3y)/(−1 + 3x). Moreover, using equation (12.3.3), 6(1 + 3y) 6(1 + 3y)(−1 + 3x) 1 )= . F11 (F2 )2 − 2F12 F1 F2 + F22 (F1 )2 = y =− 3 3 (−1 + 3x) (−1 + 3x)2 (F2 ) (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 − 25y 9 y (5y 4 )3

3. (a) With F (x, y) = 2x 2 + xy + y 2 , y = −F1 /F2 = −(4x + y)/(x + 2y) = −4 at (2, 0). Moreover, y = −(28x 2 + 14y 2 + 14xy)/(x + 2y)3 = −14 at (2, 0). The tangent has the equation y = −4x + 8. (b) y = 0 requires y = −4x. Inserting this into the original equation gives the two points. 4. If we deﬁne F (x, y) = 3x 2 − 3xy 2 + y 3 + 3y 2 , the given equation is F (x, y) = 4. Now, F1 (x, y) = 6x−3y 2 and F2 (x, y) = −6xy+3y 2 +6y, so according to (12.3.1), y = −(6x−3y 2 )/(−6xy+3y 2 +6y).

12.4
1. (a) and (b) are easy. (c) Deﬁning F (x, y, z) = exyz − 3xyz, the given equation is F (x, y, z) = 0. Now, Fx (x, y, z) = yzexyz − 3yz, Fz (x, y, z) = xyexyz − 3xy, so (12.4.1) gives, zx = −Fx /Fz = −(yzexyz − 3yz)/(xyexyz − 3xy) = −yz(exyz − 3)/xy(exyz − 3) = −z/x. (Actually, the equation exyz = 3xyz has two constant solutions. From xyz = c we ﬁnd zx much easier.) √ 3. (a) Equation (∗) is here P /2 L∗ = w. Solve for L∗ . (b) The ﬁrst-order condition is now Pf (L∗ ) − CL (L∗ , w) = 0 (∗)

Differentiate (∗) partially w.r.t. P keeping in mind that L∗ depends on P . To ﬁnd the partial derivative of Pf (L∗ ) w.r.t. P use the product rule to get 1 · f (L∗ ) + Pf (L∗ )(∂L∗ /∂P ). The partial derivative of CL (L∗ , w) w.r.t. P is CLL (L∗ , w)(∂L∗ /∂P ). So, all in all, f (L∗ ) + Pf (L∗ )(∂L∗ /∂P ) − CLL (L∗ , w)(∂L∗ /∂P ) = 0. Then solve for ∂L∗ /∂P . Differentiating (∗) w.r.t. w gives, Pf (L∗ )(∂L∗ /∂w) − CLL (L∗ , w)(∂L∗ /∂w) − CLw (L∗ , w) = 0. Then solve for ∂L∗ /∂w. 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. (b) Taking the logarithm of both sides, we get (1 + c ln y) ln y = ln A + α ln K + β ln L. Differentiation 1 ∂y α c ∂y ln y + (1 + c ln y) = . Solving for ∂y/∂K yields the given answer. with respect to K gives y ∂K y ∂K K ∂y/∂K is found in the same way.
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12.5
3. With F (K, L) = AK a Lb , FK = aF /K, FL = bF /L, FKK = a(a − 1)F /K 2 , FKL = abF /KL, and FLL = b(b−1)F /L2 . Also, −FK FL (KFK +LFL ) = −(aF /K)(bF /L)(a+b)F = −ab(a+b)F 3 /KL. Moreover, KL (FL )2 FKK − 2FK F2 FKL + (FK )2 FLL = −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 ) + y2xy = 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 ] = y 3 [(x/y)3 + x/y] x 2 f11 + 2xyf12 + y 2 f22 = x 2 (6x) + 2xy(2y) + y 2 (2x) = 6x 3 + 4xy 2 + 2xy 2 = 3 · 2f (x, y) ∂f xy 3 − x 3 y + x 3 y − xy 3 ∂f +y = = 0 = 0 · f, ∂x ∂y (x 2 + y 2 )2

4. Using the results in Example 11.2.1(b), x so f is homogeneous of degree 0.

8. From (∗), with k = 1, we get f11 = (−y/x)f12 and f22 = (−x/y)f21 . With f12 = f21 we get f11 f22 − (f12 )2 = (−y/x)f12 (−x/y)f12 − (f12 )2 = 0.

12.7
1. (a) and (f) are easy. In (b) you can use Euler’s theorem, as in (e) below. √ √ √ √ √ √ √ 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, by Euler’s theorem, H is not homogeneous of any degree. 2. (a) f (tx1 , tx2 , tx3 ) = 1 (tx1 tx2 tx3 )2 1 1 = + + 4 + (tx )4 + (tx )4 (tx1 ) tx1 tx2 tx3 3 2 1 1 1 1 = tf (x1 , x2 , x3 ), so f is homogeneous of degree 1. + + x1 x2 x3 t

t 6 (x1 x2 x3 )2 4 4 4 t 4 (x1 + x2 + x3 ) −μ/ (b) x(tv1 , tv2 , . . . , tvn ) = A δ1 (tv1 )− + δ2 (tv2 )− + · · · + δn (tvn )− = − (δ v − + δ v − + · · · + δ v − ) −μ/ = (t − )−μ/ A δ v − + δ v − + · · · + δ v − −μ/ = A t 1 1 2 2 n n 1 1 2 2 n n μ A δ v − + δ v − + · · · + δ v − −μ/ = t μ x(x , x , x ), so x is homogeneous of degree μ. t n n 1 2 3 2 2 1 1

12.8
1. In both (a) and (b) we use the approximation f (x, y) ≈ f (0, 0) + f1 (0, 0)x + f2 (0, 0)y. √ 1 , so f1 (0, 0) = (a) For f (x, y) = 1 + x + y, f (0, 0) = 1, and f1 (x, y) = f2 (x, y) = √ 2 1+x+y 1 1 f2 (0, 0) = 1/2, and the linear approximation to f (x, y) about (0, 0) is f (x, y) ≈ 1 + 2 x + 2 y. x e . Here, f (0, 0) = 0, (b) For f (x, y) = ex ln(1 + y), f1 (x, y) = ex ln(1 + y) and f2 (x, y) = 1+y f1 (0, 0) = e0 ln 1 = 0 and f2 (0, 0) = 1. That yields f (x, y) = ex ln(1 + y) ≈ 0 + 0 · x + 1 · y = y.
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40 3.

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TOOLS FOR COMPARATIVE STATICS
1/(1−β)−1 β/(1−β)

1 ∂g ∗ = (1 + μ)(1 + ε)α ∂μ 1−β ∂g ∗ 1 (1 + μ)(1 + ε)α = 1−β ∂ε

(1 + ε)α =

1 (1 + μ)(1 + ε)α 1−β

β/(1−β)

(1 + ε)α ,

(1 + μ)α(1 + ε)α−1 . See the text.

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.

12.9
2. We can either use the deﬁnition of the differential, (12.9.1), or the rules for differentials as we do here. 2 2 2 2 (a) dz = d(x 3 ) + d(y 3 ) = 3x 2 dx + 3y 2 dy (b) dz = (dx)ey + x(dey ). Here dey = ey dy 2 = 2 2 2 2 ey 2ydy, so dz = ey dx + 2xyey dy = ey (dx + 2xydy). 2xdx − 2ydy 1 . (c) dz = d ln u, where u = x 2 − y 2 . Then dz = du = u x2 − y2 √ √ √ 5. d(U eU ) = d(x y), and so eU dU + U eU dU = dx y + (x/2 y)dy. Solving for dU yields the answer.

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 : ∂y2 ∂y1 2 ∂y2 2 2 ∂y1 − 9y2 = 0 (ii) 3x1 + 6y1 − =0 (i) 3 − ∂x1 ∂x1 ∂x1 ∂x1 Solve for the partials and see the text. (An alternative, in particular if one needs all the partials, is to use total differentiation:
2 2 2 (i) 3 dx1 + 2x2 dx2 − dy1 − 9y2 dy2 = 0, (ii) 3x1 dx1 − 2dx2 + 6y1 dy1 − dy2 = 0

Letting dx2 = 0 and solving for dy1 and dy2 leads to dy1 = Adx1 and dy2 = Bdx1 , where A = ∂y1 /∂x1 and B = ∂y2 /∂x1 .) 4. Differentiation 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 on standard form, we get −S (Y )YM + I (r)rM = 0 aYM + L (r)rM = 1 Cramer’s rule (or use ordinary elimination) gives 0 I (r) 1 L (r) I (r) = = S (Y )L (r) + aI (r) −S (Y ) I (r) a L (r)

YM

and

rM =

S (Y ) S (Y )L (r) + aI (r)

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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41

5. Differentiation w.r.t. x yields, y + ux v + uvx = 0, u + xux + yvx = 0. Solving this system for ux and vx we get u2 − y 2 2xy − 1 u2 − y 2 xy − uv ux = = , vx = = yv − xu 2yv yv − xu 2yv where we substituted xu = −yv and uv = 1 − xy. Differentiating ux w.r.t. x ﬁnally yields uxx = ∂ 2uux 2yv − (u2 − y 2 )2yvx (u2 − y 2 )(4uv − 1) ∂ 2u ux = = = ∂x 2 ∂x 4y 2 v 2 4y 2 v 3

9. (a) Differentiation yields 2uvdu + u2 dv − du = 3x 2 dx + 6y 2 dy, and eux (udx + xdu) = vdy + ydv. At P these equations become 3du + 4dv = 6dy and dv = 2dx − dy. Hence du = 2dy − (4/3)dv = −(8/3) dx + (10/3) dy. So ∂u/∂y = 10/3, ∂v/∂x = 2. (b) See the text. 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, du dX = g(u) + g (u)(ϕ (N) − u), = g(u) + Ng (u) dN dN Differentiating g(u) + g (u)(ϕ (N) − u) w.r.t. N gives d 2X du du du ) + g (u) (ϕ (N) − u) + g (u)(ϕ (N) − = g (u) dN dN dN dN 2 1 2 = g ϕ(N)/N ϕ (N) − ϕ(N)/N + g ϕ(N)/N ϕ (N) N 5. (a) Take the natural logarithm, ln E = ln A − a ln p + b ln m, and then differentiate. (b) See the text. u= ϕ(N ) N

11. 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, 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, yields (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 , = F (L) ∂w 1 ∂p = , ∂B F (L) ∂L F (L) − LF (L) , = ∂w pF (L)F (L) ∂L F (L) =− ∂B pF (L)F (L)

(b) We know that p > 0, F (L) > 0, and F (L) < 0. Also, F (L) = (wL + B)/p > 0. Hence, it is clear that ∂p/∂w > 0, ∂p/∂B > 0, and ∂L/∂B > 0.
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To ﬁnd the sign of ∂L/∂w, we need 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.

13 Multivariable Optimization
13.1
3. FK = −2(K − 3) − (L − 6) and FL = −4(L − 6) − (K − 3), so the ﬁrst-order conditions yield −2(K − 3) − (L − 6) = 0.65, −4(L − 6) − (K − 3) = 1.2. The solution is (K, L) = (2.8, 5.75). 4. (b) First-order conditions: Px = −2x + 22 = 0, Py = −2y + 18 = 0. It follows that x = 11 and y = 9.

13.2
3. Note that U = (108−3y−4z)yz. Then ∂U/∂y = 108z−6yz−4z2 = 0 and ∂U/∂z = 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 z. See Theorem 13.6.3.) 7. Solve the constraint for z: z = 4x + 2y − 5. Then minimize P (x, y) = x 2 + y 2 + (4x + 2y − 5)2 w.r.t. x and y. The ﬁrst-order conditions are: P1 = 34x + 16y − 40 = 0, P2 = 16x + 10y − 20 = 0, with solution x = 20/21, y = 10/21. Since P11 = 34, P12 = 16, and P22 = 10, we see that the second-order conditions for minimum are satisﬁed.

13.3
3. (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 and (c). (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 . ∗ At (t ∗ , x ∗ ), Vtx = 0, Vxx < 0, and Vtt = h(x ∗ )[g (t ∗ ) − 2rg (t ∗ ) + r 2 g(t ∗ )]e−rt . Because g (t ∗ ) = ∗ rg(t ∗ ), we obtain Vtt = h(x ∗ )[g (t ∗ ) − r 2 g(t ∗ )]e−rt < 0. Thus (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. The two conditions in (c) are satisﬁed. (Note that in part (c) of this problem in the text, the second condition should be h (x ∗ ) < 0.) Obviously, h (x ∗ ) = −(1 + x ∗ )−2 < 0. √ √ 1 ∗ Moreover, g (t ∗ ) = √ e t ( t ∗ − 1) = r 2 (1 − 2r)e1/2r < r 2 e1/2r , which is true when r > 0. ∗ t∗ 4t 5. (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.) 2x −x 4 2y − 2x 2 y 2 , and f22 (x, y) = . , f12 (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. Since f11 f22 − (f12 )2 = 0 at all the stationary points, the second-derivative test tells us nothing about the stationary points. See the text.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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43

13.4
2. (a) See the text. (b) The new proﬁt function is π = −bp2 − dp 2 + (a + βb)p + (c + βd)p − α − β(a + c) ˆ a + c + β(b + d) . and the price which maximizes proﬁts is easily seen to be p = ˆ 2(b + d) c2 a+c c a2 a ∗ , and π(p ∗ , q ∗ ) = + − α. Moreover, p = ˆ with ,q = (c) If β = 0, then p ∗ = 2d 4b 4d 2(b + d) 2b (ad − bc)2 (a + c)2 ≥ 0. Note that the difference is 0 when − α, and π(p ∗ , q ∗ ) − π(p) = ˆ ˆ π(p) = ˆ ˆ 4bd(b + d) 4(b + d) ∗ = q ∗ , so the ﬁrm wants to charge the same price in each market anyway. ad = bc, in which case p 3. Imposing a tax of t per unit sold in market area 1 means that the new proﬁt function is π(Q1 , Q2 ) = π(Q1 , Q2 ) − tQ1 . The optimal choice of production in market area 1 is then Q1 = (a1 − α − t)/2b1 (see the text), and the tax revenue is T (t) = t (a1 − α − t)/2b1 = [t (a1 − α) − t 2 ]/2b1 . This quadratic 1 polynomial has maximum when T (t) = 0, so t = 2 (a1 − α). 4. (a) Let (x0 , y0 ) = (0, 11.29), (x1 , y1 ) = (1, 11.40), (x2 , y2 ) = (2, 11.49), and (x3 , y3 ) = (3, 11.61), so that x0 corresponds to 1970, etc. (The numbers yt are approximate, as are most subsequent results.) 1 1 We ﬁnd that μx = 4 (0 + 1 + 2 + 3) = 1.5, μy = 4 (11.29 + 11.40 + 11.49 + 11.61) = 11.45, and 1 2 + (1 − 1.5)2 + (2 − 1.5)2 + (3 − 1.5)2 ] = 1.25. Moreover, we ﬁnd σ σxx = 4 [(0 − 1.5) xy = 0.13125, ˆ and so a = σxy /σxx = 0.105 and b = μy − aμx ≈ 11.45 − 0.105 · 1.5 = 11.29. ˆ ˆ (b) With z0 = ln 274, z1 = ln 307, z2 = ln 436, and z3 = ln 524, we have (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 σxx = 1.25. 1 Moreover, μz = 4 (5.61 + 5.73 + 6.08 + 6.26) = 5.92 and σxz ≈ 0.2875. Hence c = σxz /σxx = 0.23, ˆ ˆ d = μz − cμx = 5.92 − 0.23 · 1.5 = 5.575. ˆ (c) With ln (GNP) = 0.105x + 11.25, GNP = e11.25 e0.105x = 80017e0.105x . Likewise, FA = 256e0.23x . The requirement that FA = 0.01 GNP implies that e0.23x−0.105x = 80017/25600, and so 0.125x = ln(80017/25600). Thus x = ln(80017/25600)/0.125 = 9.12. Since x = 0 corresponds to 1970, the goal would have been reached in 1979. 5. (a) See the text. (b) Firm A’s proﬁt is now πA (p) = px −5−x = p(29−5p+4q)−5−29+5p−4q = 34p − 5p 2 + 4pq − 4q − 34, with q ﬁxed. This quadratic polynomial is maximized at p = pA (q) = 1 2 5 (2q + 17). Likewise, ﬁrm B’s proﬁt is now πB (q) = qy − 3 − 2y = 28q − 6q + 4pq − 8p − 35, with p ﬁxed. This quadratic polynomial is maximized at q = qB (p) = 1 (p + 7). For (c) and (d) see 3 the text. y
4 (3, 3)

4
Figure SM13.5.2

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
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(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. [0, (II) x = 4, y ∈ [0, 4]. Then f (4, y) = y 3 − 36y + 91. The function g(y) = y 3 − 36y + 91, y ∈√ 4] √ 2 − 36 = 0 at y = 12. Possible extreme points along (II) are therefore (4, 0), (4, 12), has g (y) = 3y and (4, 4). (III) y = 4, √ ∈ [0, 4]. Then f (x, 4) = x 3 − 36x + 91, and as in (II) we see that possible extreme points x 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, and f (3, 3) = 0, f (0, 0) = 27, f (4, 0) = f (0, 4) = 91, f (4, 12) = f ( 12, 4) = 91 − 24 12 ≈ 7.7, f (0, 0) = 27. The conclusion follows. (b) The constraint set S = (x, y) : x 2 + y 2 ≤ 1 are all 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 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 ignore 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 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) √ 9 1 1 1 1 Now, f ( 2 , 0) = − 4 , f (− 2 , ± 2 3) = 4 , f (1, 0) = 0, and f (−1, 0) = 2. The conclusion 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 of each of the ﬁve edges in Fig. A13.5.3. (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.
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(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 points found on the edges labeled (I) and (V), we conclude that the maximum value of f is 27/8, which is achieved at (3/4, 0). 5. (a) f1 (x, y) = e−x (1 − x)(y − 4)y, f2 (x, y) = 2xe−x (y − 2). It follows that the stationary points are (1, 2), (0, 0) and (0, 4). (Make sure you understand why!) 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 0 B 0 −4 4 C 2e−1 0 0 AC − B 2 8e−2 > 0 −16 < 0 −16 < 0 Type of point Loc. min. point Saddle point Saddle point

(b) We see that f (1, y) = e−1 (y 2 − 4y) → ∞ as y → ∞. This shows that f has no global maximum point. Since f (−1, y) = −e(y 2 − 4y) → −∞ as y → ∞, f has no global minimum point either. y (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 (III) 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 (II) (IV) (1, 2) 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 x point of f in the interior of S is (1, 2). The function value at this point (I) 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), which has its least value for y = 2 and its greatest value for y = 0 and for y = 4. (Note that y ∈ [0, 4] for all points (x, y) on the line segment (II).) The values are f (5, 2) = −20e−5 ≈ −0.1348 and f (5, 0) = f (5, 4) = 0. (iii) On 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 =− = = 0 when x = 1. (d) y = − 1 −x (y − 2) 2xe 2x(y − 2) f2 (x, y)

13.6
1. (a) fx (x, y, z) = 2−2x = 0, fy (x, y, z) = 10−2y = 0, fx (x, y, z) = −2z = 0. The conclusion follows.
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(b) fx (x, y, z) = −2x − 2y − 2z = 0, fy (x, y, z) = −4y − 2x = 0, fz (x, y, z) = −6z − 2x = 0. 1 From the two last equations we get y = − 2 x and z = − 1 x. Inserting this into the ﬁrst equation we get 3 −2x + x + 2 x = 0, and thus x = 0, and then y = z = 0. 3 4. To calculate fx is routine. To differentiate f w.r.t. y and z we use (9.3.7) and (9.3.6). The derivative of z t2 z t2 y2 y e dt w.r.t. y, keeping z constant is according to (9.3.7), −e . The derivative of y e dt w.r.t. z, keeping y constant is according to (9.3.6), 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 point. See the text.
2 2 2

13.7
1 2. (a) First-order conditions: πK = 2 pK −1/3 − r = 0, πL = 2 pL−1/2 − w = 0, πT = 1 pT −2/3 − q = 0. 3 3 −1/3 = 3r/2p, L−1/2 = 2w/p, and T −2/3 = 3q/p. Raising each side of K −1/3 = 3r/2p to the Thus, K 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) See the text.

4. 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 variables to the right-hand side, suppressing the fact that the partials are evaluated at (K ∗ , L∗ ): pFKK dK ∗ + pFKL dL∗ = dr − FK dp pFLK dK ∗ + pFLL dL∗ = dw − FL dp We use Cramer’s rule to express the differentials dK ∗ and dL∗ in terms of dp, dr, and dw. Putting = FKK FLL − FKL FLK = FKK FLL − (FKL )2 , we get dK ∗ = 1 p2 dr − FK dp dw − FL dp F −FKL −FK FLL + FL FKL pFKL dw dp + LL dr + = pFLL p p p

In the same way dL∗ = 1 p2 pFKK pFLK −FLK F −FL FKK + FK FLK dr − FK dp dp + dr + KK dw = dw − FL dp p p p

We can now read off the required partials. (Note that there is an error in the expression for ∂K ∗ /∂p in the answer to this problem in the text. In the numerator, replace FKK by FLL . (c) See the text. (Recall that FLL < 0 follows from (∗∗) in Example 13.3.3.)
∗ ∗ ∗ ∗ 5. (a) (i) R1 (x1 , x2 ) + s = C1 (x1 , x2 ) (marginal revenue plus subsidy equal marginal cost) ∗ ∗ ∗ ∗ (ii) R2 (x1 , x2 ) = C2 (x1 , x2 ) + t = 0 (marginal revenue equals marginal cost plus tax). For (b) see the text. (c) Taking the total differentials of (i) and (ii) yields ∗ ∗ (R11 − C11 )dx1 + (R12 − C12 )dx2 = −ds, ∗ ∗ (R21 − C21 )dx1 + (R22 − C22 )dx2 = dt

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∗ ∗ Solving for dx1 and dx2 yields, after rearranging, ∗ dx1 =

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47

−(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
∗ ∗ ∗ ∗ R − C11 −R22 + C22 ∂x1 −R12 + C12 ∂x2 R − C21 ∂x2 ∂x1 = 11 0, = > 0, = 21 < 0, ∂t D ∂s D ∂t D ∂s D

where the signs follow from the assumptions in the problem and the fact that D > 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 ˆ conditions for maximal proﬁt: π1 = P1 −0.2Q1 −0.1Q2 = 0, π2 = 90−0.1Q1 −0.2Q2 = 0. If we need ˆ to have Q1 = 400, the ﬁrst-order conditions reduce to P1 − 80 − 0.1Q2 = 0 and 90 − 40 − 0.2Q2 = 0. It follows that P1 = 105. 3. (a) Stationary points where P1 (x, y) = −0.2x −0.2y +47 = 0 and P2 (x, y) = −0.2x −0.4y +48y = 0. It follows that x = 230 and y = 5. Moreover, P11 = −0.2 ≤ 0, P12 = −0.2, and P22 = −0.4 ≤ 0. Since also P11 P22 − (P12 )2 = 0.04 ≥ 0, (230, 5) maximizes proﬁts. (b) With x + y = 200, and so y = 200 − x, the new proﬁt function is π(x) = f (x, 200 − x) = −0.1x 2 + 39x + 1000. This function ˆ is easily seen to have maximum at x = 195. Then y = 200 − 195 = 5. 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) is only satisﬁed when x = 0. If x = 0, then (ii) is satisﬁed and (i) is only satisﬁed when y = 0. Thus, in addition to (0, 0), any other stationary point must satisfy 8x 2 − 5xy + 1 = 0 and 2y 2 − 5xy + 1 = 0. Subtracting the second from the ﬁrst yields 8x 2 = 2y 2 , 0, or y = ±2x. Inserting y = −2x into 8x 2 − 5xy + 1 = 0 yields 18x 2 + 1 = √ which has no solutions. 2 − 5xy + 1 = 0 yields x 2 = 1 , and so x = ± 1 2. We conclude that the Inserting y = 2x into 8x 2 √ √ √ √2 1 1 stationary points are: (0, 0) and ( 2 2, 2), (− 2 2, − 2) 5. (a) The ﬁrst-order conditions for (K ∗ , L∗ , T ∗ ) to maximize π are: πK = pa/K ∗ − r = 0, πL = pb/L∗ − w = 0, πT = pc/T ∗ − q = 0. Hence, K ∗ = ap/r, L∗ = bp/w, T ∗ = cp/q. (b) π ∗ = pa ln(ap) − pa ln r + pb ln(bp/w) + pc ln(cp/q) − ap − bp − cp = −pa ln r plus terms that do not depend on r. So ∂π ∗ /∂r = −pa/r = −K ∗ . (c) See the text.
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7. (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 where 2x − y − 3x 2 = 0 and −2y − x = 0. The last equation yields y = −x/2, which inserted into the second equation gives x( 5 − x) = 0. It follows that there are two stationary points, 6 (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 Lok. max. point

(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. In particular, one must 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, this is a (global) maximum point for f in S. fmax = 125/432. 8. (a) We ﬁnd f1 (x, y) = x − 1 + ay, f2 (x, y) = a(x − 1) − y 2 + 2a 2 y. Stationary points require that x −1 = −ay and a(x −1) = y 2 −2a 2 y. These two equations yield −a 2 y = y 2 −2a 2 y, and so a 2 y = y 2 . Hence 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.) (b) Note that the partial derivative of f w.r.t. a, keeping x and y constant is ∂f/∂a = y(x − 1) + 2ay 2 . Evaluated at x = 1 − a 3 , y = a 2 , this partial derivative is also a 5 , thus conﬁrming the envelope theorem. (c) See the text. ˆ ˆ 9. For (a)–(c), see the text. (d) ∂ 2 π/∂p2 = −2b, ∂ 2 π/∂q 2 = −2γ , and ∂ 2 π/∂p∂y = β + c. The direct ˆ partials of order 2 are negative and = (∂ 2 π/∂p2 )(∂ 2 π/∂q 2 ) − (∂ 2 π/∂p∂q)2 = 4γ b − (β + c)2 , so ˆ ˆ ˆ the conclusion follows.

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 implies x = y. So the solution is (x, y) = (50, 50) with λ = 250. 5. The budget constraint is 2x + 4y = 1000, so with L(x, y) = 100xy + x + 2y − λ(2x + 4y − 1000), the ﬁrst-order conditions are L1 = 100y + 1 − 2λ = 0 and L2 = 100x + 2 − 4λ = 0. From these equations, by eliminating λ, we get x = 2y, which inserted into the constraint gives 2x + 2x = 1000. So x = 250 and y = 125. 7. The problem is: max −0.1x 2 − 0.2xy − 0.2y 2 + 47x + 48y − 600 subject to x + y = 200. With L(x, y) = −0.1x 2 − 0.2xy − 0.2y 2 + 47x + 48y − 600 − λ(x + y − 200), the ﬁrst-order conditions
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are L1 = −0.2x − 0.2y + 47 − λ = 0 and L2 = −0.2x − 0.4y + 48 − λ = 0. Eliminating λ yields y = 5, and then the budget constraint gives x = 195. 9. (a) With L(x, y) = 100 − e−x − e−y − λ(px + qy − m), Lx = Ly = 0 when 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, then solving for ln λ, yields 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
1 1 5 1 1 1 1 3. (a) Solving x +2y = a for y yields y = 2 a − 2 x, and then x 2 +y 2 = x 2 +( 2 a − 2 x)2 = 4 x 2 − 2 ax + 4 a 2 . This quadratic function certainly has a minimum at x = a/5. (b) L(x, y) = x 2 + y 2 − λ(x + 2y − a). The necessary conditions are L1 = 2x − λ = 0, L2 = 2y − 2λ = 0, implying that 2x = y. From the constraint, x = a/5 and then y = 2a/5, λ = 2a/5. The value function f ∗ (a) = (a/5)2 + (2a/5)2 = a 2 /5, so df ∗ (a)/da = 2a/5, which is also the value of the Lagrangian multiplier. Equation (2) is conﬁrmed. (c) See the text. √ x 4. (a) With L(x, y) = √+ y − λ(x + 4y − 100), the ﬁrst-order conditions for (x ∗ , y ∗ ) to solve the problem 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 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 The increase in the maximum utility as 100 is increased to 101, is thus U − U ∗ = 0.25 = λ. (In general, the increase in utility is approximately equal to the value of the Lagrange multiplier.) √ (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) px ∗ = pa +α/λ and qy ∗ = qb +β/λ give 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. (If we think of a and b as kind of existence minimum of the two goods, the assumption pa + qb < m means that the consumer can afford to buy (a, b).) (b) With the U ∗ given in the answer, since α + β = 1, 1 ∂U ∗ α β = = λ > 0. Moreover, = + m − (pa + qb) ∂m m − (pa + qb) m − (pa + qb) ∂U ∗ −βa −a α α ∂U ∗ ∗ −αa α = − = −aλ − , and − x = = − + m − (pa + qb) p p ∂m ∂p m − (pa + qb) p m − (pa + qb) α α ∂U ∗ ∂U ∗ ∗ −λ(a + ) = −aλ − , so =− x . The last equality is shown in the same way. λp p ∂p ∂m
1 6. f (x, T ) = x 0 [−t 3 + (αT 2 + T − 1)t 2 + (T − αT 3 )t] dt = x 0 [− 4 t 4 + (αT 2 + T − 1) 1 t 3 + 3 1 1 1 1 2 (T − αT 3 ) 2 t ] = − 6 αxT 5 + 12 xT 4 + 6 xT 3 . In the same way, g(x, T ) = 1 xT 3 . The solution of 6 (∗) is x = 384α 3 M, T = 1/4α, with f ∗ (M) = M + M/16α. (The easiest way to solve the problem 1 is to note that because 1 xT 3 = M, the problem reduces to that of maximizing M + 2 MT − αMT 2 6 for T ≥ 0. For T = 0, the expression is equal to M, and its maximum is attained for T = 1/4α.) T T

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Alternatively, eliminating the Lagrange multiplier from the ﬁrst-order conditions f1 = λg1 and f2 = λg2 , we eventually obtain T = 1/4α. The Lagrange multiplier is λ = 1 + 1/16α. Clearly, ∂f ∗ (M)/∂M = λ, which conﬁrms (2).

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. Since (0, 0) does not satisfy the constraint, from these equations we get x 2 = y 2 . Inserted into the constraint this yields x 2 = 4, and so x = ±2, and the solution candidates are: (2, 2), (2, −2), (−2, 2), (−2, −2). Here f (2, 2) = f (−2, −2) = 12 and f (−2, 2) = f (2, −2) = −12. So (2, 2) and (−2, −2) solves the maximization problem, and (−2, 2) and (2, −2) solves the minimization problem, because the extreme value theorem ensures that solutions exist. (f is continuous and the constraint curve is a closed bounded set (a circle).) (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. From these equations we get 2λx + 3λy = 3λx + 6λy, or λ(3y + x) = 0. Here λ = 0 is impossible, so x = −3y. Inserted into the constraint we have (3, −1) and (−3, 1) as the only possible solutions of the maximization and minimization problems, respectively. The extreme value theorem ensures that solutions exist. (The objective function is continuous and the constraint curve is a closed bounded set (an ellipse, see (5.5.5)). 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) yields λ = 1 − 1/x (why can we be sure that x = 0?), and equation (ii) shows that y = 0 or λ = 1/4. If y = 0, then x 2 = 16−4y 2 = 16, so x = ±4, and 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, the second is the maximum point (while (iii) and (iv) are the minimum points). (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. √ √ √ 4 1 1 1 1 From (iii), y = 1 − 2 x 2 = 1 − 2 2. Thus, (x2 , y2 ) = ( 2, 1 − 2 2) and (x3 , y3 ) = (− 4 2, 1 − 2 2) √ are candidates. Now f (x1 , y1 ) = f (0, 1) = ln 2 + 1 ≈ 1.69, f (x2 , y2 ) = f (x3 , y3 ) = ln (2 + 2 ) + √ √ √ 3 1 (1 − 2 2 )2 = ln (2 + 2 ) + 2 − 2 ≈ 1.31. Hence, the minimum points for f (x, y) (subject to x 2 + 2y = 2) 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. 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 solution is (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 the optimum is found. (b) According to (14.2.3) the approximate change is λ · 1 = 1.
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14.4
4. The minimum is 1 at (x, y) = (−1, 0). Actually, this problem is quite tricky. The Lagrangian can be written in the form L = (x + 2)2 + (1 − λ)y 2 + λx(x + 1)2 . The only stationary point which satisﬁes the constraint is (0, 0), with λ = −4, and with f (0, 0) = 4. (In fact, L2 = 0 only if λ = 1 or y = 0. For λ = 1, L1 = 3(x + 1)2 + 2 > 0 for all x. For y = 0, the constraint gives x = 0 or x = −1. But x = −1 gives L1 = 2, so x = 0 is necessary for a stationary point.) Yet at (−1, 0) both g1 (−1, 0) and g2 (−1, 0) are 0, and the Lagrange multiplier method fails. The given problem is to minimize (the square of) the distance from (−2, 0) to a point on the graph of g(x, y) = 0. But the graph consists of the isolated point (−1, 0) and a smooth curve, as illustrated in Fig. SM14.4.4. 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

Figure SM14.4.4

14.5
4. U11 (x, y) = a(a − 1)x a−2 ≤ 0, U22 (x, y) = a(a − 1)y a−2 ≤ 0, and U12 (x, y) = 0, so U is concave. 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. Hence, ax a−1 = λp and ay a−1 − λq. Eliminating λ we get (x/y)a−1 = p/q, and so x = y(p/q)1/(a−1) . Inserted into the budget constraint we get px + qy = py(p/q)1/(a−1) + qy = ypa/(a−1) q −1/(a−1) + qy = yq −1/(a−1) [p a/(a−1) + q a/(a−1) ] = m, and so y = mq 1/(a−1) /[pa/(a−1) + q a/(a−1) ]. A similar expression is obtained for x.

14.6
1. (a) Lx = 2x − λ = 0, Ly = 2y − λ = 0, Lz = 2z − λ = 0. It follows that x = y = z, etc. See the text. 3. (a) The Lagrangian is L = α ln x+β ln y+(1−α−β) ln(L− )−λ(px+qy−w −m), which is stationary when: (i) Lx = α/x ∗ − λp = 0; (ii) Ly = β/y ∗ − λq = 0; (iii) L = −(1 − α − β)/(L − ∗ ) + λw = 0. From (i) and (ii), qy ∗ = (β/α)px ∗ , while (i) and (iii) yield ∗ w = wL − [(1 − α − β)/α]px ∗ . Insertion into the budget constraint and solving for x ∗ yields the answer in the text. The corresponding values for y ∗ and ∗ follows. The assumption m ≤ [(1 − α − β)/α]wL ensures that ∗ ≥ 0. (b) See the text. 6. 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 (i) and
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(ii), λ(x − 2y) = 0. If λ = 0, then (ii) and (iii) yield the contradiction μ = 1 and μ = 0. Therefore x = 2y instead. Substituting this value for x into the constraints gives 6y 2 + z2 = 1, 3y + z = 1. Thus z = 1 − 3y and 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 constraints determine geometrically the curve which is the intersection of an ellipsoid (see Fig. 11.4.2) and a plane. The continuous function x + y does attain a maximum and a minimum over this closed bounded set.) 7. (a) With a Cobb–Douglas utility function, Uk (x) = αk U (x)/xk , so from (6) (with j = 1), we have pk /p1 = Uk (x)/U1 (x) = αk x1 /α1 xk . Thus pk xk = (ak /a1 )p1 x1 . Inserted into the budget constraint, we have p1 x1 + (a2 /a1 )p1 x1 + · · · + (an /a1 )p1 x1 = m, which implies that p1 x1 = a1 m/(a1 + · · · + an ). Similarly, pk xk = ak m/(a1 + · · · + an ) for k = 1, . . . , n. a−1 a−1 (b) From (6) (with j = 1), we get xk /x1 = pk /p1 and so xk /x1 = (pk /p1 )−1/(1−a) or pk xk /p1 x1 = (pk /p1 )1−1/(1−a) = (pk /p1 )−a/(1−a) . The budget constraint gives

p1 x1 1 + (p2 /p1

)−a/(1−a)

+ · · · + (pn /p1

)−a/(1−a)

= m. So p1 x1 = n −a/(1−a) mp1

n

pi i=1 −a/(1−a)

.

Arguing similarly for each k, one has xk = mpk

−1/(1−a)

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 conditions are: 3 (i) L1 = 1 − 2λx = 0; (ii) L2 = 4 − 4λy = 0; (iii) L3 = 3 − 2 λz = 0. It follows that λ = 0, 3 and so x √ 1/2λ, y = 1/λ, z = 9/2λ. Inserting these values into the constraint yields λ2 = 9/b, so = √ 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. See the text. 4. 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): x = 0. From (ii), 2y(1 − 2λ) = 0, so y = 0 or λ = 1/2. If (A.1), 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, 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, P6 = (− 15/4, 0, 1/8) with λ = 1. By computing the values of the criterion function, f (0, 0, 1/2) = 1/2, f (0, 0, −1/2) = −1/2, √ √ √ √ f (0, 6/4, 1/4) = 5/8, f (0, − 6/4, 1/4) = 5/8, f ( 15/4, 0, 1/8) = 17/16, f (− 15/4, 0, 1/8) = 17/16, we obtain the conclusion in the text. (c) See the text. 5. The Lagrangian is L = rK + wL − λ(K 1/2 L1/4 − Q), so necessary conditions are: 1 1 (i) LK = r − 2 λK −1/2 L1/4 = 0, (ii) LL = w − 4 λK 1/2 L−3/4 = 0, (iii) K 1/2 L1/4 = Q. From
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1 (i) and (ii) we get by eliminating λ, L = 2 (rK/w). Inserting this into the constraint and solving for K yields the answer given in the text.

14.8
2. (a) Conditions (2)–(3): (i) 2x − 1 − 2λx = 0; (ii) 4y − 2λy = 0; (iii) λ ≥ 0, but λ = 0 if x 2 + y 2 < 1. (b) From (ii), y(2 − λ) = 0, so either (I) y = 0 or (II) λ = 2. (I) y = 0. If λ = 0, then from (i), x = 1/2 and (x, y) = (1/2, 0) is a candidate for optimum (since it satisﬁes all the Kuhn–Tucker conditions). If y = 0 and λ > 0, then from (iii) and x 2 + y 2 ≤ 1, x 2 + y 2 = 1. But then x = ±1, and (x, y) = (±1, 0) are candidates, with√ = 1/2 and 3/2, respectively. λ √ 2 = 3/4, so y = ± 3/2. So (−1/2, ± 3/2) are (II) λ = 2. Then from (i), x = −1/2 and (iii) gives y the two remaining candidates. For the conclusion, see the text. 3. (a) The Kuhn-Tucker conditions are: (i) −2(x − 1) − 2λx = 0; (ii) −2yey − 2λy = 0; 2 (iii) λ ≥ 0, with λ = 0 if 4x 2 + y 2 < a. From (i), x = (1 + λ)−1 , and (ii) reduces to y(ey + λ) = 0, 2 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 1 1 2(x − 1) = − 1 = √ − 1 > 0. value of λ and check that it is > 0. From equation (i) we get λ = −2x x a Conclusion: The only point that satisﬁes the Kuhn–Tucker conditions is (x, y) = (1, 0) if a ≥ 1 and √ 1 ( a, 0) if 0 < a < 1. The corresponding value of λ is 0 or √ − 1, respectively. In both cases it follows a from Theorem 14.8.1 that we have found the maximum point, because L is concave in (x, y), as we can −2 − 2λ 0 L11 L12 = . see by studying the Hessian 2 0 −ey (2 + 4y 2 ) − 2λ L21 L22 √ √ √ (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 (not 2, as it says in the answer section of the ﬁrst printing of the book). 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 242 in the book) (f ∗ ) (1+ ) = lim+ h→0 2

f ∗ (1 + h) − f ∗ (1) h

and

(f ∗ ) (1− ) = lim− h→0 f ∗ (1 + h) − f ∗ (1) h

exist and are equal. The right derivative is obviously 0, since f ∗ (a) = 1 for all a ≥ 1. To ﬁnd the left √ derivative we need to calculate limh→0− (2 1 + h − (1 + h) − 1)/ h. A straightforward application of l’Hôpital’s rule shows that this limit is also 0. Hence (f ∗ ) (1) exists and equals 0.

14.9
2. (a) The admissible set is the shaded region in Fig. A14.9.2 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 ﬁrst-order 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
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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 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. 3. (a) The feasible set is shown in Fig. A14.9.3 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 > 0, so since from (iii) λ1 ≥ 0, we must have λ1 > 0 and also 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 assume that x + y = 0. Then y = −x, and since x 2 + y 2 = 1, we get x = 2 2 (recall that we √ 1 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 we must have a ≤ −1 in this case. Equation (i) gives λ1 = 1 + λ2 /2x = 1 − a/4x = 1 − a/2 2. (II) Then consider the case x + y > 0. Then λ2 = 0, and we get 1 − 2λ1 x = 0 and a − 2λ1 y = 0, which gives x = 1/(2λ1 ) and y = a/(2λ1 ). Since (x, y) must lie on the circle, we then get 1 = x 2 +y 2 = √ 1 a 1 + a 2 /4λ2 , and therefore 2λ1 = 1 + a 2 . This gives x = √ and y = √ . Because 1 2 1+a 1 + a2 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. 4. (a) 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. Complementary slackness requires in addition, (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, 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 .
1 (b) We write the problem as max xey−x − 2ey subject to y ≤ 1 + 2 x, x ≥ 0, y ≥ 0. The Lagrangian is y−x − 2ey − λ(y − 1 − x/2), so the ﬁrst-order conditions (14.9.4) and (14.9.5) are: L = xe

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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); 1 (iii) λ ≥ 0 with λ = 0 if y < 1 + 2 x. 1 1 From (i) we have x ≥ 1 + 2 λex−y ≥ 1, so (iv) (x − 1)ey−x = 2 λ. Suppose λ > 0. Then (iii) 1 1 implies (v) y = 1 + 2 x > 0. From (ii) and (iv) we then have xey−x = 2e + λ = ey−x + 2 λ and so 1 1 λ = 2ey−x − 4e = 2e(e− 2 x − 2), by (v). But then λ > 0 implies that e− 2 x > 2, which contradicts x ≥ 0. 1 When λ = 0, (iv) gives x = 1. If y > 0, then (ii) yields ey−1 = 2e, and so y − 1 = ln(2e) > 2 x when x = 1. Thus feasibility requires that y = 0, so we see that (x, y) = (1, 0) is the only point satisfying all the conditions, with λ = 0. ∗ ∗ ∗ 5. A feasible triple (x1 , x2 , k ∗ ) solves the problem iff there exist numbers λ and μ such that (i) 1−2x1 −λ ≤ 0 ∗ ∗ ∗ ∗ + λ + μ ≤ 0 (= 0 if k ∗ > 0); (= 0 if x1 > 0); (ii) 3 − 2x2 − μ ≤ 0 (= 0 if x2 > 0); (iii) −2k ∗ ∗ (iv) λ ≥ 0 with λ = 0 if x1 < k ∗ ; and (v) μ ≥ 0 with μ = 0 if x2 < k ∗ . ∗ ∗ If k ∗ = 0, then feasibility requires x1 = 0 and x2 = 0, and so (i) and (iii) imply that λ ≥ 1 and ∗ > 0. Next, if μ = 0, then x ∗ ≥ 3/2 and λ = 2k ∗ > 0. So μ ≥ 3, which contradicts (iii). Thus, k 2 ∗ ∗ ∗ ∗ x1 = k ∗ = 1/4, contradicting x2 ≤ k ∗ . So μ > 0, which implies that x2 = k ∗ . Now, if x1 = 0 < k ∗ , 1 ∗ ∗ ∗ then λ = 0, which contradicts (i). So 0 < x1 = 2 (1 − λ). Next, if λ > 0, then x1 = k ∗ = x2 = 1 1 1 2 (1 − λ) = 2 (3 − μ) = 2 (λ + μ). But the last two equalities are only satisﬁed when λ = −1/3 and 1 1 ∗ ∗ μ = 5/3, which contradicts λ ≥ 0. So λ = 0 after all, 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. (The Lagrangian is concave in (x1 , x2 , k). See FMEA, Theorem 3.2.4.) 1 1 6. A minus sign has disappeared in the objective function which should be: −(x + 2 )2 − 2 y 2 . −x ≤ y ≤ 2/3, and so ex ≥ 3/2. (a) See Fig. A14.9.6 in the text. Note that for (x, y) to be admissible, e 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 are: 3 −x = 0; (ii) −y + λ − λ −x < y; (i) −(2x + 1) + λ1 e 1 2 = 0; (iii) λ1 ≥ 0, with λ1 = 0 if e (iv) λ2 ≥ 0, with λ2 = 0 if y < 2/3. From (i), λ1 = (2x + 1)ex ≥ 3/2, because of (a). From (ii), λ2 = λ1 − y ≥ 3/2 − 2/3 > 0, so y = 2/3 because of (iii). Solution: (x ∗ , y ∗ ) = (ln(3/2), 2/3), with λ1 = 3[ln(3/2) + 1/2], λ2 = 3 ln(3/2) + 5/6. The Lagrangian is concave so this is 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.

Review Problems for Chapter 14 3. (a) Interpretation of the ﬁrst-order condition p(x ∗ ) = C1 (x ∗ , y ∗ ) − x ∗ p (x ∗ ): How much is gained by selling one ton extra of the ﬁrst commodity? p(x ∗ ), because this is the price obtained for one ton. How much is lost? First, selling one ton extra of the ﬁrst commodity incurs the extra cost C(x ∗ + 1, y ∗ ) − C(x ∗ , y ∗ ), which is approximately C1 (x ∗ , y ∗ ). But since presumably p (x) < 0, producing one ton extra leads to a decrease in income which is approximately −x ∗ p (x ∗ ) (the number of tons sold times the decrease in the price. So what we loose by increasing production by 1 ton (C1 (x ∗ , y ∗ ) − x ∗ p (x ∗ )) is approximately what we gain (p(x ∗ )). The other ﬁrst-order condition, q(y ∗ ) = C2 (x ∗ , y ∗ ) − y ∗ q (y ∗ ) has a similiar interpretation. (b) See the text. If we assume that the restriction is x + y ≤ m, we have to add the condition λ ≥ 0, with λ = 0 if x + y < m. ˆ ˆ
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4. See the text. If one were to ﬁnd the partial derivatives of x and y w.r.t. p as well, it would be better to calculate differentials, which would yield the equations (i) y dp + p dy = 24 dw − w dx − x dw and (ii) U1 dp + p(U11 dx + U12 dy) = U2 dw + wU21 dx + wU22 dy, and then solve these equations for dx and dy in terms of dp and dw.
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 9 3

y = ± b − 4 . This gives two candidates: (x1 , y1 ) = (4/3, b − 4 ) and (x2 , y2 ) = (4/3, − b − 4 ). 9 9 9 √ 2 = 4b, i.e. x = ±2 b. This gives two further candidates: (x , y ) = (II) If y = 0, then from (iii), x 3 3 √ √ (2 b, 0) and (x4 , y4 ) = (−2 b, 0).√ objective function evaluated at the candidates are: f (x1 , y1 ) = The √ √ √ 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 √ 1 the minimum, we have to decide which of 4b − 4 b + 1 and b − 3 is the largest. The difference is √ √ √ 2 4b − 4 b + 1 − b − 1 = 3 b − 4 b + 4 = 3 b − 2 > 0 since b > 4 . Thus the minimum occurs 3 9 3 9 3 at (x1 , y1 ) and (x2 , y2 ). The constraint x 2 /4 + y 2 = b is the ellipse indicated in the ﬁgure. 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 passes through the maximum and minimum points. (b) See the text. y 4 (x1 , y1 )

2 (x4 , y4 ) -4

-2 -2

2

(x3 , y3 ) 4 6

x

(x2 , y2 )

-4

For Problem 14.R.5

√ √ 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)/ x + y + b ] = 0, √ √ (ii) L2 = 2y − 2 − λ[ x + y + b + (x + y)/ x + y + b ] = 0. From these equations it follows immediately that 2x − 2 = 2y − 2, so x = y. See the text.

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(b) Differentiation 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 = 2 ∂a ∂a ∂a = ∂ 1 12x + 2b ∂x 12x + 2b 6x + b =− =− =− 2 + 2bx 2 + 2bx)2 ∂a 2 + 2bx)3 ∂a 6x (6x (6x 4(3x 2 + bx)3

8. (a) The Lagrangian is L = xy − λ1 (x 2 + ry 2 − m) − λ2 (−x), and the necessary Kuhn–Tucker conditions for (x ∗ , y ∗ ) to solve the problem are: (i) L1 = y ∗ − 2λ1 x ∗ + λ2 = 0; (ii) L2 = x ∗ − 2rλ1 y ∗ = 0; (iii) λ1 ≥ 0, with λ1 = 0 if (x ∗ )2 + r(y ∗ )2 < m; (iv) λ2 ≥ 0, with λ2 = 0 if x ∗ > 1; (v) (x ∗ )2 + r(y ∗ )2 ≤ m; (vi) x ∗ ≥ 1 (b) From (ii) and (vi) we see that λ1 = 0 is impossible. Thus λ1 > 0, and from (iii) and (v), (vii) (x ∗ )2 + r(y ∗ )2 = m. (I): Assume λ2 = 0. Then from (i) and (ii), y ∗ = 2λ1 x ∗ and x ∗ = 2λ1 ry ∗ , so y ∗ = 4λ2 ry ∗ . If √1 y ∗ = 0, then (ii) implies x ∗ = 0, which is impossible. Hence, λ2 = 1/4r and thus λ1 = 1/2 r. Then 1 √ √ √ y ∗ = x ∗ / r, which inserted into (vii) and solved for x ∗ yields x ∗ = m/2 and then y ∗ = m/2r. Note √ √ √ that x ∗ ≥ 1 ⇐⇒ m/2 ≥ 1 ⇐⇒ m ≥ 2. Thus for m ≥ 2, x ∗ = m/2 and y ∗ = m/2r, with √ λ1 = 1/2 r and λ2 = 0 is a solution candidate. √ (II): Assume λ2 > 0. Then x ∗ = 1 and from (vii) we have r(y ∗ )2 = m − 1, so y ∗ = (m − 1)/r √ (y ∗ = − (m − 1)/r contradicts (ii)). Inserting these values for x ∗ and y ∗ into (i) and (ii) and solving √ √ for λ1 and λ2 yields 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.)

15 Matrix and Vector Algebra
15.1
2. Here is one method: Adding the 4 equations gives x1 + x2 + x3 + x4 = 1 (b1 + b2 + b3 + b4 ), after dividing 3 by 3. From this equation, subtracting each original equation in turn gives x1 = − 2 b1 + 1 (b2 + b3 + b4 ), 3 3 x2 = − 2 b2 + 1 (b1 + b3 + b4 ), x3 = − 2 b3 + 1 (b1 + b2 + b4 ), x4 = − 2 b4 + 1 (b1 + b2 + b3 ). 3 3 3 3 3 3 Systematic elimination of the variables starting by eliminating (say) x4 is an alternative. 6. Suggestion: Solve the ﬁrst equation for y. Insert this expression for y and x = 93.53 into the third equation. Solve it for s. Insert the results into the second equation and solve for c, etc.
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15.3
1. (a) 0 3 −2 1 −1 4 1 5 = 0 · (−1) + (−2) · 1 3 · (−1) + 1·1 0 · 4 + (−2) · 5 3·4+1·5 = −2 −2 −10 17

The rest goes in the same way. Note that in (d) AB is not deﬁned because the number of columns in A is not equal to the number of rows in B. 5. (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. (b) We know from part (a) that if BA and AB are deﬁned, then B must be a 2 × 2 matrix. So let B = x y x y 1 2 x + 2y 2x + 3y 1 2 x y . Then BA = = , and AB = = z w z w 2 3 z + 2w 2z + 3w 2 3 z w x + 2z y + 2w . Hence, BA = AB iff (i) x + 2y = x + 2z, (ii) 2x + 3y = y + 2w, 2x + 3z 2y + 3w (iii) z + 2w = 2x + 3z, and (iv) 2z + 3w = 2y + 3w. The ﬁrst and last of these four equations are true if and only if y = z, and if y = z, then the second and third are true if and only if x = w − y. Hence, the matrices B that commute with A are precisely the matrices of the form B= w−y y y w =w 1 0 0 1 +y −1 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

⎛

7. (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 A = 0 is 1 1 A= . −1 −1 (c) Multiplying (i) in (a) by A and using A3 = 0 yields (ii) (a + d)A2 = (ad − bc)A. Multiplying by A once more gives (ad − bc)A2 = 0. If ad − bc = 0, then A2 = 0. If ad − bc = 0, (ii) yields (a + d)A2 = 0, and if a + d = 0, again A2 = 0. Finally, if ad − bc = a + d = 0, then (i) implies A2 = 0.

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.
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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 1 1 1 1 equality, we do not have to consider the individual elements: T3 = TT2 = T( 2 T + 2 S) = 2 T2 + 2 TS = 1 1 1 1 1 3 2 ( 2 T + 2 S) + 2 S = 4 T + 4 S.

(b) The appropriate formula is (∗) Tn = 21−n T + (1 − 21−n )S. This formula is correct for n = 1 (and for n = 2, 3). Suppose (∗) is true for n = k. Then using the two ﬁrst equalities in (a), Tk+1 = 1 1 TTk = T(21−k T + (1 − 21−k )S) = 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 elementary operations, we ﬁnd that w 2 ⎝1 1 ⎛ x 1 3 1 y 4 2 2 z ⎞ ⎛ 1 3 1 ⎠ ∼ ⎝0 −1 3c 0 1 c2 ⎞ 1 − c2 ⎠ 2c2 − 1 2 + 3c + 2 −5c

0 1 0

2 0 0

2 −1 0

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. 4. (a) After moving the ﬁrst row down to row number three, Gaussian elimination yields the matrix ⎞ ⎛ 1 2 1 b2 1 3 ⎠. Obviously, there is a unique solution iff a = 3/4. ⎝0 1 −2 2 b2 − 2 b3 3 1 0 0 3 − 4a b1 + (2a − 2 )b2 + ( 2 − a)b3
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 are an inﬁnite number of solutions. In fact, 1 3 x = −2b2 + b3 − 5t, y = 2 b2 − 2 b3 + 2t, z = t, with t ∈ .

15.7
3. Using the deﬁnitions of vector addition and multiplication of a vector by a real number, we get 3(x, y, z) + 5(−1, 2, 3) = (4, 1, 3) ⇐⇒ (3x − 5, 3y + 10, 3z + 15) = (4, 1, 3). Since two vectors are equal if and only if they are component-wise equal, this vector equation is equivalent to the equation system 3x − 5 = 4, 3y + 10 = 1, and 3z + 15 = 3, with the obvious solution x = 3, y = −3 , z = −4. 5. We need to ﬁnd numbers t and s such that t (2, −1)+s(1, 4) = (4, −11). This vector equation is equivalent to (2t + s, −t + 4s) = (4, −11), which in turn is equivalent to the equation system (i) 2t + s = 4 (ii) −t + 4s = −11. This system has the solution t = 3, s = −2, so (4, −11) = 3(2, −1) − 2(1, 4).

15.8
2. (a) See the text. (b) As λ runs through [0, 1], the vector x will run through all points on the line segment S between a and b. In fact, according to the point–point formula, the line L through (3, 1) and (−1, 2) 7 1 has the 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, (1 − λ)a + λb = (3 − 4λ, 1 + λ). Any point (x1 , x2 ) 1 on L satisﬁes x1 + 4x2 = 7 and equals (3 − 4λ, 1 + λ) for λ = 4 (3 − x1 ) = x2 − 1. Any point on the segment of this line between a = (3, 1) and b = (−1, 2) equals (3 − 4λ, 1 + λ) for some λ ∈ [0, 1].
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8. (||a|| + ||b||)2 − ||a + b||2 = ||a||2 + 2||a|| · ||b|| + ||b||2 − (a + b)(a + b) = 2(||a|| · ||b|| − a · b) ≥ 2(||a|| · ||b|| − |a · b|) ≥ 0 by the Cauchy–Schwarz inequality (2).

15.9
3. One method: (5, 2, 1) − (3, 4, −3) = (2, −2, 4) and (2, −1, 4) − (3, 4, −3) = (−1, −5, 7) are two vectors in the plane. The normal (p1 , p2 , p3 ) must be orthogonal to both these vectors, so (2, −2, 4) · (p1 , p2 , p3 ) = 2p1 − 2p2 + 4p3 = 0 and (−1, −5, 7) · (p1 , p2 , p3 ) = −p1 − 5p2 + 7p3 = 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 7. (a) 1 4 1 −2 1 0 5 1 4 1 1 4 1 ← ∼ ∼ ∼ 2 2 8 ← −4 0 1 −1 0 −6 6 −1/6 0 1 −1 The solution is x1 = 5, x2 = −1. ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 2 2 −1 2 ← 1 −3 1 0 −2 −3 1 −3 1 0 (b) ⎝ 1 −3 ∼ ⎝0 1 0⎠ ← ∼ ⎝2 2 −1 2 ⎠ ← 8 −3 2 ⎠ 1/8 4 −1 1 3 4 −1 1 ← ⎛ ⎞ 0 13 −4 1 ⎛ 3 ⎞ 1 −3 1 0 1 −3 1 0 ∼ ⎝0 1 −3/8 1/4 ⎠ 1 −3/8 1/4 ⎠ −13 ∼ ⎝ 0 0 7/8 −9/4⎞ 8/7 −4 1 ← ⎛0 ⎛ 0 13 ⎞ 1 0 −1/8 3/4 ← 1 −3 1 0 ← ⎠← ⎝0 ⎠ 3 ∼ ⎝ 0 1 −3/8 ∼ 1/4 1 −3/8 1/4 3/8 1/8 0 0 1 −18/7 0 0 1 −18/7 ⎛ ⎞ 1 0 0 3/7 ∼ ⎝ 0 1 0 −5/7 ⎠. The solution is x1 = 3/7, x2 = −5/7, x3 = −18/7. 0 0 1 −18/7 (c) ∼ 1 5 1 0 3 1 4 1 0 0 −5 ∼ ← 0 0 1 0 3 −14 1 0 0 1 4 0 −19 0 −1/14 19/14 −1/14 0 0

3 4 1 19/14

← ∼ −3

The solution is x1 = (1/14)x3 , x2 = −(19/14)x3 , where x3 is arbitrary. (One degree of freedom.) 10. (a) See the 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) is maximized at λ = 0. Only in the special case where 6p1 − 2p2 − 6p3 = 0 can the two plants both remain in use.
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61

11. 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 .

16 Determinants and Inverse Matrices
16.1
8 −1 5 −2 −16 + 5 11 = = , 3. (a) Cramer’s rule gives x = −6 + 1 5 3 −1 1 −2 (b) and (c) are done in the same way. 7. (b) Note that because c1 is the proportion of income consumed, we can assume that 0 ≤ c1 ≤ 1. Likewise, 0 ≤ c2 ≤ 1. Because m1 ≥ 0 and m2 ≥ 0, we see that D > 0 (excluding the case c1 = c2 = 1). (c) Y2 depends linearly on A1 . 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. 3 8 1 5 7 15 − 8 =− . = y= −5 5 3 −1 1 −2

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) 1 3 2 = 3 − 2 − 0 − 0 − 4 − (−1) = −2 1 2 1 (c) 5 of the 6 products have 0 as a factor. The only product that does not include 0 as a factor is the product of the terms on the main diagonal. The determinant is therefore adf . (d) a 0 c 0 e 0 b 0 = aed + 0 + 0 − bec − 0 − 0 = e(ad − bc) d

1 1 −1 3. (a) The determinant of the coefﬁcient matrix is |A| = 1 1 −1 = −4. −1 −1 −1 The numerators in (16.2.4) are (verify!) 2 0 −6 −1 1 −1 1 −1 = −4 , −1 1 1 −1 2 0 −6 1 −1 = −8 , −1 1 1 −1 −1 1 −1 2 0 = −12 −6

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).
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6. (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⎛ (b) below. ⎞ ⎛ ⎞ ⎛ ⎞ in 1 −1 0 Y A0 (b) We write the system as ⎝ −b 1 b ⎠ ⎝ C ⎠ = ⎝ a ⎠. Then Cramer’s rule yields −t 0 1 T d A0 −1 a 1 d 0 1 −1 −b 1 −t 0 1 −1 −b 1 −t 0 1 −1 1 −b −t 0 0 b a − bd + A0 1 = , C= 0 1 − b(1 − t) b 1 A0 a t (a + A0 ) + (1 − b)d d = 0 1 − b(1 − t) b 1 1 A0 0 −b a b −t d 1 1 −1 0 −b 1 b −t 0 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. Note how systematic elimination is much more efﬁcient.)

16.3
1. Each of the 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 abcd. (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 shown here: 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.

16.4
10. (a) and (b) see the text. (c) We have (In − A)(In + A) = In · In − AIn + In A − AA = In − A + A − A2 = In − A2 , and this expression equals 0 if and only if A2 = In .
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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63

12. The description in the answer in the text amounts to the following: 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 answer. (b) One possibility is to expand by the second row or the third column (because they have both two zero entries). But it is easier ﬁrst to use elementary operations to get a row or a column with one more zero. For instance in this way: 1 0 2 −2 2 −1 −1 0 3 0 0 −1 −2 4 11 3 ← 3 ← 2 1 0 = 0 0 = −1 0 0 2 −1 −5 4 0 −6 5 3 0 −6 5 4 11 −5 11

−1 = −5 4

0 −6 5

11 −5 −5 ← 11 ← = −1 −6 5

−4

11 −60 55

−60 = −(−330 + 300) = 30 55

When computing determinants one can use elementary row operations as well as column operations, but column operations become meaningless when solving linear equation systems using Gaussian elimination. When elementary operations have produced a row or column with only one non-zero element, then it is natural to expand the determinant by that row or column. (c) See the answer in the text.

16.6
6. (b) From (a), A3 − 2A2 + A = I, or A(A2 − 2A + I) = I, so using Theorem 16.6.2, A−1 = A2 − 2A + I. The last expression can also be written (A − I)2 . (c) See the text. 9. B2 +B = 3/2 −5 −1/2 5 1 0 + = = I. One can verify directly that B3 −2B+ −1/4 3/2 1/4 −1/2 0 1 I = 0, but here is an alternative that makes use of B2 + B = I: B3 − 2B + I = B3 + B2 − B2 − 2B + I = B(B2 + B) − B2 − 2B + I = B − B2 − 2B + I = −(B2 + B) + I = 0.

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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

C11 C12 −5/2 2

C21 C22

=

5 −4

−3 , so the inverse is 2

5 −4

−3 2

=

3/2 −1

(b) If we denote the matrix by A, the adjoint is C11 adj A = ⎝ C12 C13 ⎛ C21 C22 C23 ⎞ ⎛ 1 C31 C32 ⎠ = ⎝ 2 C33 4 4 −1 −2 ⎞ 2 4⎠ −1

and the determinant is |A| = a11 C11 + a21 C21 + a31 C31 = 1 · 1 + 2 · 4 + 0 · 2 = 9, (by expansion along the ﬁrst column). Hence, ⎛ ⎞ 1 4 2 A−1 = 1 (adj A) = 1 ⎝ 2 −1 4⎠
9 9

4 (c) Since the determinant is 0 there is no inverse.

−2

−1

⎞ 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 ⎠. 2 19 8 ⎠. This gives (I − A)−1 = 62 25 4 7 16 4 7 16 4. Let B denote the n × p matrix whose ith column has the elements b1i , b2i , . . . , bni . 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 premultiplication by A−1 , it must be true that B∗ = A−1 B.) When k = r, the solution to the system is ∗ ∗ ∗ x1 = b1r , x2 = b2r , . . . , xn = bnr . 5. (a) The following shows that the inverse is 1 3 1 0 2 4 2 1 1 0 −2
3 2

⎛

−2
3 2

−1 1 : −2 0 1
1 −2

0 −3 ∼ 1 ← −1 1 −2

1 0

2 −2

1 −3

∼

1 0

2 1

1
3 2

0 ← ∼ 1 −2 −2

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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DETERMINANTS AND INVERSE MATRICES

65

(b)

1 2 ⎝2 4 3 5 ⎛ 1 ⎝0 ∼ 0 ⎛ 1 ⎝0 ∼ 0

⎛

⎞ ⎛ −2 −3 1 2 3 1 3 1 0 0 ⎠← ⎝0 ∼ 0 −1 −2 5 0 1 0 −3 0 0 1 ← 0 −1 −3 6 ⎞ ⎛ 1 2 3 1 0 2 3 1 0 0 ⎠ −1 ∼ ⎝ 0 1 3 3 0 −1 −3 −3 0 1 −2 1 0 −1 0 0 1 2 −1 0 −1 ⎞ ⎛ ← 0 −3 −5 0 2 1 0 0 ⎠← ⎝0 1 0 ∼ 1 3 3 0 −1 2 −1 0 0 1 −3 3 0 0 1

0 1 0

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

(c) We see that the third row is the ﬁst row multiplied by −3, so the matrix has no inverse.

16.8
1 1. (a) The determinant |A| of the coefﬁcient matrix is |A| = 2 1 The determinants in (16.8.2) are (verify!) −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 = −1. (b) The determinant |A| of the coefﬁcient matrix is 0 1 1 1 0 1 0 1 The determinants in (16.8.2) are (verify!) 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. 3. According to Theorem 16.8.2, the system has nontrivial solutions iff the determinant of the coefﬁcient equal to 0. Expansion according to 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 iff 3abc − a 3 − b3 − c3 = 0.
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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DETERMINANTS AND INVERSE MATRICES

16.9
4. The equation system is obtained directly from (16.9.4). Review Problems for Chapter 16 3. 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 row 2 to row 4 in the middle determinant. If we expand the last determinant by the row 1, we end up with an easy 3 × 3 determinant to evaluate. The equation becomes (a − b)2 (4x 2 − (a + b)2 ) = (a − b)2 (2x + (a + b))(2x − (a + b)) = 0. The conclusion follows. 5. (a) Expanding by column 3: |A| = q −1 q − 2 1 −p q −1 − (2 − p) = 1 −p 2 − p = (q − 2) 2 −1 2 −1 2 −1 0 (q − 2)(−1 + 2p) − (2 − p)(−q + 2) = (q − 2)(p + 1), but there are many other ways. q +1 0 q −1 q +1 q −1 |A + E| = = (1 − p)[q + 1 − 3(q − 1)] 2 1 − p 3 − p = (1 − p) 3 1 3 0 1 = 2(p − 1)(q − 2). For the rest see the text.

8. (a) Note 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 A = ⎝3 3 ⎛ 3 4 3 ⎞ ⎛ 3 1 3⎠ = ⎝0 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, which is equal to I3 if we choose b = −3/10. It follows that ⎛ ⎞ ⎛ 3 ⎛ ⎞ 3 3 ⎞ 1 0 0 7 −3 −3 10 10 10 3 3 ⎠ = 1 ⎝ −3 A−1 = (I3 + 3U)−1 = I3 − (3/10)U = ⎝ 0 1 0 ⎠ − ⎝ 3 7 −3 ⎠ 0
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

0

1

10 3 10

10 3 10

10 3 10

10

−3

−3

7

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LINEAR PROGRAMMING

67

11. (a) Gauss-elimination yields ⎛ a ⎝2 1 1 1 0 ⎞ ⎛ 4 2 ← 1 2 ⎠ ∼ ⎝2 a 2 −3 a ← a ⎛ 1 ∼ ⎝0 0 ⎛ 1 ∼ ⎝0 0 ⎞ −2 −a 0 −3 a 2 ⎠← 2 1 a 1 4 2 ← ⎞ 0 −3 a 1 a 2 + 6 −2a + 2 ⎠ −1 1 3a + 4 −a 2 + 2 ← ⎞ 0 −3 a −2a + 2 ⎠ 1 a2 + 6 2 + 3a − 2 −a 2 + 2a 0 −a

It follows that the system has unique solution iff −a 2 + 3a − 2 = 0, i.e. iff a = 1 and a = 2. If a = 2, the last row consists only of 0’s so there ia an inﬁnite number of solutions, while if a = 1, there are no solutions. (b) If we perform the same elementary operations as in (a) on the associated extended matrix, we get ⎛ 1 ⎝0 0 0 1 0 −3 a2 + 6 −a 2 + 3a − 2 ⎞ b3 ⎠, b2 − 2b3 b1 − b2 + (2 − a)b3

We see that there are inﬁnitely many solutions iff all elements in the last row are 0, i.e. iff a = 1 and b1 − b2 + b3 = 0, or when a = 2 and b1 = b2 . 13. For once we use “unsystematic elimination”. Solve the ﬁrst equation for y, the second for z, and the fourth for u, using the expression found for y. Insert all this into the third equation. This gives: a(b − 2)x = −2a +2b +3. There is a unique solution provided a(b −2) = 1. We easily verify the solutions in the text. 15. We prove the result for 3 × 3 matrices that differ only in the ﬁrst row: a11 + x a21 a31 a12 + y a22 a32 a13 + z a11 = a21 a23 a33 a31 a12 a22 a32 a13 x a23 + a21 a33 a31 y a22 a32 z a23 a33

(∗)

We expand the ﬁrst determinant in (∗) by the ﬁrst row, where C11 , C12 , C13 are the complements of the three entries in the ﬁrst row, and we get (a11 + x)C11 + (a12 + y)C12 + (a13 + z)C13 = [a11 C11 + a12 C12 + a13 C13 ] + [xC11 + yC12 + zC13 ] The sums in square brackets are the two last determinants in (∗).

17 Linear Programming
17.1
3. The set A is the shaded set in Fig. SM17.1.3. (a) The solution is obviously at the point P in the ﬁgure because it has the largest x2 coordinate among
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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LINEAR PROGRAMMING

all points in A. P is the point 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). x2 −2x1 + x2 = 2

4

P

3x1 + 2x2 = c

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

(b) The point in A with the largest x1 coordinate is obviously Q = (8, 0). (c) One of the lines 3x1 + 2x2 = 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 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 to the left. The line that has the smallest value of c and still has a point in common with A, is the point P in the ﬁgure. (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 obtained when the line “covers” 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 must occur at the same point as maximizing 3x1 + 2x2 subject to (x1 , x2 ) ∈ A.)

17.2
1. (a) See Fig. A17.1.1(a) in the text. When 3x1 + 2x2 ≤ 6 is replaced by 3x1 + 2x2 ≤ 7 in Problem 17.1.1, the feasible set expands because the steepest line is moved 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
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

17

LINEAR PROGRAMMING

69

1 is 3 · 2 + 4 · 2 = 8 = 40/5, and the difference in optimal value is u∗ = 5/4. 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 the text.

17.3
1. (a) From Fig. A17.3.1(a) in the text it is clear as c increases, the dashed line moves out farther and farther to the northeast. The line that has the largest value of c and still has a point in common with the feasible set is the point P , which has coordinates (x, y) = (0, 3). (b) In Fig. A17.3.1(b), 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 point P , which has coordinates (u1 , u2 ) = (0, 1). The associated minimum value is 20u1 + 21u2 = 21. This is the maximum value in the primal problem, so the answer to question (c) is yes. 2. Actually not much to add. From the easily produced ﬁgure we can read off the solution. 3. See Fig. SM17.3.3 and the text. x2 2x1 + x2 = 16 x1 + 2x2 = 11 5
∗ ∗ (x1 , x2 ) = (7, 2)

x1 + 4x2 = 16 x1

5 400x1 + 500x2 = konst.

10

Figure SM17.3.3

17.4
∗ ∗ 2. (a) The problem is similar to Problem 17.3.3. See the answer in the text. Note that 300x1 +200x2 = 2800. (b) The dual problem is

min (54u1 + 48u2 + 50u3 )

subject to

⎧ ⎪ 6u1 + 4u2 + 5u3 ≥ 300 ⎨ 3u + 6u2 + 5u3 ≥ 200 ⎪ 1 ⎩ u1 , u2 , u3 ≥ 0

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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17

LINEAR PROGRAMMING 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

∗ ∗ The optimal solution of the primal is x1 = 8, x2 = 2. Since they are both positive, the ﬁrst two constraints in the dual is satisﬁed with equality at the optimal triple (u∗ , u∗ , u∗ ). Since the second constraint in the 2 3 1 ∗ ∗ primal is satisﬁed with strict inequality: 4x1 + 6x2 = 44 < 48, u∗ = 0. So 6u∗ + 5u∗ = 300 and 1 3 2 3u∗ + 5u∗ = 200. It follow that u∗ = 100/3, u∗ = 0, and u∗ = 20, with 54u∗ + 48u∗ + 50u∗ = 2800. 2 3 1 2 3 1 3 1 (c) See the text.

17.5
3. (a) See the text. (b) The dual is given in the text. Look at Fig. SM17.5.3. 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 criterion is 300 · 100 + 500 · 450 = 255 000. By complementary slackness, the constraints in the dual problem must in optimum be satisﬁed with equality. Since the second constraint in the primal in the optimum has a slack (20 · 100 + 10 · 450 < ∗ ∗ ∗ ∗ ∗ 8000), then y2 = 0. Hence 10y1 + 20y3 = 300, 20y1 + 20y3 = 500. It follows that the solution ∗ ∗ ∗ of the dual problem is y1 = 20, y2 = 0, y3 = 5. The minimum 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, the maximum in the primal problem would increase ∗ by y1 = 20. (The numbers y1 , y2 , and y3 are the shadow prices for the resources in the primal.) Increasing the costs in factory 1 by 100 will therefore increase the maximum in the primal by 100 × 20 = 2000. (We assume that the optimal point in the primal does not change.) Since the maximum in the primal is the minimum in the dual, it follows that the minimum costs in the dual will increase by 2000 if the cost per hour in factory 1 increases by 100. Review Problems for Chapter 17
© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

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71

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, and the line that has the largest value of c and still has a point in common with the feasible set is the point (0, 8), which is the solution to (D). (b) We see that when x1 = 0 and x2 = 8, the second and fourth constraint in (D) are satisﬁed with strict inequality, so in the optimum in (P), y2 = y4 = 0. Also, since x2 = 8 > 0, the second constraint in (P) is in optimum satisﬁed with equality, i.e. 2y1 − y3 = 1. But then we see that the objective function in (P) 1 is 16y1 + 6y2 − 8y3 − 15y4 = 16y1 − 8y3 = 8(2y1 − y3 ) = 8. If we put y3 = b, y1 = 2 (1 + b), and we 1 conclude that (y1 , y2 , y3 , y4 ) = ( 2 (1+b), 0, b, 0) must be a solution of (P) provided b is chosen such that 1 y1 = 2 (1 + b) ≥ 0, i.e. b ≥ −1, and y3 = b ≥ 0, and the ﬁrst constraints in (P) is satisﬁed. (The second 1 constraint we already know is satisﬁed with equality.) The ﬁrst constraint reduces to − 2 (1+b)−2b ≥ −1, 1 1 or b ≤ 1 . We conclude that ( 2 (1 + b), 0, b, 0) is optimal provided 0 ≤ b ≤ 5 . 5 (c) The objective function in (D) is now kx1 + x2 . If k ≥ 0, there is no solution. The condition for (0, 8) to be the solution is that k is less or equal to the slope of the constraint −x1 + 2x2 = 16, i.e. k ≤ −1/2. 4. (a) See the text. (b) With 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), 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 and if 0.04x1 + 0.03x2 < 100 λ3 = 0 λ4 = 0 if 0.05x1 < 100 if 0.025x1 + 0.05x2 < 100 if 0.08x2 < 100 (i) (ii) (iii) (iv) (v) (vi)

(c) The Kuhn–Tucker conditions are sufﬁcient for optimality since the Lagrangian is easily seen to be concave in (x1 , x2 ) for a ≥ 0. If (x1 , x2 ) = (2000, 2000/3) is optimal, then (i) and (ii) are satisﬁed with equality. Moreover, the inequalities in (iv) and (vi) are strict when x1 = 2000 and x2 = 2000/3, so λ2 = λ4 = 0. Then (ii) gives λ1 = 25000/3. It remains to check for which values of a that λ3 ≥ 0. From (i), 0.05λ3 = 500 − 4000a − 0.04(25000/3) = 500/3 − 4000a ≥ 0 iff a ≤ 1/24.

© Knut Sydsæter, Arne Strøm, and Peter Hammond 2008

Student’s Manual Essential Mathematics for Economic Analysis

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