ECO206 final exam

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Item ID: 6639

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D rI eUNIVERSITY OF TORONTO ad lo
Faculty of Arts and Science n ow
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6639

ECO206Y1Y (Microeconomic Theory)
Instructor: Victor Couture and Rebecca Lindstrom
Item

Final Examination
August 2011

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Duration: 180 minutes (3 hours)
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Examinations Aids: te
Non-Programmable Calculators
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This examination paper consists of 16 pages and 8 questions. Please bring any discrepancy to the attention of an invigilator. The number in brackets at the start of each question is the number of points the question is worth.
Answer all questions.
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To obtain credit, you must give arguments to support your answers. er Score

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1. Lude has $12, 000. He plans to bet on a football game. Assume no ties can occur.
Experts think that team A is more likely to win, and the odds at the local gambling house are 3 to 1 in favor of team A, so that for $0.25 one can buy or sell a ticket that
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29 nothing if A wins. Lude, however, thinks that the two pays $1 if team B wins and
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teams are equally likelyI to win. Lude is an expected utility maximizer, and his utility er ad when he has a wealth of w is ln(w). lo Item

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(a) [3] Find an expression for Lude’s budget constraint.
Solution: There are two state of nature: team A wins (state A) and team
B wins (state B). Let wA be Lude’s wealth if team A wins, wB be his wealth if team B wins and T be the number of tickets that he purchase.
We have that wA = 12000 − 0.25T and wB = 12000 − 0.25T + T . Isolating
T = (12000 − wA )/0.25 in the expression for wA and plugging it into the
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expression for wB we obtain the budget constraint 3wA + wB = 48000.
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(b) [2] Find an expression for Lude’s expected utility function. te I

Solution: Lude believes that team A wins with probability 0.5 and that team
B wins with probability 0.5, so his expected utility is 0.5ln(wA ) + 0.5ln(wB ).
(c) [4] How many tickets does Lude buy or sell?
Solution: At an optimum we have that M UA /M UB = p976 B , which we can
A /p
2
write as 0.5wB = 3 and solve for wB = 3wA . Plugging wB in the budget
0.5wA
1
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constraint we can find wA = 48000/6 = 8000. dThen using our expression er a for wA in part a) we solve 8000 = 12000 − lo
0.25T to find that Lude buys n w
T = 16000 tickets.
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(d) [4] After Lude finishes his transaction at the gambling house, team A loses a star player and the ticket price moves to $0.40. Lude can buy new tickets or sell those he already has. He now believes that team A has a 30% chance of winning.
Suppose that when the game is finally played, team A wins. How much wealth does he end up with?
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Solution: Lude bought 16000 tickets, which cost him $4000, so he has612000 −
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4000 = 8000 dollars left. The value of his tickets is now 16000 ∗ ID = 6400,
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m te so his new wealth is 8000 + 6400 = 14400. We can find the budget constraint
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as before. We have that wA = 14400 − 0.4T and wB = 14400 − 0.4T + T .
Isolating T = (14400 − wA )/0.4 in the expression for wA and plugging it into the expression for wB we obtain the budget constraint (3/2)wA + wB =
(5/2)14400. Lude’s expected utility is now 0.3ln(wA ) + 0.7ln(wB ). At an optimum we have that M UA /M UB = pA /pB , which we can write as 0.3wB =
0.7wA
3/2
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and solve for wB = (7/2)wA . Plugging wB in the budget constraint we can find (3/2)wA + (7/2)wA = (5/2)14400 and solve it for wA = 7200, which is Lude’s wealth team A wins.
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2. A competitive firm produces a single output using a single input. In 2010, the cost of the input was $1 and the price of the output was $6, and the firm used 3 units of

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page 3 input to produce 2 units of output. In 2011, the cost of the input was $3 and the price of the output was $10, and the firm used 1 unit of input to produce 1 unit of output.
The firm’s technology is the 7same in 2010 and in 2011.
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(a) [3] With this information only, can you tell whether the firm was profit-maximizing rI de in 2010? oa nl
Solution:w If a firm is profit-maximizing in 2010, then at 2010 prices its 2010
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input and output choices must lead to a higher profit than its 2011 input and output choices, i.e. we must have that 6(2) − 1(3) = 9 is larger than
6(1) − 1(1) = 5 (which is true). So its 2010 choice is consistent with profitmaximization, but we cannot tell whether it is profit-maximizing (so the answer is that we cannot tell). Note that it possible to provide a perfectly valid graphical argument to arrive at the same conclusion.
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(b) [3] With this information only, can you tell whether the firm is profit-maximizing
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Solution: If the firm is profit-maximizing in 2011, then at 2011 prices its 2011 input and output choices must lead to a higher profit than its 2010 input and output choices, i.e. we must have that 10(1) − 3(1) = 7 is larger than
10(2) − 3(3) = 11 (which is not true). So the firm’s 2011 choice is not
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consistent with profit-maximization, and we can tell 2that the firm is not
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profit-maximizing in 2011. Again, it is also possible to provide a perfectly
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(c) [3] Suppose that you are told that the firm’s technology exhibits increasing returns wn Do to scale (irs). Now, can you tell whether the firm was profit-maximizing in 2010?
Explain carefully.
Solution: With increasing returns to scale, the profit-maximizing problem has not solution (and therefore a firm with irs cannot be profit-maximizing); profits can be increased indefinitely just by increasing the scale of production.
To show this, consider a profit function like π = pf (y) − wx. By scaling up
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66 its operation by a factor t, the value of the firm’s production (pf (tx)) will
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of irs), and its cost (wtx = twx) will increase by a factor t. So a firm with irs technology can always increase its profits by increasing the scale of production, and therefore the 2010 choice of y = 2 and x = 3 cannot be profit-maximizing. 3. Firm Z uses two factors to produce airplanes using the technology y = min{x1 , 2x2 } where y is the number of airplanes, and x1 and x2 are the quantities of factor 1 (airplane bodies) and factor 2 (airplane wings), respectively. The two factors are bought in competitive factors markets where the price of factor 1 is 9 per unit and the price of

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page 4 factor 2 is 5 per unit. Because firm Z has limited warehouse space, it cannot use more than 10 units of factor 1 in production.
If firm Z produces some positive number of airplanes (y > 0), then it incurs an ad76
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ditional cost of A whereDA ≥ 0. If firm Z produces no airplanes (y = 0), it does not
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(a) [5] Suppose the market for airplanes is competitive with a very large number of
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firms, all choosing to supply some y > 0. These firms are all profit maximizers and have the same technology (and warehouse constraints) as firm Z. Let A = 90.
In the long run, what is the price of airplanes?
Solution: In the long run, firms enter if they can make positive profit, and exit if they make negative profit. Note that here, since there are no truly fixed costs, producing y = 0 (being “inactive”) and exiting are equivalent, so they
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Suppose there is a LR equilibrium with price p where some firm is active but te I chooses some y < 10 (in other words, y ∈ (0, 10)). For this range of y, the marginal cost is constant at 19. From the fact that the firm does not exit (or equivalently, chooses y > 0), we know that oader ID: 2976

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π = (p − 19)y − 90 ≥ 0

⇒

p≥

90
+ 19 >919 = M C
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y
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p = 9 + 19 = 28

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Marks distribution
• Give [2] for calculating M C = 19, even if don’t state for what range of y
• Give [2] for arguing that firms will not choose y ∈ (0, 10) in LR eqm
• Give [1] for solving for p = 28 [1]

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(although this equality holds only approximately in general, we ID assume can em that this it holds generally since the number of firms is “very large”). It see
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is indeed profit-maximizing for the firm (this is not necessary since we know that some firms are active but that they cannot have chosen y ∈ (0, 10)), note that if the firm reduces output by some amount ∆, then the fall in cost is 19∆, while the fall in revenue is 28∆. Thus, this would reduce the firm’s profit (making it negative rather than zero). Hence, choosing y = 10 (and making zero profit) is indeed profit-maximizing for the firm. Hence, p = 28 must be the LR equilibrium price.

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π = (p − 19)10 − 90 = 0

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But since price is above the constant marginal de cost, it is easy to see that oa the firm can increase profit by expanding output. This is a standard IRS nl ow argument. Hence, choosing y ∈ (0, 10) cannot maximize the firm’s profit.
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Hence, no profit-maximizing firm will choose y ∈ (0, 10) in LR eqm.
Following the argument above, all firms that produce something at a given price p in a LR eqm must choose y = 10. If firms neither exit nor enter (in
LR eqm) we assume that

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(b) [5] Now instead suppose that firm Z is a profit-maximizing monopolist in the market for airplanes. Let A = 0. Graphically, illustrate the extent of social welfare loss from monopoly pricing (deadweight loss) in this market. You should
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rI enecessary to consider two cases (ideally, drawing two graphs – or
Solution: It is d oa a messylgraph with two sets of MR curves in). wn Caseo1: If the MR curve cuts the MC curve at y < 10, then we have the
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usual constant MC with positive (“triangle” shaped) DWL.
Case 2: If the MR curve cuts the p = 19 line to the right of y = 10, then the monopolist will choose to supply y = 10. This is the efficient level of output
(no quantity distortions). Hence, there is no DWL. The only effect is the price increase (which will now be above MC).
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Marks distribution
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• Give [2] for Case 1: must draw constant MC and triangular DWL. Give
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• Give [3] for Case 2: state MR conditions for this case to occur [1], argue no DWL [1] because quantity does not change compared to efficient benchmark [1].
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4. Be-For-Tea is the only tea producer in Atlantis, a closed island 2economy in the West
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Pacific. Be-For-Tea is a profit maximizing firm and can erI produce tea at a constant ad marginal cost of 20 dollars per pound. o nl

Atlantis has two provinces: North and South. In the w
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PN (y) = 100 − y,

0 ≤ y ≤ 100

where P is the price of tea in dollars per pound and y is the amount of tea (in pounds).
In the poorer South, the inverse market demand for tea is given by
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PS (y) = 40 − 2y,

0 ≤ y ≤ 20

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where P and y are defined similarly. Although Be-For-Tea does not I know each consumer’s individual demand, it does know the market demand in each of the provinces as given above.
(a) [4] Suppose it is impossible for consumers to trade tea across the North-South provincial border. Write down Be-For-Tea’s profit maximization problem and calculate its total profit from tea sales in Atlantis.
Solution: Since there is no possibility for consumer arbitrage, the monopolist can charge different prices in different markets (i.e., market segmentation/3rd degree price discrimination). The problem is given by

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max{(a − by)y − 20y : y ≥ 0} y page 6

so we have y ∗ =

a − 20 with 2b

100 − 20 76
9= 40,
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∗
yS = de a 2(2) = 5, lo ∗ yN =

p∗ = 60,
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∗ πN = (60 − 20)40 = 1, 600

p∗ = 30,
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∗ πS = (30 − 20)5 = 50

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Total

Marks distribution
• Give [1] for realizing (implicitly or explicitly) that the firm chooses market
(province) specific prices
• Give [2] for writing down a correct profit maximization problem and taking correct FOC (don’t 39 penalize if fail to state non-negativity con6
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straint on y). Since MC is constant, can write either one or two profitD mI (equivalent). maximization problems e It
• Give [2] for each correct profit/sum of profits. Don’t penalize if they fail to add them up to the total profit.
(b) [6] Suppose consumers can trade tea freely across the North-South provincial border. Write down Be-For-Tea’s profit maximization problem and calculate its
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total profit from tea sales in Atlantis.
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Solution: Here, the monopolist should set the same price in both markets. Other erwise, purchases would ever only be made in the d oamarket with the lower price. nl Invert the demand functions
First, we need to find the market demand curve. ow so that we can sum them. Then, we get D

yN (P ) = 100 − p,
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yS (P ) = 20 − p,
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0 ≤ p ≤ 100
0 ≤ p ≤ 40

The market demand is given by y(P ) =

120 − 3 p
2
100 − p,

0 ≤ p ≤ 40
40 < p ≤ 100

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⇔

P (y) =

100 ID y
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80 3
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0 ≤ y ≤ 60
60 ≤ y ≤ 100

Note that it is continuous but kinked, so the MR curve is discontinuous. If you draw the graph, it is obvious that the monopolist wants to price on the top piece (serving only consumers in the North) because the MR is always negative on the piece where both markets are served.
To see this formally, note that the marginal revenue curve is
M R(y) =

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100 − 2y
80 − 4 y
3

0 ≤ y ≤ 60
60 ≤ y ≤ 100

An elasticity argument will work as well (this part is always inelastic). But
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for y ≥ 60, M R(y) = 80 − 3 y ≤ 80 − 4 60 = 0, so the monopolist never prices
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page 7 on this piece of the demand curve. Hence, we can write the monopolist’s problem 76
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max{(100 − y)y − 20y : y ≥ 0}

y
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= 40, l= n
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p∗ = 60,

π ∗ = (60 − 20)40 = 1, 600

This is the same as that for the North only above. It’s easy to confirm
0 ≤ y ∗ ≤ 60, so we are indeed on the part of the demand curve where both markets are served.
Marks distribution
• Give [1] for realizing (implicitly or explicitly) that the firm is constrained to choosing a single price 39
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• Give [1] for correct market supply / inverse market supply
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• Give [2] for arguingtem will not supply on the part of demand curve that I where south are served (graphically or algebraically); an elasticity argument could work as well
• Give [1] for calculating total profit
Note: If students don’t pay attention to the fact that the demand curve is
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76 kinked, they may want to set 80 − 3 y = 20 ⇒ y ∗ = 45. 9This is incorrect. In
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er ad o nl 5. Kai and Lina are playing a two-player, two-action wgame with simultaneous moves
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represented in the following payoff matrix

Lina
Kai

R
S

T a, 0 b, 3

U c, d
2, 4
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where a, b, c and d are some real numbers. m te

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(a) [2] For what values of a, b, c and d is (R, T ) a Nash equilibrium in this game?
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Solution: No profitable deviation for Kai requires a ≥ b. No profitable deviation for Lina requires 0 ≥ d.
Marks: [1] for each condition, penalize [0.5] each if use strict inequality.
(b) [2] For what values of a, b, c and d is playing S a dominant strategy for Kai? em It
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Solution: If b ≥ a and 2 ≥ c.
Marks: [1] for each condition, don’t penalize if use strict inequality.
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(c) [2] For what values of a, b, c and d is playing U a dominated strategy for Lina?

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Solution: There are no such values. Since 3 < 4, U is a best response for Lina when Kai plays S, so U cannot be dominated strategy.
Marks: [1] for saying no such values, [1] for saying why.

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(d) [8] Assume a = 0, b = 3, c = 5 and d = 1. Draw Kai’s and Lina’s best response functions and find all Nash equilibria in this game.
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Solution: Denote Kai’s strategy by (p, 1 − p) and Lina’s strategy with (q, 1 − q).
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Kai’s best response is to play R or (p = 1) if
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5(1 − q) ≥ 3q + 2(1 − q) ⇒ q ≤

1
2

So his best response function is

1
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p = BRK (q) = [0, 1]


0

if q < if q = if q >

1
2
1
2
1
2

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Lina’s best response is to play T
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3(1 − p) ≥ p + 4(1 − p) ⇒ 3 ≤ 4

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Since 3 ≤ 4 is never true, T is never a best response for Lina. This is because playing U is a (strictly) dominant strategy. Thus, Lina’s best response function is
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q = BRL (p) = 0 for all p 297
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The BR mapping must have p on one axis and q onrone axis, and depict both de oa best response functions (Lina’s is a line, Kai’slhas two kinks).
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• Give [2] for calculating Kai’s “critical” point q = 1/2, and noticing that
Lina has a dominant strategy (“line” best response function).
• Give [4] for drawing the correct best-response functions. The axes and functions must be fully labelled for full marks. There is no need to write down the BR functions, but if no drawing is made, [1] can be given for
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writing down the BR functions.
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• Give [2] for stating the NEs. mI lo

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(e) [7] Now suppose Kai and Lina play a different game. This game is as follows:
First, Kai chooses between the two-player, two-action simultaneous move games
A and B. Then, Kai and Lina play either Game A or Game B depending on Kai’s choice. Game A can be represented in the following payoff matrix
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Lina
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R
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3, 2
2, 0

U
0, 1
0, 0

Game B can be represented in the following payoff matrix

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Y
2, 4
1, 2

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1, 3
0, 1

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ID er ad lo Find all subgame perfect equilibria (SPE) of this game. wn Do
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Solution: Start from the second stage of the game and use backwards induction.
• If Game A is played, there are exactly two NE in this part: (R, T ) and
(S, U ), giving Kai payoffs 3 and 0, respectively.
• If Game B is played, there is are exactly one NE in this part: (W, Y ), giving Kai payoff 2.
If (R, T ) is the eqm of Game A, 9 then Kai chooses Game A (since 3 > 2). If
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(S, U ) is the eqm of Game A, 6 then D
Hence, there are two SPE: I m e
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(A, ((R, T ), (W, Y )))

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and

(B, ((S, U ), (W, Y )))

It may be easier to think about the equilibrium as “no regret at any point where I had to move”.
Note: There is no reason to be super strict exactly how these strategy profiles
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• Give [1] each for each NE in games A andlB, for a total of [3] wn Do and [3] for the second
• Give [2] for the first SPE that is found,
– i.e., if only one SPE is given, the maximum mark is [5]
6. Tanya owns a small plot of land. The only use of Tanya’s plot of land is to grow rice.
If a worker exerts effort x working on Tanya’s land, then the amount of rice produced in one year (in tons) is given by
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y = f (x) =

ln x if x ≥ 1
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Since the land plot is very small, it is not possible for more than one person to work on it. The market for rice is competitive, with a market price of 40 dollars per ton.
Since Tanya cannot work on her land herself, she is considering asking Ivan to work for her. As Tanya’s plot is in a very isolated area, Ivan is the only worker that could work on the plot. Ivan’s cost of exerting effort x is given by em 66

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c(x) = 5x2 ,

x≥0

Ivan seeks to maximize his utility U (x) = M − c(x), where M is his money income (in dollars). If Ivan does not work for Tanya, he can earn k dollars per year in an effortless call centre job. The market for call centre employees is competitive. If Ivan works on
Tanya’s plot, he cannot take the call centre job.

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(a) [4] What is the efficient level of effort, xE , for Ivan to exert working on Tanya’s land plot?
Solution: The total 976 surplus of Ivan working on Tanya’s land is given by
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T S(x) = 40f (x) − c(x) − k er d oa l
If Ivan n ow works x ≥ 1 units on Tanya’s land, then the efficient level of x is given
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by

x∗ = arg max{40 ln x − 5x2 − k} = 2 x It is efficient for Ivan to exert effort at all only if
40 ln x∗ −5(x∗ )2 −k = 40 ln 2−5(2)2 −k ≥ 0
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k ≤ 40 ln 2−20 (≈ 7.7259)

Hence, the efficient effort level 63
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D mI te 40 ln 2 − 20 (≈ 7.7259) if kI≤

2
0

otherwise

⇐ should do call centre job instead

Marks distribution

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• Give [2] for interior solution (irrespective of whether 7k is included in TS
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function or not)
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• Give [2] for arguing that zero will be optimalr if k is above the cut-off e ad value lo n w

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(b) [8] Tanya wants to maximize the profit earned from her land plot. She cannot force Ivan to work for her or force him to exert any particular level of effort.
However, she may hire Ivan as a worker and pay him a wage w(y) that depends on the amount of rice he produces. Assume k = 5.

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(i) If Tanya wants Ivan to exert effort xE from (a), write down the incentive compatibility constraint (ICC) and participation constraint (PC) in her profit
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maximization problem.
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(ii) Suggest a wage schedule w(y) that is profit maximizing for Tanya. Explain em carefully why your proposed wage schedule is profit maximizing. Calculate
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Tanya’s profit.
Solution: Note that xE = 2.
(i) The ICC is w(ln(xE )) − 5(xE )2 ≥ w(ln(x)) − 5x2 or, w(ln(2)) − 20 ≥ w(ln(x)) − 5x2
The PC is

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w(ln(xE )) − 5(xE )2 ≥ 5 or, w(ln(2)) − 20 ≥ 5

for all x for all x

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(ii) There is an infinite number of possible wage schedules. Here, we consider some linear wage schedule w(y) = ay + b for some numbers a and b. Since w(y) is differentiable in y (for y > 0), we can write the ICC as the FOC
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⇒

FOC:

a
− 10x = 0 x Now, for incentive compatibility,

a a − 10xE = − 20 = 0
E
x
2

⇒ a = 40

If Tanya wants to maximize her profit, she should leave no surplus for
Ivan, i.e., the PC holds with 9 equality. So
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w(ln(2)) − 20 = (40 ln(2) + b) − 20 = 5
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⇒ b = 25 − 40 ln 2(≈ −2.7259)

That is, the wage schedule is w(y) = 40y + 25 − 40 ln 2.
Another wage schedule that would accomplish the same thing is w(y) =

25
0

if y = ln 2 otherwise ID

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If student chooses this schedule, s/he still needs to show carefully why it ad lo satisfies IC and binds PC. wn These wage schedules must be profitDo maximizing because (i) Ivan will choose the efficient level of effort, maximizing total surplus of the relationship, and (ii) Tanya can extract all total surplus of the relationship.
Hence, there is no way she could be better off under some alternative wage schedule.
Tanya’s profit is 40 ln 2 − 25 ≈ 2.7259
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Marks distribution: [3] for part (i), [5] for part (ii)
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Part (i)
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• Constraints can be stated either in terms of xE or directly in terms of 2
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• Give [2] for ICC. If already assumed w(.) is differentiable and stated the
FOC rather than the general form, give [1].
• Give [1] for PC. Don’t penalize if write with equality.
Part (ii)
• Give [1] for stating a wage schedule
• Give [1] for showing that the chosen wage schedule satisfies ICC
• Give [1] for showing that it binds PC
– if doing the linear wage schedule solving for constants, these conditions hold by construction, so no need for any additional reasoning

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• Give a total of [2] for arguing that the proposed wage schedule must be profit maximizing. This is because the efficient effort is induced so that total surplus is 76 maximized [1], and because Tanya gets all surplus (worker
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2surplus) [1]. gets no extra
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(c) [5] Keep the assumptions from (b). Ivan suggests that he and Tanya should share ad the revenue lo n from the rice sales instead, with a share β for himself and a share ow 1 − β forDTanya (with 0 ≤ β ≤ 1). Show that this wage schedule cannot maximize
Tanya’s profit.

Solution: Revenues are 40y, so this schedule is w(y) = 40βy. Ivan now solves
40β
− 10x = 0 x x
√
It is easy to see that the x chosen by Ivan is x = 2 β, meaning that the
39
6 efficient effort level xE = 2 D 6 be chosen only if β = 1. But if β = 1, then will I e Tanya gets nothing so hermprofit are clearly below the previous one. If β < 1,
It
then Ivan’s effort level is inefficiently low, so the total surplus must be below that in the previous question. This means that even if Tanya could extract all surplus, her profits would be lower than before. Hence, her profits must be lower so it cannot maximize Tanya’s profit. max{(40β ln x) − 5x2 − 5}

⇒

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• Give [1] for noting that Ivan gets 40βy, or thatDTanya gets 40(1 − β)y rI de
(explicitly or implicitly) oa nl efficient level of effort only if w • Give [2] for showing that Ivan will choose
Do
β=1
• Give a total of [2] for arguing that the scheme cannot be profit maximizing: this is because either the efficient effort is not chosen [1], or Tanya fails to extract some of the surplus [1]
– other correct and clearly written arguments can also get full marks
39
66 and
7. Ashley and Mary-Kate are two utility maximizers who like to eat strawberries
D
chocolates. Ashley’s utility from eating s strawberries and c chocolates ismI e given by
It

uA (s, c) = s + ln c
Mary-Kate’s utility from eating s strawberries and c chocolates is given by uM (s, c) = 3s + 2c
Whenever they have access to strawberries and chocolates, Ashley and Mary-Kate may trade these with each other but not with anyone else. Since they have access to a sharp knife, they can trade chocolate and strawberry pieces of any size (i.e., there is no need to trade only whole strawberries and chocolates).
(a) [3] Suppose the total endowment in this pure exchange economy is 5 strawberries and 3 chocolates. Find the set of Pareto efficient allocations (i.e., the Pareto set).

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Solution: For interior solutions, we set M RSA = M RSB , so we have
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cA =

3
2

That is, forder interior solutions, the chocolates are shared equally while the the oa strawberries can be shared in any fashion. nl In addition, there are some corner solutions (these are easiest to show in an ow D
Edgeworth box). They are first those allocations where Ashley consumes no strawberries and between 0 and 3 chocolates, and second, where Mary-Kate
2
consumes no strawberries and between 0 and 3 chocolates.
2
Marks distribution
• The Pareto set can be stated either in words or algebra, or clearly marked
39
in an Edgeworth box
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• Give [1] for the interiorID of PE allocations set m te • Give [1] each for the two segments of corner solutions, for a total of [2]
I

(b) [3] Now suppose that Ashley initially has all of the 5 strawberries while MaryKate has all of the 3 chocolates. Normalize the price of chocolates to 1. In the competitive equilibrium of this pure exchange economy, what is the price of strawberries? How many strawberries does Mary-Kate eat? 76
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Solution: We normalize pc = 1 so that ps is the number of strawberries traded
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er for each chocolate. There are two main ways of ad solving the problem: l Diagram:
If the diagram has been drawno correctly, it will be obvious wn Do that we only need to care about the interior solutions and that the BC must coincide with Mary-Kate’s utility function. Hence, the price ratio must be the same as the negative of Mary-Kate’s MRS, or, ps 3
=
pc
2
9

3 which gives ps = 3/2 if we normalize pc = 1. Putting c on y-axis and s on
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x-axis (can do reverse), we see that (sA , 3/2) must be on the line D with slope mI t
−3/2 through (5, 0), giving sA = 4 so that Mary-Kate eats oneestrawberry.
I
Using market clearing conditions : As long as Ashley eats some strawberries, her demand for chocolates is given by cA (ps ) = ps and she spends the rest of her income buying strawberries. For Mary-Kate, eating one chocolate always gives the same utility as 3/2 strawberries, so she buys only chocolates if ps > 2/3 and only strawberries if ps < 2/3. If ps = 2/3, she’s indifferent.
From here, we can argue that the only price where the market can clear (on the interior) is such that ps = 2/3.
Note that Mary-Kate’s total wealth is 3, and since she spends 3/2 buying 3/2 chocolates, she has 3/2 left to spend on strawberries. Since they cost 2/3, she gets exactly one strawberry.
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• Give [3] if both price and number of strawberries provided is correct, as long as there is some evidence of any type of reasoning (it does not matter if it is super messy / sort of incomprehensible).
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Partial marks D
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• Give [1] er realizing that the budget line must coincide with Mary-Kate’s d for oa utility function, [1] for one strawberry, [1] for the correct price. nl w
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Noteothat if the Edgeworth box is drawn sufficiently neatly, it is possible to just read the quantity off the diagram. Don’t penalize if this is the “method” used. The diagram in (a) may be used.
(c) [4] Now suppose that the total endowment (5 strawberries and 3 chocolates) originally belonged to Ashley and Mary-Kate’s mother. The mother prefers that
Mary-Kate eats 3 strawberries. However, since the girls’ babysitter already gave
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2.5 strawberries to Ashley and only32.5 to Mary-Kate, the mother can only de66
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cide how to distribute the chocolates. How should the mother distribute the 3 mI te Mary-Kate eats 3 strawberries in the competitive chocolates to make sure that
I
equilibrium? er ad

lo wn Do

Solution: Again, it’s easiest to do this in a graph. The solution will be on the interior. The only possible eqm allocation where Mary-Kate eats 3 strawberries
3
3 is (3, 2 ) for Mary-Kate and (2, 2 ) for Ashley.
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We now to find an initial endowment of chocolates e 2for Ashley so that a
3 D budget a line with slope −3/2 goes through (2, erI and (2.5, e) (read from
)
2 ad Ashley’s origin). This can hold only for e = 0.75, i.e., the mother should give lo wn
Ashley 0.75 chocolates, while Mary-Kate gets 2.25 chocolates. o :

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Marks distribution
• Give [1] for writing down / labelling the only possible eqm allocation
• Give [2] for finding the number of chocolates to be given to one person, and [1] for the number of chocolates to be given to the second person
– This can be done graphically if the graph is sufficiently neat. Don’t
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penalize for this.
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– If not actually solve for the endowment, can still get [1] mark for saying
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8. Wen lives alone on an otherwise deserted tropical island. His time endowment is twenty-four hours per day. Every morning, he can effortlessly pick ten oysters on the beach. If he wants to eat more than ten oysters, then he needs to spend some time searching for them. If he spends h hours searching for oysters, then he will find h2 oysters, but this is only true for the first four hours of searching. If he searches for oysters for more than four hours in a day, then he can find 3 oysters per hour for the remaining hours. In the tropical heat, oysters spoil quickly and cannot be saved for the next day. In other words, they must be eaten on the day that they are found.
Wen’s utility from eating y oysters and enjoying

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hours of leisure is given by

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u(y, ) = ln(y − y ) + ln( − ¯)
¯

where y ≥ 0 and ¯ ≥ 0 are some real numbers. Note that for Wen’s utility function to
¯
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be well-defined, he needs to eat y > y oysters and enjoy > ¯ hours of leisure. Wen is
¯
ID a utility maximizer. der oa nl
(a) [1] What ow the economic interpretation of y and ¯?
¯
D is

Solution: They can be interpreted as minimum survival quantities. If Wen chooses only y = y , then his utility is negative infinity. However, if he chooses
¯
y > y , then his utility is a finite number. Same for .
¯
Marks: [1] for any vaguely economic explanation; [0] for a purely mathematical explanation
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(b) [4] In a diagram, draw Wen’s production possibilities set (that is, the set of feasible
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combinations of oysters and leisure). Label the diagram clearly.
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It the horizontal axis and y on the vertical axis (as in
Solution: Here, I put h on

the book). Alternatively, we may put on the horizontal axis, in which case the image is “mirrorred”. First note that if Wen doesn’t work, he still gets 10 oysters, so (0,10) is on the PPF. If Wen works 4 hours, he gets 10 + 42 = 26 oysters, so (4,26) is also on the PPF. The function connecting these points
6
97 is convex (piece of quadratic parabola). Finally, if Wen2works 24 hours, then he gets 10 + 42 + 3(20) = 86 oysters, so (24,86) isrID on the PPF. A line also de with slope 3 connects this point to (4,26). Theoproduction possibilities set is a nl the area below this line, in the first quadrant, and to the left of 24. ow D

Marks distribution
• Give [1] for marking (0,10).
• Give a total of [1] for the piece between (0,10) and (4,26): [0.5] for shape
(convex), [0.5] for the fact that it connects to (4,26).
• Give a total of [1] for the piece between (4,26) and (24,86): [0.5] for shape
(straight line), [0.5] for the fact that it connects to (24,86).
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• Give [1] for marking the set of feasible allocations, e.g., byID shading or em verbal description.
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(c) [5] Suppose y = 20 and ¯ = 6. How many oysters will Wen eat in a day?
¯
Solution: The solution may be either on the linear or the convex part. However, if you draw a utility curve of approximately the right shape (it’s a regular, well-behaved CD but asymptotic to y = 20 and h = 18) in the diagram, it is obvious that the solution must be on the linear part. The equation for the
PPF on the linear part, i.e., on the line through (4,26) and (24,86), can be written y = 3h + 14. Plugging the time constraint h + = 24 into the utility function, we can write u(y, h) = ln(y − 20) + ln(18 − h)

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The easiest way to solve this problem is to solve
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max{u(y, h) s.t. y = 3h + 14} h,y and verify thatDh ≥ 4. Plugging in the constraint, we have rI de oa maxwnl
{ln(3h − 6) + ln(18 − h)} ⇔ max{ln 3 + ln(h − 2) + ln(18 − h)} ho h
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with FOC

1
1
−
=0
h − 2 18 − h

⇒

h = 10

Since Wen works 10 hours, he eats 10 + 42 + 3(6) = 16 + 18 = 44 oysters.
Alternatively, we can use the tangency condition MRS = MRTS directly.
This gives
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1
y − 20
1
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18−h
mI e− 1 = − 18 − h = − 3 y−20 It
Now, plugging in y = 3h + 14, we can solve for h
(3h + 14) − 20
1
=
18 − h
3

⇒

h = 10
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Marks distribution
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• Give [2] for arguing that solution must be on ad linear part (either grapho the nl ically, or otherwise) w Do
• For a total of [2], give [1] for writing down the maximization problem and
[1] for the correct technology constraint OR [1] for the tangency condition and [1] for the correct technology constraint
• Give [1] for solving for y = 44
– if the PPF (constraint in the utility maximization problem) is not drawn correctly in (b) and this affects the calculations in this 9ques3
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tion, a maximum of [3] may still be given
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mI
– if the mistake in (b) significantly simplifies the probleme(e.g., PPF is t drawn with no kinks), a maximum of [2] may be givenI
– if write y = 3h + m where m = 14 but the PPF is drawn correctly in
(b), a maximum of [4] may be given

End of examination
Total pages: 16
Total marks: 100

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