Solutions to Exercises 1

A Short Course in Intermediate Microeconomics with Calculus

Solutions to Exercises1

c
2013
Roberto Serrano and Allan M. Feldman
All rights reserved

This file is intended to be used by instructors only.

1
We thank EeCheng Ong and Amy Serrano for their superb help in working out these solutions. We also thank
Rajiv Vohra for contributing some nice improvements to our previous version.
Solutions to Exercises 2

Chapter 2 Solutions

1.(a) Our consumer prefers a cup of coffee with one teaspoon (6 g) of sugar to a cup of coffee.
with two teaspoons (12 g) of sugar, i.e., 6 12. However, she is indifferent between a cup
of coffee with n grams of sugar and a cup of coffee with n + 1 grams, i.e., n ∼ n + 1. By

transitivity, if n ∼ n + 1 and n + 1 ∼ n + 2, then n ∼ n + 2. With repeated applications of

transitivity, if 6∼7, 7∼8, ..., and 11∼12, then 6∼12. However, 6∼12 contradicts our
first statement, 6 12.

1.(b)Vote 1:xversusy.
Person 1 votes for x, Person 2 votes for y, and Person 3 votes for x. Therefore x > y.

Vote 2: yversusz.
Person 1 votes for y, Person 2 votes for y, and Person 3 votes for z. Therefore, y is chosen over z.

Vote 3:xversusz.
Person 1 votes for x, Person 2 votes for z, and Person 3 votes for z. Therefore, z wins over x.

Now if x is related to y and y is related to z, then by transitivity we should have x related to z. But in Vote 3 we see

z x, which means majority voting violates transitivity.

2.(a) The indifference curve corresponding to u= 1 passes through the points (0.5,2), (1,1), and
(2,0.5). The indifference curve corresponding to u= 2 passes through the points (0.5,4),
(1,2), (2,1), and (4,0.5).

2.(b) The MRS equals 1 along the ray from the origin x 2=x1 and it equals 2 along the ray
from the origin2= 2x1 .

3.(a) The indifference curves are downward-sloping parallel lines with a slope of -1 and the
arrow pointing northeast.

3.(b) The indifference curves are upward-sloping with the arrow pointing northwest.
Solutions to Exercises 3

3.(c) The indifference curves are vertical with the arrow pointing to the right.

3.(d) The indifference curves are downward-sloping and convex with the arrow pointing north.
east.

4.(a) The indifference curves are horizontal; the consumer is neutral about x. 1and likes2 .

4.(b) The indifference curves are downward-sloping parallel lines with a slope of -1; the con-
sumer considers1andx2to be perfect substitutes.

4.(c) The indifference curves are L-shaped, with kinks along the ray from the origin. 2=x1 ; 1
2
the consumer considers1andx2to be perfect complements.

4.(d) The indifference curves are upward-sloping and convex (shaped like the right side of a U);
the consumer likes2 , but dislikes1 i.e., good 1 is bad for the consumer.

∂u(x 1 ,x2 ) ∂(3x12x24)

5.(a)MU1= ∂x 1
= ∂x 1
= 6x1 x24

∂u(x 1 ,2 ) ∂(3x12x24)
5.(b)MU 2= ∂x 2
= ∂x 2
= 12x12x23

MU1 6x1 x24 x2

5.(c)MRS= MU2
= = 2x1
12x12x23

5.(d) MRS= x2 = 4 =1
2x1 2·2

5.(e)MRS= x2 = 2 = 1
2x1 2.8 8
His MRShas decreased. As he spends more and more time fishing, he is increasingly loath
to give up hammock time for an extra hour of fishing.

5.(f) Last week, u(x 1 , x2 ) = 3x12 x42= 3·22·443·22·28= 3·210.

This week,u(x1 , x2= 3x12x42= 3·82·24= 3·26·243·210.
Thus, he is as happy this week as he was last week.
Solutions to Exercises 4

6.(a) The MRS is the amount of money I am willing to give up in exchange for working an
hour.

6.(b) Since work is a burden, I would need to receive a positive amount of money for every hour.
I work. Therefore, my indifference curves are upward sloping, and the slope of an upward
The sloping curve is positive. Given that MRS = - Indifference Curve Slope, the MRS should

be negative.

6.(c) If I’m working only an hour a day, a low hourly wage, say, $10, would be sufficient to
induce me to work another hour. If I’m already working 12 hours a day, I would need a
very high hourly wage, say, $100, to induce me to work another hour. Thus the slope of
the indifference curve is increasing as the hours of work increase. Given that MRS = -
Indifference Curve Slope, the MRS is decreasing as the hours of work increase.
Solutions to Exercises 5

Chapter 3 Solutions

1.(a) Let p 1= 2p1andp2 =p2 . 1

2
The new budget line equation is p. 1 x1+p 2 x2Mor 2p1 x1+p2 x2=M. 1
2
p1 p1
The slope of the budget line has changed frompto-p = -2p1
1
= -4p1
p2
.
2 2 2p2

1.(b) Let p 1= 2p1andM = 3M.

The new budget line equation is p. 1 x1+p2 x2=M or 2p1 x1+p2 x2= 3M.
p1
The slope of the budget line has changed from -pto− p p1 = -2p1
p2
.
2 2

2.(a) Budget constraint: 3x 1+ 2x2= 900. Horizontal intercept at 300 and vertical intercept at
450.

x2 2
p1 3
2.(b) Tangency condition: MRS= ⇔ 2x1 x2 = 2 ⇔ x2= 3x1
p2
Pluggingx2= 3x1into the budget constraint gives us the optimal consumption bundle,
(x∗1 , x∗2 ) = (100,300).

3.(a) George's budget constraint is p a a+pb b=M. Given two affordable bundles, we have two
equations:

3·10 +pb-30 = M

3.15 + pb-15 = M

Solving the two equations simultaneously yieldsb= 1 and M = 60.

3.(b) Since for George apples and bananas are perfect substitutes in a 1:1 ratio, he will spend
His entire allowance on the cheaper good. Thus he will consume 0 apples and 60 bananas.

4.(a) Thex 1intercept is 27, thex2intercept is 12, and the kink is at (20,2).

4.(b) For Peter, pumpkins and cider are perfect substitutes. His indifference curves are linear.
1 optimal consumption bundle is (0,12).
with a slope of− 3His
Solutions to Exercises 6

4.(c) Thex 1intercept is 11, thex2intercept is 4, and the kink is at (4,2).

4.(d) For Paul, pumpkins and cider are perfect complements. His indifference curves are L-
shaped, with kinks at (2,3), (4,6), etc. His optimal consumption bundle is (2,3).

5.(a) Budget constraint: c 1+ 1+π c2=M⇔c1+ 1.01 c2= 50 ⇔c1+c2= 50

1+i 1.01

5.(b) Tangency condition: MRS= 1+i c2 1.01

⇔ c1 = 1.01 ⇔ c2=c1
1+π
Pluggingc2=c1into the budget constraint gives us the optimal consumption bundle,
(c∗1 , c∗2(25,25).

5.(c) Budget constraint: c 1+ 1+π c2=M⇔c1+ 1.10 c250

1+i 1.00
Tangency condition: MRS= 1+i c2 1.00 1.00
⇔ c1 = 1.10 c2=
1+π 1.10 c1
Plugging2= 1.00 c1the budget constraint gives us the optimal consumption bundle,
1.10
(c∗1 , c∗2) = (25,22.73).

6.(a) Budget constraint: c1 + 1+π c2=M1+ 1 1.05 c2= 100+ 1 100

1+i 1+i M2⇔ c1+ 1.10 1.10
c 1+ 1.05 c2equals 190.91
1.10
Thec1 -intercept isM1+ 1 M2190.91, which shows the amount of consumption if
1+i
Sylvester chooses to consume everything today and nothing tomorrow.
Thec2 -intercept is 1 (1 + i)M1+M2 = 200, which shows the amount of consumption
1+π
if Sylvester chooses to consume nothing today and everything tomorrow.
The slope is− 1+π1+i 1.10 reflecting the relative price of current consumption.
=−1.05-1.048,
The zero savings point is M1 , 1+π =M(100,95.24),
2 which is where Sylvester neither saves
nor borrows, i.e., he consumes exactly his income in each period.

6.(b) Budget constraint: c 1+ 1.05 c2= 190.91

1.10
Tangency condition: MRS= 1+i 2c1 c2 2c2 1.10 1 1.10
1+π⇔ c21 = c1 = 1.05 c2= 2 1.05 c1
Plugging2= 1 1.10 c1into the budget constraint gives us the optimal consumption
2 1.05
bundle, (c∗1 , c∗2 ) = (127.27,66.67). Therefore, Sylvester is a borrower.
BundleSis en dessous et à droite du point de zéro économies. La courbe d'indifférence de Sylvester.

going through bundleSis tangent to the budget line.

Solutions to Exercises 7

6.(c) The budget line pivots counterclockwise at the zero savings point, and now has a slope of

-1.
Budget constraint: c1+ 1.05 c2= 100 + 1 100 ⇔c1+c2= 195.24
1.05 1.05
Tangency condition: MRS= 1+i 2c2 1.05 1
⇔ c1 = 1.05 ⇔ c2=c1 2
1+π
Plugging2=c1into12the budget constraint gives us the optimal consumption bundle,
(c∗1 , c∗2(130.16, 65.08).

6.(d) BundleS is below and to the right of the zero savings point, and also below and to the
right of bundleS. Sylvester's indifference curve going through bundleS is tangent to the
new budget line, which is less steep than the old budget line. His new indifference curve is
above his old indifference curve; he is better off than before.
Solutions to Exercises 8

Chapter 4 Solutions

1.(a) Budget constraint: p 1 x1+p2 x2=M

Tangency condition: MRS = p1 x2 p1 p1
p2 ⇔ x1 = p2 ⇔ x2= ( )x1 p2
p1
Pluggingx2= p2
x1into the budget constraint yields the following:

p1
p1 x1+p2 x1=M
p2

2p1 x1=M
M
x1=
2p1

1.(b) Good 1 is normal (an increase in M results in an increase in x)1 ) and ordinary (an increase
inp1results in a decrease in x1 Goods 1 and 2 are neither substitutes nor complements.
of one another (p2does not appear in the demand function for good 1).

2.(a) First, find the original consumption bundle.

Budget constraint: x1+x2= 10
p1 x2
Tangency condition: MRS = ⇔ x1 = 1 ⇔ x2=x1
p2
Thus the original consumption bundle is (x∗1 , x∗
2 ) = (5,5).

Next, find the new consumption bundle given p 12.5.

Budget constraint:5 x1+x
2 2
= 10
Tangency condition: MRS = p1x2 5 5
⇔ x1 = 2 ⇔ x2=x1 2
p2
Pluggingx2= 5x1into
2
the budget constraint gives us the new consumption bundle,
(x∗1∗ , x∗2∗(2,5).
Hence the total effect on the demanded amount of good 1 is 2−5 =−3 units.

2.(b) To calculate the substitution effect, we allow M to increase such that utility is unchanged.
We must identify the point where MRS = p1 or xx2 = 52, andu(x
p2 1 , x2 ) =x1 x2= 5·5 = 25.
1

Sincex2=x1=5225we can solve for x1 ;x1= 10. √

x1
√ units and the income effect is 2−10 units. √
Hence the substitution effect is 10−5
Solutions to Exercises 9

3. With the Giffen good on the horizontal axis, the Hicks substitution effect bundle is to the
southeast of the original bundle, and the final bundle is to the northwest of the original
bundle. See Solutions-graphs file.

4.(a) His budget constraint is 0.20x + 0.05y = 2.

Given his preferences, he must consume equal quantities of x and y, i.e., x = y.
Therefore, his budget constraint can be rewritten as 0.20x + 0.05x = 0.25x = 2.
Thus∗ =y∗ = 8.

4.(b) His budget constraint is now 0.25x + 0.08x = 0.33x = 2.

Thusx∗∗ =y∗∗ = 200 .
33
He will be paying 200 (0.05 + 0.03) = 16 in taxes.
33 33

4.(c) The demand functions are x = y = M .

px +py
The goods are normal (higher income results in higher consumption), ordinary (higher price
results in lower consumption), and complements of one another (higher price of one results
in lower consumption of the other).

5.(a) Sammy’s budget constraint is x + y = 2.

His tangency condition is MRS= px y = 1 ⇔ y = x.
⇔ x py
Thus his optimal consumption bundle is (x∗ ,and∗ ) = (1,1).

Sammy's budget constraint is now 2x + y = 2.

y px
His tangency condition is MRS= ⇔ x = 2 ⇔ y = 2x.
py
Plugging 2x into his budget constraint gives us his optimal consumption bundle.
(x∗∗,y∗∗) = 21,1 .

5.(c) In order for him to be as well off as he was originally, we must identify the point such that
MRS= px or y = 2, andu(x, y) = xy = 1·1 = 1. 1
Since y = 2x =, we can solve for
py x x
xandy;x= 2. In order√ to afford consumption bundle (x, y) =
1
√2
1 √
√ 2,2 ,

Sammy requires an allowance of 2x + y = 2. 1 √ 2 = 2 √2≈$2.83. His parents would

√ +
2
have to increase his allowance by $0.83.
Solutions to Exercises 10

5.(d) All the answers are the same because v is an order-preserving transformation of u.

6.(a) The x-intercept is 8, and the y-intercept is 5. The budget line is horizontal between (0,5)
and (3,5), and is downward-sloping with a slope of -1 beyond (3,5).

6.(b) His budget constraint is y = 5 for x ≤ 3 and x + y = 8 for x ≥ 3. We shall consider the

case for x ≥ 3.
px y + 3 = 1 ⇔ y = x - 3.
His tangency condition is MRS= ⇔ x
py
Plugging y = x - 3 into his budget constraint gives us his optimal consumption bundle.
(x∗ ,y∗ (5.5,2.5).
Solutions to Exercises 11

Chapter 5 Solutions

1. Budget constraint: pC + wL = 24w

Tangency condition: MRS= w C = w
p ⇔ L 1 C=wL
Plugging p = 1 and C = wL into the budget constraint yields the following L = 12.

Thus∗ =T−L∗ 24 minus 12 equals 12.

2. The budget line is downward-sloping between 0, wT+MandT,

p
Mand vertical atT. The
p
optimal bundle is T,M. See
p
Solutions-Graphs file.

3.(a) The budget line has a kink at the zero-savings point. The slope is steeper to the right of
the zero savings point, and flatter to its left.

3.(b) The budget line has a kink at the zero-savings point. This time the slope is flatter to
the right of the zero-savings point, and steeper to its left. An indifference curve has two
tangency points with the budget line, each one at either side of the zero-savings point.

4.(a) Budget line equation: c 1+ 1.05 c21.05

= 100 +1.05 100 1
is equivalent to c1+c2= 195.24
1 = 195.24, which shows the amount of consumption if they
Thec1 -intercept is 100 +1.05 100
chose to consume everything today and nothing tomorrow.
1
Thec2 -intercept is1.05 [(1.05)100+100] = 195.24, which shows the amount of consumption
if they chose to consume nothing today and everything tomorrow.
The slope is− 1.051.05
-1, which reflects the relative price of current consumption.
100
The zero savings point is 100,1.05(100,95.24), which is where they neither save nor
borrow, i.e., they consume their income each period.

4.(b) Mr. A's tangency condition: MRS = 1+i c 1.05

⇔ 2c1 = 1.05 c2= 2c1
2
1+π
Plugging2= 2c1into the budget constraint gives us Mr. A’s optimal consumption bundle,
(c∗1 , c∗2) = (65.08,130.16). Thus Mr. A is a lender.
Solutions to Exercises 12

Mr. B's tangency condition: MRS = 1+i 2c 1.05 1

⇔ c1 = 1.05 2=c1
2
1+π 2
1
Plugging2=c1into2 the budget constraint gives us Mr. B’s optimal consumption bundle,
(c∗1 , c∗2) = (130.16,65.08). Thus Mr. B is a borrower.

4.(c) Recall that the savings supply curve is s(i) = M 1−p1 c1 (i). Therefore, we need to derive
the demand for current consumption, c,1 (i).
First, rewrite the budget constraint.1 + 1105
. c = 100+ 100 as (1+i)c
+i 2 1+I
1
1 +1.05c2= 100(2+i)

For Mr. A, rewrite his tangency condition. c2 = 1+i as 1.05c2= 2(1 +i)c1 .
2c1 1.05
Plug 1.05c2= 2(1 +i)c1into the budget constraint and solve for c1 ;c1= 100 2+i .
3 1+i
Thus Mr. A’s savings supply curve isA (i) = 100− 100 2+i orsA (i) = 100 1+2i This
3 1+i 3 1+i
is an increasing function of i; s = 33.33 for i = 0 and s = 50 for i = 1. Mr. A is always a

lender.

For Mr. B, rewrite his tangency condition. 2c2 = 1+i 1

as 1.05c2(1 +i)c
c1 1.05 2 1.

Plug 1.05c2(1 +12 i)c1into the budget constraint and solve for c1 ;c1= 200
3
2+i
1+i
.
Thus Mr. B's savings supply curve isB (i) = 100− 200 2+i orsB (i) = 100−1+i This
3 1+i 3 1+i
is an increasing function of i; s = -33.33 for i = 0 and s = 0 for i = 1. For the most

relevant values of i, i.e., any i < 1, Mr. B is a borrower.

The aggregate savings supply curve isA (i) +sB 100 I , an upward-sloping curve
1+i
starting at the origin. Mr. A’s saving exceeds Mr. B’s borrowing; the economy saves
overall. See Solutions-Graphs file.

4.(d) Budget line equation: c 1+ 1.05 c21.10

= 100 +1.10 100 1

Mr. A’s tangency condition: c2 = 1.10 1.10

2c1 1.05 ⇔ c2= 2 1.05 c1
Plugging2= 2 1.10 c1into the budget constraint gives us Mr. A’s optimal consumption
1.05
bundle, (c∗1 , c∗2 (63.64, 133.33).
Mr. B’s tangency condition: 2c2 = 1.10 1 1.10
c1 1.05 ↔c2= 2 1.05 c1
Plugging2= 1 1.10 c1into the budget constraint gives us Mr. B’s optimal consumption
2 1.05
bundle, (c∗1 , c∗2 (127.27,66.67)
Mr. A's initial utility was 29.71, and his current utility is 30.09; Mr. A is better off than
before. Mr. B's initial utility was 29.71, and his current utility is 29.36; Mr. B is worse off.
Solutions to Exercises 13

than before. Lenders benefit from a rise in interest rates, while borrowers suffer.

5. One possible savings function in which the consumer switches from being a borrower to
a saver at a given interest rate. See Solutions-Graphs file. While there is a great deal of
arbitrariness in the shape of the savings curve, take into account the following observations.
Hint: Why must the savings supply curve be strictly increasing when the consumer is a
borrower, but not necessarily when he is a saver? Why can’t a saver ever become a borrower
in response to a rise in the interest rate?

6. The saver’s budget constraint is c1+ 1+π c21+i 1

=M1+ 1+IM2 A decrease in π causes the budget
line to rotate clockwise on the x-intercept while an increase in causes the budget line to
rotate clockwise on the zero savings point.
In both cases, the substitution effect causes c1to fall and2to rise, while the income effect
causes both1andc2to rise. In the first case, you can't predict the direction of change in
savings either for a borrower or a saver. In the latter case, it is ambiguous for a saver, but
a borrower will definitely borrow less.
Solutions to Exercises 14

Chapter 6 Solutions

1.(a) Leah's budget constraint is 4b + 2c = 200. Her optimal consumption bundle is (25, 50).
Her utility is 25·50 = 1,250.

1.(b) The price of cream rises to $2.50 a pint. Her budget constraint is now 4b + 2.50c = 200.
Her new consumption bundle is (25,40).

1.(c) Let the subsidy be such that the new price of berries is p. b-s. With the tax and the
subsidy, her budget constraint would be (4 −s)b+ 2.50c= 200. If her utility in (a) is
maintained, then bc= 1,250. With a Cobb-Douglas function, c ∗ = 200 1
· 2.50 = 40. Thus
2
b= 1,250 = 31.25. Plugging the values for b and c into the budget constraint gives us
40
(4−s)31.25 + 2.50·40 = 200 or s = 0.80. Thus the subsidy should be $0.80 a pint or 20
percent.

2.(a) Rachel's budget constraint is m + 3c = 45. Her optimal consumption bundle is (15,10).
Her utility is 15.102+ 100 = 1,600.

2.(b) Rachel's new budget constraint is m + 4c = 45 + R or m + 4c = 45 + c. With a Cobb-Douglas

function,m∗∗ = 1
45+c
· 1 andc
3
∗∗ = 2 45 and abovec Solving the two equations simultaneously gives
3 · 4
us the new consumption bundle, (18,9). Her utility is 18·92+ 100 = 1,558<1,600.

3. William views the two goods as perfect complements while Mary views them as perfect
substitutes. William is always made worse off by the tax, while Mary would be made worse
off by the tax only if the original price of good x were less than the price of good y.

4.(a) Louis's budget constraint is 32c+s= 80. His optimal consumption bundle is (2,16). His

utility is 10·24·16 = 2,560.

4.(b) Louis's new budget constraint is 16c + s = 80. His new consumption bundle is (4, 16).

new utility is 10.44·16 = 40,960.

Solutions to Exercises 15

4.(c) With Hicks, the hypothetical budget line has the same slope as the new budget line, but
is tangent to the original indifference curve. Therefore, the hypothetical bundle fulfills the
following:
4s = 16
c 1 s = 4c
u(ch, sh ) =u(c∗ , s∗ 10(c)h)4h 2,560
Solving the two equations simultaneously yields (ch , sh (2.2974,9.1896).
At the new prices (pc , ps) = (16,1), the hypothetical bundle (ch, sh ) costs 16·2.2974 +
9.1896 = 45.948, and the new bundle costs 80. Thus the income effect is 34.052.

5.(a) The couple's budget constraint is 100x+y= 33,000. Their optimal consumption bundle.

is (x∗ ,and∗ ) = (220,11,000) and their utility isu= 220211,000 = 5.324 · 108 The cost of
the program is $3,000.

5.(b) The couple’s budget constraint is 80x+y= 30,000. Their optimal consumption bundle is

(x∗ ,and∗ Utility is U = (250, 10,000) and their utility is U = 250210,000 = 6.25·108 The cost of the
program is 250·20 = $5,000.

Since the second program yields a higher utility than the first program, the couple will
choose the second program.

6. Before the policy, the consumer in Group solves the following problem:

maxu(xi , yi ) subject topx xI+py yi=Mi

Group A solves: maxu(xA , andA ) subject to 10xA+ 10 yearsA= 500

Group B solves: maxu(xB , andB ) subject to 10xB+ 10 yearsB= 400
Group C solves: maxu(xC , andC ) subject to 10xC+ 10 yearsC= 300
Clearly,x∗i=y∗ ∗ ∗ ∗ ∗ ∗ ∗
i ;xA =yA = 25,xB =yB = 20, andx C=yC = 15.

Post-policy, the consumer solves the following problem:

maxu(xI, andi ) subject to (px−sx )xi+ (py−sy )yi=Mi−Ti

Group A solves: maxu(xA , andA subject to 9xA+ 9yA500 minus 68

Group B solves: maxu(xB , yB ) subject to 9xB+ 9yB= 400−40
Solutions to Exercises 16

Group C solves: maxu(xC , andC ) subject to 9xC+ 9yC300 minus 12

Clearly,x∗∗
I =y∗∗ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗
i ;xA =yA = 24,xB =yB = 20, andxC =yC = 16.

Therefore, the welfare of the median consumer (Group B) is unchanged. Lower-income

consumers (Group C) are better off and higher-income consumers (Group A) are worse off.
Solutions to Exercises 17

Chapter 7 Solutions

1. Consider two points, (x1 , x2 ) and (x1 , x 2 ), such that u(x1 , x2 10 andu(x1 , x 2 ) = 5. The

vertical distance between the two points is x2-x 2 .

Rewriteu(x1 , x2 ) =v(x1 ) +x2= 10 asx2=−v(x1 ) + 10, and rewriteu(x1 , x 2 ) =v(x1 ) +

x 2= 5 asx 2=-v(x1 ) +5. Thus the vertical distance between the two points is x2-x 2= 5.

2. Let the old bundle bex∗ = (x∗ ∗

1 , x2 ), the new bundle bex
∗∗ = (x∗∗, x∗∗), and the hypothetical
1 2

bundle bey= (y1 , and2 ) Since indifference curves are parallel under quasilinear preferences,
yes directly above∗∗and the hypothetical bundle can be rewritten as y = (x∗∗
1 , and2 ).

The change in utility is u(x)∗∗)−u(y) = [v(x∗1∗) +x∗2∗ ]−[v(x∗1∗) +y2 ] =x∗∗

2 -y2 .
The change in dollars is [p1 x∗1∗ +p2 x∗2∗ ]−[p1 y1+p2 y2 ] = [p1 x∗∗ ∗∗ ∗∗
1 +p2 x2 ]−[p1 x 1 +p2 y2 ] =

p2 (x∗∗ ∗∗
2 -y2 ) =x2 -y2 .
Thus the change in utility equals the change in dollars.

3. From Figure 7.5, Consumer 1’s surplus = ab, and12 Consumer 2’s surplus = cd. Thus the12
1
sum of Consumer 1’s surplus and Consumer 2’s surplus = ab+cd. 1
2 2
Consumers’ surplus in the graph at the far right can be decomposed into two triangles with
1
areasabandcd. 1
Therefore 1
Consumers’ surplus = ab+cd= 1Consumer 1’s surplus +
2 2 2 2
Consumer 2’s surplus.

4.(a) Case 1: p = 0
$2.5 million 1 1
2 2

Government Revenue minus Cost = pXp-1,000,000 = 0 - 1,000,000 = -$1 million

Thus Net Social Benefit = $1.5 million.

Case 2:p= 5− √ 5
2
X= 1,000,000−200,000 5− √ 5= 100,000 5 + √5
2
Consumers' Surplus 1 √ 5− √ 5 = 25,000 5 + √5
2
=
2 100,000(5 + 5) · 5− 2
Solutions to Exercises 18

$1.309 million
Government Profit = 0
Thus Net Social Benefit = $1.309 million.

4.(b) The government solves the following problem:

pXpp(1,000,000 - 200,000p)
By the First Order Condition, ∂R = 1,000,000 − 400,000p = 0 and p = 2.5.
∂p

Case 3: p = 2.5
X= 1,000,000−200,000·2.5 = 500,000
$625,000 1
2
Government Revenue minus Cost = 2.5·500,000−1,000,000 = $250,000
Thus Net Social Benefit = $0.875 million.

4.(c) Solve the following problem:

max Net Social Benefit = Consumers’ Surplus + Government Revenue minus Cost

1
=X2p (5−p) +pXp-1,000,000
1
= (1,000,000−200,000p)(5−p) +p(1,000,000−200,000p)−1,000,000.
2

Simplifying this expression, it is easy to see that this function is maximized at p = 0.

5. Given the demand function x1= M andM= 18,x1= 9 andp1. 9 = 1,x = 9.

Atp
2p1 p1 x 1 1
1

Atp1= 2.25,x1= 4. Thus whenp1rises from 1 to 2.25:

x1 =9 x1 =9 9
Loss in Consumer’s surplus = p1 (x1 )dx1= dx1
x1 =4 x1 =4 x1
9
= [9 ln(x1 )]49= 9 ln = 7.2984.
4
Solutions to Exercises 19

6. Correction. There is an error in the textbook, where Carter's utility function is shown as
u(x, y) = 10x + x3The
1
3
correct utility function is u(x, y) = 10x − x3+y. Also, 1
3
assume throughout this problem that py= 1, and that Carter’s income is at least 12.
Here is the solution for the corrected problem:

6.(a) Carter's tangency condition is MRS = 10 - x2 = px

=px Therefore his demand function
1 py
forxisx = 10 - px
py
= √ 10 minus px . Whenpx= 1, he consumes x = 3. (He also consumes
M−xunits of goody.)

6.(b) His inverse demand function for x is p x= 10−x2 . Whenpx = 1 and Carter is consuming
3 units of x, his consumer’s surplus from his consumption of x is
x=3
Consumer’s surplus = px (x)dx - pxx
x=0

x=3 3 3 1
= 10 minus x2dx - 3 = 10x - x -3 = 21 3- 3 = 18.
x=0 0

6.(c) Whenp xrises to 6 whileyremains at 1, Carter will consume x = 10 - px= 4 √= 2. √

(This will cost)x x= 12, which is why we assumed that M≥12.) His consumer’s surplus
from his consumption of this is now

x=2
Consumer's surplus 10 minus x2dx−6·2
x=0

2
1 8 8 16
= 10x−x 33-12 = 20 - -12 = 8 - 3 = .
0
3 3
Solutions to Exercises 20

Chapter 8 Solutions

1.(a) dy 1
=x−1/2
dx 2
d2 y
dx2
=−4 x1−3/2<0 forx= 0

1.(b)y=x 1/2 x=y2

⇔
C(y) = wx = 1·y 2=y2
C(y) 2
AC(y) =
y
= yy =y
MC(y) = dC(y)
dy
= 2y

1.(c) The supply curve is the MC(y) curve if p ≥ min AC(y), and MC(y) is rising.
First, determine minAC(y): min AC(y) = AC(0) = 0.
dMC(y)
Second, check whether MC(y) is rising: dy
= 2 >0.
Thus for p ≥ 0, the supply curve is given by p = MC(y) ⇔ p = 2y ⇔ y∗ (p) = p. 1
2

2 1 1 2
1.(d)π=p·y−w·x(y) =p 21 p−w 21p= 10 -10^(-1)·10 = 25 2 2

3
2.(a)y= 5x1/3-30 x=51 y + 6
⇔
3 3
C(y) = wx = 1 51 y + 6 = 51 y + 6
3
AC(y) = C(y) = ( 5y + 6 )
1

y y
dC(y) 1 2
MC(y) = = 3
5y +6
dy 5

2.(b) The supply curve is the MC(y) curve, if p ≥ min AC(y), and MC(y) is rising.
dAC(y)
First, determine minAC(y): dy
= 0 ⇔y= 15. Therefore, minAC(y) =AC(15) =
48.6.
dMC(y) 1
5y + 6 > 0.
Second, check whether MC(y) is rising: = 6
dy 25
Thus for p < 48.6, the supply curve is y = 0.
3 2
And for p ≥ 48.6, the supply curve is given by p = MC(y) ⇔ p = 1 ∗
5 5 y + 6 (p) =
5 5p
3 -30.
Solutions to Exercises 21

3.(a)x=y2
dx = 2y
dy
d2 x =2>0
dy2

)x( fd
3.(b)MP(x) = dx
= 2 √1 x
f(x) √
AP(x) =
x
= xx = √1
x

3.(c)VMP(x) = p·MP(x) = 10· 1 = 5

2√ x √x

V AP(x) = p · AP(x) = 10
√x

3.(d) The input demand curve is the VMP(x) curve, if max V AP(x) ≥ w, and VMP(x) is
falling.
First, determine maxV AP(x): max V AP(x) = V AP(1) = 10.
dVMP(x)
Second, check whether VMP(x) is falling: dx
=−2x5−3/2<0 forx= 0.
Thus forw >10, the input demand curve is x= 0.
And for w≤10, the input demand curve is given by w=VMP(x)⇔w= 5
√x ⇔
x∗ (w) 25
w2
.

3.(e)π = p·y(x) - w·x = p 25 1/2 25 = 10 25 1/2 25 = 25

w2 -w w2 12 -1 12

)x( fd
4.(a)MP(x) = dx
= 13 2
√3x +1
AP(x) = f(x)
x
= √31 + 1
3
x

4.(b)VMP(x) = p · MP(x) = 6 · 1 2 +1=2 2 +1

3 √3x √3x

In AP(x) = p·AP(x) = 6 1
√3x + 1 =2 3
√3x +1
3

4.(c) The input demand curve is the VMP(x) curve if max V AP(x) ≥ w, and VMP(x) is
falling.
First, determine maxV AP(x): max V AP(x) = V AP(1) = 8.
dVMP(x)
Second, check whether VMP(x) is falling: dx
=−3x4−4/3<0 forx= 0.
Thus for w > 8, the input demand curve is x = 0.
And for w ≤ 8, the input demand curve is given by w = VMP(x) ⇔ w = 2 2 +1
√3x ⇔
3
x∗ (w) = 4
w−2
forw >2. If w ≤ 2, the firm would demand an infinite amount of input.
Solutions to Exercises 22

5. In a (y1 , and2 In the first quadrant, a typical isofactor curve is concave to the origin (using the same
amount of input, the more additional units of output1the firm wants to produce requires
to give up more units of output2 The isorevenue curves are downward-sloping straight
lines of slope -p1 /p2 The solution to the revenue maximization problem, conditional on a
The level of input x is found at the tangency of the highest possible isorevenue line with the fixed.
isofactor curve. The solution to this revenue maximization problem yields the conditional
output supply functions1 (1 , p2 , x) andy2 (p1 , p2 Finally, the profit maximization
the problem is thus written:

maxπ=p
x 1·y1 (p1 , p2 , x) +p2·y2 (p1 , p2 , x)−wx

Solving the maximization problem yields the input demand function, x(p1 , p2

6.(a)C(y1 , y2 ) =wx=y12+y22+y1 y2
∂C(y)
M C 1 (y1 ) = ∂y 1
= 2y1+y2
∂C(y)
M C 2 (y2 ) = ∂y 2
= 2y2+y1

6.(b) The supply curves are the MC i (yi ) curves, subject to the non-negative profit condition.
The supply curves are given by p1= 2y1+y2andp2= 2y2+y1 Solving the two equations
1 1
simultaneously, the supply curves are1∗ (p1 , p2) = (2p1- p32 ) andy2∗ (p1 , p2(2p2- p 1 ). 3

6.(c)y 1∗ (1,1) = (2·1−1)

1
3
= 1
3
y2∗ (1,1) = (2·1−1)
1
3
= 1
3
1 1 1
Thus π = p1 y1+p2 y2-wx= +-(31 )2+(13 )2+(1 1
3 3)( 3 ) = . 3

6.(d)y 1∗ (2·1−2) = 130

y2∗ (1,2) = (2·2−1)
1
3
=1
Thus π = p1 y1+p2 y2-wx = 0 + 2 - 0 2+ 12+ 0 = 1.
Solutions to Exercises 23

Chapter 9 Solutions

Isoquants represent a firm's production of output, and output is a cardinal measure.

Indifference curves represent a consumer’s utility level, and utility is an ordinal measure.

2. Profit-maximizing firms produce up to the point where Price = Marginal Cost. If the price
rises, then the marginal cost must rise to maintain the equality. Since the firm produces
where the marginal cost curve is rising, a rise in the marginal cost implies a rise in output.

3.(a) Tangency condition: TRS= w1 x2 1

⇔ x1 = 2 ⇔ x1= 2x2
w2
Pluggingx1= 2x2into the production function gives us the conditional factor demands,
√ 2 andx∗(y) =
x∗(y) = 2y y2
.
1 2 √
2

3.(b)C(y) =w 1 x1+w2 x2= 1·2y+ 2· √ 2 y2

√ = 2 √2y2
2

3.(c) The supply curve is the MC(y) curve, if p ≥ min AC(y), and MC(y) is rising.

First, derive AC(y) and MC(y).

√
AC(y) = C(y) = 2 2y2 = 2 √2y
y y
MC(y) = dC(y)
dy
= 4 √2y
Next, determine minAC(y): AC(0) = 0.
Then, check whether MC(y) is rising: dMC(y)
dy
= 4 √2>0.
Thus for p ≥ 0, the long-run supply curve is given by p = MC(y) ⇔ p = 4 2y ⇔ √
p
y ∗ (p) = √ .
4 2

4.(a) Since f(tL, tK) = tLtK = t2 LK tLK =tf(L, K) for t > 1, this technology shows
10 10 > 10
increasing returns to scale.
The isoquants are symmetric hyperbolas; the isoquants get closer and closer to each other
away from the origin.
Solutions to Exercises 24

4.(b) Tangency condition: TRS= wL K 10

⇔ L = 100 L = 10K
wK
Plugging L = 10K into the production function gives us the conditional factor demands.
L∗ (y) = 10 √
yandK∗ (y) = y. √

4.(c)C(y) = w L L+wK K = 10 · 10y + 100 ·√y = 200 y √ √

√ ∗ (1) = 1 = 1, and √C(1) = 200 1 = 200.

4.(d) If y = 1, then L ∗ (1) = 10 1 = 10,K √
√
If y = 2, then L∗ (2) = 10 2,K∗ (2) √ = 200 2.
= 2, and C(2) √

C(y) 200 √ y 200

4.(e)AC(y) = y
= y
= √y

MC(y) = dC(y) 100

dy
= √y

Both AC(y) and MC(y) are decreasing hyperbolas, and MC(y) < AC(y). There is no
long-run supply curve.

5.(a) Since the inputs are perfect substitutes, and input 2 is cheaper than input 1, the cost
minimizing technique is to use only input 2. Hence, the conditional factor demands are
x∗1 (y) = 0 and x∗2 (y) = y.
ThusC(y) = w1 x1+w2 x2= 2·0 + 1·y=y.

5.(b) First, derive AC(y) and MC(y).

C(y)
AC(y) =
y
= yy = 1
MC(y) = dC(y)
dy
=1
Therefore, the long-run supply curve is y = 0 for p < 1, and y ∈ [0, ∞) for p = 1.

5.(c) Ifw 2= 2, the firm's conditional factor demands are x1∈[0, y] and x2∈[0, y] such that
x1+x2=y.
2y
Thus C(y) = 2x1 (y) + 2x2 (y) = 2[x1 (y) + x2 (y)] = 2y, AC(y) = y
= 2, and MC(y) = 2.
Therefore, the long-run supply curve is y = 0 for p < 2, and y ∈ [0, ∞) for p = 2.

wi xj 1
6.(a) Tangency condition: TRS x ,x i j
= wj ⇔ xi
= 1 ⇔ x=x
I jfor i = 1, 2, 3, j = 1, 2, 3,
i=j.
Solutions to Exercises 25

Pluggingi=xjinto the production function gives us the conditional factor demands,

x∗1 (y) = x∗ ∗
2 (y) = x3 (y) = y
5/3 .

6.(b)C(y) = w 1 x1+w2 x2+w3 x3= 1·y5/3+ 1·y5/3+ 1·y5/3= 3y5/3

6.(c) The supply curve is the MC(y) curve, if p ≥ min AC(y), and MC(y) is rising.

First, derive AC(y) and MC(y).

C(y) 5/3
AC(y) =
y
= 3yy = 3y2/3
MC(y) = dC(y)
dy
= 5y2/3
Next, determine minAC(y): AC(0) = 0.
dMC(y) 10 >0 fory= 0.
Then, check whether MC(y) is rising: dy
= 3 √3y

Thus for p ≥ 0, the long-run supply curve is given by p = MC(y) ⇔ p = 5y2/3⇔ y∗ (p) =
( 5p )3/2 .
Solutions to Exercises 26

Chapter 10 Solutions

f ( x 1,x2 ) [243+1(x−9)
1
3 ]x 2 [243+1(x−9)
1
3 ]
1.(a)AP(x1 1) = x1
= 3
x1
= 3
x1
∂f(x 1,x2 )
M P ( x 1 1) = ∂x 1 = (x1-9)2 x2= (x1-9)2

f ( x 1 ,x2 ) 1
1.(b)AP(x 2 |x10) = x2 = 243 + (x310-9)3
∂f(x 1,x2 ) 1
M P ( x 2 |x10) = ∂x 2 = 243 + (x310-9)3

1.(c) In (a), AP(x 1 andMP(x1 vary with x1 . In (b), AP(x2 |x10) =MP(x2 |x01), and both
average product and marginal product curves are constant for a given level of input 1.

2.(a) Ifw 1rises,x∗1falls, and falls.

2.(b) Ifw 2falls,x∗1remains unchanged, and rises.

2.(c) Ifprises,x ∗1rises, and rises.

1/4 1/4 1/4 1/4

x2 =x1 1 , thereforex1=y4 .
3.(a) Since y = x1
ThereforeCS (y) = w1 x1+w2 x2= 1·y4+ 2·1 =y4+ 2.

3.(b) The supply curve is the MC S (y) curve, ifp≥minAV C(y), andMCS (y) is rising.
First, derive AV C(y) and MCS(y).
AVC(y) = InC(y)
= y4 =y3
y y
dCS(y)
MC S (y) = dy
= 4y3
Next, determine minAV C(y):AC(0) = 0.
dMCS (y)
Then, check whether MCS (y) is rising: dy
= 12y2greater than or equal to 0.

Thus for p ≥ 0, the short-run supply curve is given by p = MC. S (y) ⇔ p = 4y3
⇔
y ∗ (p) = p 1/3 .
4

4.(a) Since y = LK = L·1 , therefore L = 10y.

10 10
ThereforeCS (y) = w1 x12 x2= 10·10y + 100·1 = 100y + 100.
Solutions to Exercises 27

C S(y) 100
4.(b)ATC(y) = y
= 100 + y
AVC(y) = VC(y) 100y
y
= y
= 100
dCS(y)
MC S (y) = dy
= 100
ATC(y) is a decreasing hyperbola that approaches 100. Both AVC(y) and MCS (y) are
constant at 100. Therefore the short-run supply curve is infinitely elastic at p= 100.

5.(a)ATC(y) C S(y)
= 100 + 10 - 2y + y2
y y
AVC(y) = VC(y)
= 10y - 2y
2 +y3
= 10 - 2y + y2
y y
MC S (y) = dCS(y) = 10 - 4y + 3y2
dy
ATC(y) is a parabola starting at (0,∞) with a minimum around y= 4. AVC(y) is a
Parabola starting at (0,10) with a minimum at (1,9).S (y) is a parabola starting at
2 that MC(y) intersects AVC(y) at min AVC(y)
(0,10) with a minimum aty= . Notice
3
and ATC(y) at min ATC(y).

5.(b) The supply curve is the MC S (y) curve, ifp≥minAV C(y), andMCS (y) is rising.
Thus for p < 9, the short-run supply curve is y = 0.

And for p≥9, the short-run supply curve is given by p=MCS (y) ⇔ p = 10 - 4y + 3y2⇔
√ 3p - 26
y ∗ (p) = 2
3
+ 3
.

6. If the output price is below minATC(y), then profit is negative. However, if the output
The price is below minATC(y) but above minAVC(y), the firm recoups some of the fixed cost.

if it produces output. Therefore, in the short run, the firm produces output as long as the
output price is above minAV C(y).
Solutions to Exercises 28

Chapter 11 Solutions

1.(a) Assuming coffee is a normal good, an increase in consumers' income shifts the demand
curve right. Hence, the equilibrium price and quantity both increase.

1.(b) Un aumento en los precios de los factores incrementa los costos marginales; la curva de oferta se desplaza hacia la izquierda. Por lo tanto,

the equilibrium price increases and the quantity decreases.

1.(c) A technological improvement lowers marginal costs; the supply curve shifts right. Hence,
the equilibrium price decreases and the quantity increases.

2.(a) Tangency condition: TRS= w1 x2 4

⇔ x1 = 1 ⇔ x2= 4x1
w2
Pluggingx2= 4x1into the production function gives us the conditional factor demands,
x∗1 (y) = yandx
1
2
∗ (y) = 2y.
2
1 4y.
Therefore C(y) = w1 x1+w2 x2= 4·y + 1·2y =
2
Since AC(y) = MC(y) = 4, the long run supply curve is given by p = MC(y) ⇔ p = 4.

2.(b) Since p = 4, y = 1,000,000 - 1,000 · 4 = 996,000.

3.(a) Tangency condition: TRS= wK L 256

⇔ 2K = 1 ⇔ L= 512K
wL
Plugging L = 512K into the production function gives us the conditional factor demands,
K ∗ (h) =64handL
1 ∗ (h) = 8h.

Therefore C(h) = wK K+w LL = 256·64 h + 1·8h 1= 12h.

Since AC(y) = MC(y) = 12, the long-run supply curve is given by p = MC(y) ⇔ p = 12.

The individual and market supply are infinitely elastic at p = 12, that is, for p < 12, hS= 0,

and for p = 12, the supply is any non-negative amount.

3.(b) The competitive equilibrium is the intersection of the market demand and market supply.
Thus = 12 and∗ 3,000. Each firm earns zero profit because of constant returns to
scale. The individual amount produced by each firm is therefore indeterminate; all we know
is the aggregate amount produced by the industry, that is, 3,000 units.
Solutions to Exercises 29

3.(c) Under the new competitive equilibrium, p= 12 and h ∗∗ = 2,000. Again, each firm earns
zero profit and is indifferent between producing any non-negative amount.
Since profits are zero, firms are indifferent between staying in or out of the market.
fore, the number of producers is also indeterminate. The total amount of good exchanged
in the market, however, is perfectly determined by demand.

4.(a) The representative firm's supply curve is the MC(y) curve, if p ≥ min AC(y), and MC(y)

is rising.
First, derive AC(y) and MC(y).
AC(y) = C(y) 99 1
y
= 2y −y+y
2
2

MC(y) = dC(y) =-y + 3y 2

dy
dAC(y) 99 1
Next, determine minAC(y): dy =−2y2-+ 2y=
2 0 ⇔ y= 3.
Therefore, minAC(y) = 99 1
2·3 − 2· 3 + 32= 24.
dMC(y) =−1 + 6y > 0 at y = 3.
Then, check whether MC(y) is rising: dy
Thus for p < 24, the representative firm's supply curve is y = 0.
And for p ≥ 24, the representative firm's supply curve is given by p = MC(y) ⇔ p =

−y + 3y2⇔ y∗ (p)
1
6
1 + 1√+ 12p.

4.(b) In the long run, the number of firms adjusts to drive the market to the zero-profit equilibrium.

librium. The long-run market supply curve is horizontal at p = min AC(y) = 24. Each
firm producesi= 1
6
1 + 1√+ 12·24 = 3. At equilibrium, market supply equals market
demand:y= 1,140−10·24 = 900. Therefore, there are 300 firms in the market.

1 1
5.(a)π(y) = R(y) - C(y) = 140y -2 y 2+ 40y + 2,450 = 100y - y 2-2,450 2

dπ(y)
5.(b) dy
= 100−y= 0 ⇔y∗ = 100

1
5.(c)π(100) = 100·100− · 10022-2,450 = 2,550
Solutions to Exercises 30

5.(d) Since Dakota is earning positive profits, new firms will enter the industry in the long run.
Therefore, the number of firms in the rocking horse industry will rise.

6.(a) In the long run, Dakota will produce where ∗∗ =MC(y∗∗) =AC(y∗∗) = minAC(y).
First, derive AC(y) and MC(y).
AC(y) = C(y) 2,450
y
=y+12 40 + y
MC(y) = dC(y)
dy
=y + 40
Then, set MC(y) = AC(y) and solve for y∗∗.

MC(y) = AC(y)

1 2,450
y + 40 = y + 40 +
2 y
y 2= 2,450·2

y = 4,900

y ∗∗ = 70

Alternatively, find minAC(y) and solve for y.∗∗ .

dAC(y) 2,450
minAC(y) dy
= 1
2 − y2
= 0 ⇔y∗∗ = 70

2,450
6.(b) In the long run, p ∗∗ =AC(70) = ·70 +12 40 + 70
110.
Alternatively,∗∗ =MC(70) = 70 + 40 = 110.

6.(c)π ∗∗ =π(70) = 110·70− 1

2 ·702+ 40·70 + 2,450 = 0

6.(d) The long-run market supply curve is infinitely elastic at p ∗∗ = 30. Recall that the less
elastic side of the market bears the greater burden of the tax. Therefore, the tax burden
is borne entirely by the consumers. The firm’s profits are unaffected by the tax. Note that
Dakota may drop out of the market, but whether it stays in or out, its profit will be zero.
Solutions to Exercises 31

Chapter 12 Solutions

p
The markup is the fractional amount by which price exceeds cost, and equals MC -1 =
1 1
1−1/ -1 = −1As elasticity decreases, the markup increases. Therefore, the markup (and
thus price) will be higher for the group with the lower elasticity of demand, which is Group
B.

2. Given R(h) = p·h = (50 50 h

- )h,MR(h) h
= 50−25 Given C(h) = 12h, MC(h) = 12.
Corleone will produce where:
MR(h) = MC(h)
h
50− = 12
25
h= 950

Therefore, p = 50 - 950 = 31, and π = R(h) - C(h) = 31·950 - 12·950 = 18,050.

50

1
3. SinceR1=p1·y1= (100−y1 )y11= 100−2y1 , and sinceR2=p2·y2(75 - y2 )y2 , 2
M i s t e r 2= 75−y2 . SinceC(y1 , and2 ) = (y1+y2 )2 ,MC1=MC2= 2(y1+y2 The monopolist
will produce whereMR1=MC1andMR2=MC2 , giving us two equations and two
unknowns:

M R 1=MC1

100 minus 2y1= 2(y1+y2 )

50 = 2y1+y2 (1)

M R 2=MC2

75 - y 2= 2(y1+y2 )

75 = 2y1+ 3y2 (2)

Subtracting equation (1) from equation (2), we get 2y.2= 25⇔y2= 12.50. Plugging
y212.50 into equation (1) gives us 50 = 2y1+ 12.50⇔y118.75. Finally, pluggingy1
Solutions to Exercises 32

intop1 (y1 ) andy2intop2 (y2 ), we find that p1100 minus 18.75 equals 81.25 and p2= 75−2 (12.50) = 1

68.75

4.(a) Given R B=pB·xB(100 - xB )xB MRB= 100−2xB , and givenRF=pF ·xF =

(30−2xF)xF,MRF= 30 - 4xF. Given C(xB , xF) =xB+xFB=MCF1. The
The monopolist will produce where MRB=MCBandMRF=MCF:

M R B=MCB

100−2xB= 1

xB49.50

M i s t e r F=MCF

30−4xF= 1

xF7.25

Therefore, pB= 100−49.50 = 50.50 andpF 30 - 2(7.25) = 15.50. Total profits are

π=RB+RF-C(xB F50.50(49.50) + 15.50(7.25) - (49.50 + 7.25) = 2,555.37.

4.(b) The businesses’ surplus is the area of the triangle below the demand curve. Band above
the horizontal line atpB= 50.50. The families’ surplus is the area of the triangle below
the demand curve xFand above the horizontal line atpF= 15.50. The sum of the two
Surpluses is the consumers' surplus.
1
Consumers' surplus = (49.50)(100−50.50) 1
+ (7.25)(30−15.50) = 1,277.68.
2 2
The producer's surplus is total profits.
π = 2,555.37

4.(c) The market demand from aggregating the two groups is:

⎧ 0 ifp≥100
⎪⎪
x= ⎪⎪⎪⎨ 100−p if 100 ≥ p ≥ 30

⎪⎪ 115−pif330 ≥p≥0
⎪⎪ 2
⎪⎩
Solutions to Exercises 33

The kink occurs at x = 70. The marginal revenue curves are obtained as follows:
For x < 70, R(x) = p·x = (100−x)x, so MR(x) = 100−2x.
Forx≥70,R(x)=p·x= 230 minus x, soMR(x)
2 times x = 230−4x .
3 3
Since MC(x) = 1, and the monopolist's solution is given by MR(x) = MC(x), we have
either 100−2x= 1⇔x= 49.50<70, or 230 minus=4x1⇔x= 227 70 which contradicts x ≥
3 4
70. Hence, the solution isx= 49.50,p= 50.50, andπ= 50.50(49.50)−49.50 = 2,450.25.

1
4.(d) Consumers' surplus = (49.50)(100−50.50) + 0 = 1,225.12.
2
Producer’s surplus =π= 2,450.25.
Society is worse off after this change is introduced; families cannot afford telephone services,
and Horizon Telephone earns lower profits.

1 100 - y. Given C(y) = y.2+ 10y,MC(y) =

5.(a) Given R(y) = p·y = (100 - y)y, MR(y) =
2
2y + 10. Sue will produce where:

MR(y) = MC(y)

100 - y = 2y + 10

3y = 90

y ∗ = 30

5.(b)p ∗ = 100 - · 3012= 85.

5.(c)π ∗ =R(y∗ )−C(y∗ ) = 85·30−(302+ 10·30) = 1,350.

1
5.(d) Consumers’ surplus = (30)(100−85) = 225.
2

6.(a) If Sue is forced to price at marginal cost:

p(y) = MC(y)

1
100 - y = 2y + 10
2
Solutions to Exercises 34

5
y= 90
2
y ∗∗ = 36 >30 =y∗

6.(b)p ∗∗ = 100− · 3612 = 82 <85 =p∗ .

6.(c)π ∗∗ =R(y∗∗)−C(y∗∗ = 82·36−(362+ 10·36) = 1,296<1,350 =π∗ .

1
6.(d) Consumers’ surplus = (36)(100−82) = 324 >225.
2

6.(e) Total welfare is 1,296 + 324 = 1,620 which is greater than total welfare in Question 5.
(1,350 + 225 = 1,575).
Solutions to Exercises 35

Chapter 13 Solutions

1 profit maximization problems

1.(a) The inverse market demand curve is p(h) = 60− 20 h. Their
are:

maxπ1=p(h1 , h2 )h1-C1 (h1 )

h1

1 1
h-h2h1-10h
= 60 - 120 201

maxπ
h 2=p(h1 , h2 )h2-C2 (h2 )
2

1 1
= 60 −h20
1-h2h2-20h
20 2

To derive their reaction functions, h1 (h2 ) blind2 (h1 ), we solve the maximization problems.

∂π 1 1 1
= 60-h1Invalid
10 input2-10
20 =0
∂h 1
1
h1= 500−h2
2

∂π 2 1 1
= 60−h1-h
202 -20 = 0
10
∂h 2
1
h2= 400-h1
2

1.(b) The Cournot equilibrium is the intersection of the reaction functions. Solving the reaction
functions simultaneously, we find that h∗1= 400 blind∗
2 = 200. The market price is

p∗ = 60 −(400
1 + 200) = 30. Individual profits are π∗ 30·400−10·400 = 8,000
20 1

andπ2∗ 30·200−20·200 = 2,000.

2.(a) The cartel solution is obtained from the joint maximization problem:

maxπ=p(h1 , h21+h2 )−C1 (h1 )−C2 (h2 )

h1 ,h2

1 1
= 60 − h1-h2(h1+h2 -10h1-20h2
20 20
Solutions to Exercises 36

1 1 1
22-h h
= 50h1+ 40h2-h12 1202 20 10

The first order conditions obtained from differentiation are:

∂π 1 1
= 50−h1-h
102= 0 10
∂h1

h1+h2= 500

∂π 1 1
= 40−h1-h
102= 0 10
∂h2
h1+h2= 400

Clearly, the system does not have an interior solution. Hence, either1= 0 orh2= 0.
Given the cost differential, it must be that2= 0. For completeness, we check both cases.
Case 1:h1= 0:
1
maxπ= 60 −h2h2-20h
20 2
h2

∂π 1
= 60 - h2-20
∂h 2 10= 0
h2= 400

The market price is p = 60 - · 400 = 140. Profits are π = 40 · 400 - 20 · 400 = 8,000.
20
Case 2:h2= 0:
1
maxπ = 60 - h1h1-10h1
h1 20

∂π 1
= 60−h1-10
10 = 0
∂h 2
h1= 500

The market price is p = 60 - · 500 = 135. Profits are π = 35 · 500 - 10 · 500 = 12,500.
20
Therefore, profits are maximized when∗1= 500,h∗ ∗ ∗
2 = 0, andp = 35. Then, π 12,500

and according to the agreement,1∗ =·12,50035 = 7,500 and π2∗ =·12,500

2
5
= 5,000.
Solutions to Exercises 37

2.(b) Case 1: Suppose Corleone breaks the agreement. If Chung sticks to the agreement,2= 0.
Corleone produces1= 500 and earns profits π1= 12,500, which exceeds π1∗ = 7,500 under
the agreement. Thus, Corleone has an incentive to break the agreement. Furthermore, note
that Corleone is better off under the simultaneous quantity setting model in Question 1
than he is under the agreement. In this sense, Corleone has strong reasons to complain.
about the split of cartel profits.
Case 2: Suppose Chung breaks the agreement. If Corleone sticks to the agreement,1=
500. Chung solves the following problem:

1 1
maxπ2= 60 − 500 minus h2h2-20h2
h2 20 20

∂π 2 1
= 35−h2-20
10 = 0
∂h 2
h2= 150

The market price is p = 60 - 1 1

20 500− 20 -150 = 27.50. Chung's profits are π2=
27.50·150−20·150 = 1,125, which is lower than π2∗ = 5,000 under the agreement. Thus,
Chung has no incentive to break the agreement, if we assume that, after breaking it, he
will stop receiving the check from the cartel. Otherwise, he would also have an incentive
to cheat, as the cartel agreement is not part of his reaction function either.

3.(a) The inverse market demand curve is p(y) = 10−10−5 y. Their profit maximization problems
are:

maxπ
y 1=p(y1 , and2 )y1-C1 (y1 )
1

= 10 −10−5 y1-10−5 y2y1-y12

maxπ
y 2=p(y1 , y2 )y2-C2 (y2 )
2

= 10 −10−5 y1-10−5 y2y2-5y2

Solutions to Exercises 38

Peartakesy1as given and fixed, and choosesy2to maximize profits.

∂π 2
= 10−10−5 y1-2(10−5 )y2-5 = 0
∂y2
1
y2= 250,000−y1
2
1 profit function, and chooses y accordingly.
MBIsubstitutesy2= 250,000−y1into its
2 1

1
maxπ −5 −5 2
y 1= 10 - 10 y1-10 250,000−y1y1-y1 2
1

15 1
= −5 2
2 − 2 10 y1y1-y1

∂π 1 15
= −5
∂y1 2 -10 y1-2y1= 0
200,001 15
y1=
100,000 2
y1∗ ≈3.75

Plugy1∗ = 3.75 back into y2 (y1 ) to obtainy2∗ ;y2∗ 250,000 - 3.75 = 249,998.12.
1
2
The market price is p∗ = 10 - 10−5 (3.75 + 249,998.12)≈7.50. Individual profits are
π1∗ = 7.50·3.75−3.752= 14.06 and π2∗ 624,995.31

3.(b) To calculate the Cournot equilibrium, we solve the profit maximization problems simultaneously.

We need to derive MBI's reaction function.

∂π 1
= 10−2(10−5 )y1-10−5 y2-2y1= 0
∂y1
1,000,000 1
y1= y
200,002 − 200,002 2
1
≈5− 200,000 y2

From part (a), we know that y2 (y1 = 250,000 - y1 . 1

2

Solving the reaction functions simultaneously, we find that y1C≈3.75 andy2C≈249,998.12

Thus price and profits are equivalent to part (a). This problem shows that sometimes the
The difference in costs plays a very important role in determining output and profits regardless.

of the timing of entry into the market. MBI suffers from such a large cost disadvantage
compared to Pear that MBI benefits very little from its first-mover advantage.
Solutions to Exercises 39

4.(a) Their profit maximization problems are:

maxπ1=p1 y1 (p1 , p2 )−C1 (y1 (p1 , p2 ))

p1

1 1
=p180−p1+p2-80 802 -p1+p2 2

maxπ2=p2 y2 (p1 , p2 )−C2 (y2 (p1 , p2 ))

p2

1 1
=p2160−p2+p1-160 160
2 -p2+p1 2

To derive their reaction functions, p1 (p2 ) andp2 (p1 ), we solve the maximization problems.

∂π 1 1
= 80−2p1+p2+ 80 = 0
∂p 1 2
1
p1= 80 + p2
4

∂π 2 1
= 160−2p2+p1+ 160 = 0
∂p 2 2
1
p2= 160 + p1
4

The Bertrand equilibrium is the intersection of the reaction functions. Solving the reaction
functions simultaneously, we find that p∗1= 128 andp∗ ∗
2 = 192. Output levels are1 = 80−

128+21192 = 48 andy2∗ = 160−192+21.128 = 32. Profits are π1∗ = 128·48−80·48 = 2,304

andπ2∗ = 192·32−160·32 = 1,024.

4.(b) Reuben and Simeon solve the following joint maximization problem.

maxπ=π1+π2
p1 ,p2

1 1 1 1
=p180−p1+p2-80 802 -p1+p2+p2160−p2+p1−160
2 160 −p2+p1 2 2

∂π 1 1
= 80−2p1+p2+ 80
2 + p2-80 = 02
∂p 1
Solutions to Exercises 40

1
p1= 40 + p2
2

∂π 1 1
= 160 - 2p2+p1+ 160 + p1-40 = 0
∂p 2 2 2
1
p2= 140 + p1
2

Solving the reaction functions simultaneously, we find that p∗1= 440

3 ≈146.67 andp∗2=
640
3 ≈213.33. Output levels arey1∗ 80-146.67+12-213.33 = 40 andy2∗ =160-213.33+21·
146.67 = 20. Profits are π1∗ = 146.67·40−80·40≈2,666.67 and π2∗ -2266.67
1,066.67. Thus joint profits are π∗ ≈3,733.33

5.(a) Their profit maximization problems are:

maxπ1=p1 y1 (p1 , p2 )−C1 (y1 (p1 , p2 ))

p1

1 1
=p134−p1+p2-24 343 -p1+p2 3

maxπ2=p2 y2 (p1 , p2 )−C2 (y2 (p1 , p2 ))

p2

1 1
=p240 - p2+p1-20 40 2-p2+p1 2

Jacob takes1as given and fixed, and chooses p2to maximize profits.

∂π 2 1
= 40−2p2+p1+ 20 = 0
∂p 2 2
1
p2= 30 + p1
4
1 profit function, and chooses p accordingly.
Laban substitutes2= 30 + p1into his
4 1

1 1 1 1
maxπ1=p134−p1+ 30 + p1 -24 34 -p1+ 30 + p1
p1 3 4 3 4

11 11
=p144−p1-24 12
44 +p1 12
Solutions to Exercises 41

∂π 1 11
= 44−p1+ 22 = 0
∂p 1 6
p∗1= 36

Plugp∗1= 36 back intop2 (p1 ) to obtainp∗ ∗ 1

2 ;p 2= 30 + 36 = 39. 4Output levels are

y1∗ = 34−36+31-39 = 11 andy2∗ = 40−39+12-36 = 19. Profits are π1∗ = 36·11−24·11 = 132
andπ2∗ = 39·19−20·20 = 361.

5.(b) Laban takesp 2as given and fixed, and chooses p1to maximize profits.

∂π 1 1
= 34−2p1+p2+ 24 = 0
∂p 1 3
1
p1= 29 + p2
6
1 profit function, and chooses p accordingly.
Jacob substitutes1= 29 + p2into his
6 2

1 1 1 1
maxπ2=p240−p2+ 29 + p2 −20 40 −p2+ 29 + p2
p2 2 6 2 6
109 11 109 11
=p2 p
2 − 12 p2-20 2 − 12 2

∂π 2 109 11 55
= − p 2+ =0
∂p 2 2 6 3
437
p∗2=
11 ≈39.73

Plug∗ ∗ ∗ 437 )≈35.62.1Output levels are

2 39.73 back intop1 (p2 ) to obtainp1 ;p 1= 29 + ( 6 11
y1∗ = 34−35.62 + ·39.73≈11.62
1
3
andy2∗ 40 minus 39.73 plus approximately 35.62 is12about 18.08. Profits are

π1∗ ≈135.02 and π2∗ ≈356.72.

6. For homogeneous goods markets, we rank the models in order of increasing output and
decreasing price: collusion (if successful), the Cournot and Stackelberg models, and the
Bertrand model. For differentiated goods, collusion, price leadership, and the Bertrand
model.
Solutions to Exercises 42

Chapter 14 Solutions

1.(a)

Player 2
Macho Chicken
Macho 0, 0 7, 2
Player 1
Chicken 2, 7 6, 6

1.(b) Neither player has a dominant strategy.

1.(c) The Nash equilibria are (Macho, Chicken) and (Chicken, Macho).

2.(a)

Jill
Build Don't Build
Build 1, 1 -1, 2
Jack
Don't Build 2, -1 0, 0

2.(b) Both of them have a dominant strategy, which is “Not Build.” The Nash equilibrium is
(Not Build, Not Build).

2.(c) This game resembles the Prisoner’s Dilemma. The Nash equilibrium (Not Build, Not
(Build) is not socially optimal. Both of them would be better off at (Build, Build).

3.(a) This is a sequential game where Sam is the first mover.

3.(b) Sam will take Dan’s blanket if Dan doesn’t retaliate. Dan will not retaliate if his payoff
From not retaliating, his payoff exceeds his payoff from retaliating, i.e., -10 > -10 + X ⇔ X < 0.

4.(a) This is a sequential game with payoffs occurring only when the game is terminated at
t = 1, ..., 99, and t = 100.
Solutions to Exercises 44

Chapter 15 Solutions

1.(a) The Edgeworth box is a square, and each side is 6 units long, with Michael’s origin on the
lower left hand corner. The endowment point would be (xM, andM) = (5,1); (xA , andA (1,5)

1.(b) Their indifference curves are L-shaped, with kinks where x i=yi .

1.(c) Any allocation in the area bounded by the indifference curves through the initial endowment
is a Pareto improvement. For example, atW , each of them will consume three slices of
tiramisu and three cups of espresso at the new allocation.

2.(a) To find the contract curve, set MRS g=MRSf Each MRSi=MUxi/MUyi. Setting the
M R S sequal gives:

ayg ayf
=
(1−α)xg (1−α)xf
yg y
= f (3)
xg xf

This is one equation with four unknowns. To get two more equations, write down the
market-clearing conditions:

xg+xf= 1

xf= 1−xg (4)

yg+yfequals 1

yf = 1−yg (5)

Plugging (4) and (5) into (3), we get yg=xg Therefore, the contract curve is the diagonal
of the Edgeworth box.

2.(b) The competitive equilibrium is on the contract curve. By part (a), yg=xgon the contract
curve. Therefore at the competitive equilibrium,

px ayg α
=MRSg MRSf= = .
py (1−α)xg (1−α)
Solutions to Exercises 45

3.(a) Do the totals add up in Duncan’s suggested equilibrium?

x r+x t=xr0+xt0

4+1=2+3=5

yr +yt =y0r +y0t

1+4=2+3=5

Duncan’s suggested allocation have the right totals.

3.(b) Rin’s original utility is u r0= 2·2 = 4, and his new utility isr = 4·1 = 4. Tin’s original
utility issuet03 multiplied by 32= 27, and his new utility is t= 1·42Rin's utility is unchanged.

and Tin’s utility falls. Therefore, Duncan’s suggested equilibrium allocation is not a Pareto
improvement over the endowment.

3.(c) With prices x=py= 1, Rin's budget constraint is xr+yr= 4. With this budget, his
optimal consumption bundle is (x r , andr (2,2).

3.(d) With pricesp x=py= 1, Tin's budget constraint is xt+yt= 6. With this budget
constraint, his optimal consumption bundle is (x t , andt ) = (2,4).

3.(e) Duncan's suggested equilibrium allocation and price vector is not a competitive equilibrium.
The totals add up, but consumers are not maximizing their utilities at this allocation with
these prices. Whenpx=py= 1, there is excess supply of x and excess demand for y. This
suggests that in equilibrium, either pxshould drop below 1 orpyshould rise above 1.

4.(a) Rin's budget constraint is x r+py yr= 2 + 2py Tofnidhsi desreidconsumpoitnbundel,

we set MRSr=yr /xr= 1/py . This givesxr=py yr Combining this with his budget

constraint leads to∗ ∗

r = 1 + py It follows thatr (1 + py ) / py .
Solutions to Exercises 46

4.(b) Tin’s budget constraint is x t +py yt= 3 + 3py Tofnidhsi desreidconsumpoitnbundel,

we setMRSt=yt /2xt= 1/py This gives 2xt y yt Combining this with his budget
constraint leads to∗ ∗
t = 1 + py . It follows thatt = 2(1 + py ) / py .

4.(c) In the market for x, total demand = total supply gives:

x∗r+x∗t=xr0+xt0;

(1 + py ) + (1 + py ) = 2 + 3.

3
This gives py=.
2

4.(d) Competitive equilibrium:

5 5
(x∗r , and∗
r )= ,
2 3
5 10
(x∗t, and∗
t ) = ,
2 3
3
py=.
2

5.(a) The original endowment is Pareto optimal. Shepard cannot be made better off without
making Milne worse off, i.e., there is no Pareto move from the original endowment.

5.(b) Milne’s new budget constraint is x m+py ym= 4 + 4py-4 = 4py . Setting his marginal
the rate of substitution equal to the price ratio gives MRSm=ym /3xm= 1/py . This gives

py ym= 3xm . It follows that his optimal consumption bundle is (x∗m , and∗
m ) = (py ,3).

5.(c) Shepard’s new budget constraint is x s+py ys= 0 + 0py+ 4 = 4. Setting his marginal rate
the substitution rate equal to the price ratio gives MRSs=ys= 1/py . This givespy ys=xs .

It follows that his optimal consumption bundle is (x∗s , and∗

s ) = 2,
2
py
.

5.(d) We first use the market-clearing condition to solve for the equilibrium price. We want
total demand = total supply in the market for x. This gives:

x∗m+x∗s=xm
0 +x0;
s
Solutions to Exercises 47

py+ 2 = 4 + 0.

This gives py= 2

Competitive equilibrium:
(x∗m , and∗
m ) = (2,3)

(x∗s , y∗
s ) = (2,1)

p y= 2

5.(e) At the equilibrium allocation,

M R S m=y∗ ∗
m /3x m = 3/(3·2) = 1/2.

Similarly,
M R S s =y∗ ∗
s /x s = 1/2.

SinceMRSm=MRSsthe new equilibrium allocation is Pareto optimal.

6. The first fundamental theorem says any competitive equilibrium allocation is Pareto optimal.
The second fundamental theorem says that lump-sum taxes and transfers can be used to
get an economy to any target Pareto optimal allocation. Per unit taxes and subsidies may
not accomplish the same thing. Consider Milne and Shepard in Question 5 above. Shepard
starts out with ωs = (0,0). To induce him to go to (x∗s , and∗ ) = (2,1) with straightforward
per unit subsidies, you would have to sell him one or both goods at negative net prices.
Given his utility function, he would then want to consume infinite amounts of both goods.
Solutions to Exercises 48

Chapter 16 Solutions

1.(a) The Pareto efficient allocation is the tangency of the indifference curve and the production.
function.

MRSl,x=MPl
1 1
− − 2 = 2l√
l=1

Therefore, x = 1 √= 1.

1.(b) A utility-maximizing consumer’s tangency condition is -MRS l,x = w 1

1 ⇔ w= . 2
A profit-maximizing firm solves the following problem:

√ l
maxπ= 1·x−w·l=l−
l 2

Setting the derivative of π with respect to l equal to zero yields l = 1.

Therefore, x = 1, l = 1, w = , and π =1 . 1
2 2

2.(a) The Pareto efficient allocation is the tangency of the indifference curve and the production.
function.

−MRSl,x=MPl
1 2
− − 2 = 33√l
64
l=
27
Therefore, x = 64 2/3 = 16 .
27 9

2.(b) As in Q1(b), w = . 1
2
A profit-maximizing firm solves the following problem:

max π = 1·x - w·l = l2/3-l/2

l

Setting the derivative of π with respect to l equal to zero yields l = 64/27.

Therefore, x=196 ,l= , w= , 64

andπ=
27
1
2
16 .
27
Solutions to Exercises 49

3.(a) The Pareto efficient allocation is the tangency of the indifference curve and the production.
function.

-MRSl,x=MPl
1
−(−4l) = 4l √
2l
1
l∗ =
4
Therefore, x = 1 +1=. 3
4 2

3.(b) First, use the tangency condition to find the market wage rate.

w
-MRSl,x =
1

4l∗ =w
1
w= 4 =1
4
A profit-maximizing firm solves the following problem:

maxπ= 1·x−w·l= √l+1-l

l

Setting the derivative of π with respect to l equal to zero yields l = . 1

4
x 3 1 5
2 4 4

4.(a) By symmetry of her utility function, and since the production functions are identical,
Wendy should study an equal number of chapters of economics and mathematics.
should spend two hours on each subject.

√ of economics and 2 chapters of√

4.(b) Givenl x= 2 andly= 2, she studies 2 chapters
mathematics.

4.(c) Her utility is u = 2 * 2 √- 4√= 0. √

4.(d) Wendy now studies eight hours a day, and spends four hours on each subject. Therefore,
x = 4 √= 2 and y = 4 = 2. Her
√ utility is now u = 2·2 - 8 ≈ 1.17. √
Solutions to Exercises 50

5.(a) Since the wage is normalized to 1, Robinson's firm's profit is:

π=px x+py y−l

Substituting for l from the inverse production function,

π=px x+py y - (x2+y2+ 3xy

The first order condition for x is

∂π
=px-2x - 3y = 0
∂x
The first order condition for y is

∂π
=py-2y - 3x = 0
∂y
These two equations together should yield the supply functions for x and y in terms of p.x

andpy From the first equation, we get x=p−3y. Substituting

x
2
this into the second one,
we have
px-3y =0
py-2y - 3 2
or,
2py-4y - 3px+ 9y = 0

which yields
1
y= (3p 6
5 x-2py )
as the supply function for y. Substituting this back into x =p−3ygives usx the
2
supply
function forx:
1
x= (3p (7)
5 y-2px ).

5.(b) Robinson, the price-taking consumer, chooses x, y, l to maximize u(x, y, l), subject to his
budget constraint
px x+py y=l+π.

Robinson takespx , pyand as given. To solve this utility maximization problem, we use
the budget constraint to solve for land then substitute back into the utility function. We
then maximize
Solutions to Exercises 51

3 1 3 1
u = xy + x + y - l = xy + x + y - (px x+py y−π).
2 2 2 2

This results in the following first order conditions:

∂u 3 1
=y + 1 - xp= 0
∂x 2 2
∂u 3 1
= x + 1 - yp=0.
∂y 2 2

From these we get:

y= px-2 (8)
3
x= py-2 (9)
3

as the demand functions for x and y. (The demands for x and y in this exercise are

independent of income because of quasi-linearity in l. Moreover, we have an interior solution

because the supply of labor is not assumed to have any bounds. In general, demand
functions will be more complicated functions of prices and profit than in this exercise.

An alternative way to solve the consumer’s problem is to use a Lagrange function approach;
see the Appendix to Chapter 3.

5.(c) To find an equilibrium, we solve for p xandpyfrom (6), (7), (8), and (9). The solutions
arepx=py= 5; this gives x = 1, y = 1 and, from the inverse production function, the
The demand for labor input isl= 5. The profit for the firm is 5, and, for the consumer, the
right hand-side of the budget equation is 10. The supply of labor is found by using the
budget equation, which yields supply equal to 5 (also equal to the firm's demand).

The competitive equilibrium in this one consumer economy can also be computed by first
finding the Pareto optimal allocation and then using the first welfare theorem.
The approach can provide a simpler way of computing the equilibrium but it generally works only
in a single consumer economy. Otherwise, there will be many Pareto optimal allocations).
Pareto optimality in this single consumer economy implies the maximization of u(x, y, l)
subject to the inverse production function. Substituting the latter into the utility function
results in:
Solutions to Exercises 52

3 1 1 1
xy + x + y - (x2+y2+ 3xy) = x + y - x 2-y2 ,
2 2 2 2
and the first order conditions yield x=y= 1. This is the unique Pareto optimal allocation
of consumption goods in this economy. From the inverse production function l= 5. This
is the unique Pareto optimal allocation. By the first welfare theorem, a competitive equi-
Equilibrium must result in this allocation. From (8) and (9) we get px=py= 5. Check that

The profit and budget numbers are as before.

6. Robinson solves the following profit maximization problem:

maxπ=px x+py y+pz z - wl subject to l(x, y, z).

x,y,z

He solves the following utility maximization problem:

maximize = u(x, y, z) subject tox x + py y + pz z ≤ wl + π.

x,y,z
Solutions to Exercises 53

Chapter 17 Solutions

1.(a)um (2) = 22-2·2 + 2 = 2

ul (2) = 22= 4

uc (2) = -22-1 = -5

1.(b)u m (3) = 32-2·3 + 2 = 5

ul (3) = 32= 9

uc (3) = -32-1 = -10

1.(c) We want to maximize

u=um+ul+uc=x2-2x + 2 + x2+ (−x2-1) =x2-2x + 1

The first order condition for a maximum is

∂u
= 2x−2 = 0,
∂x
but the second order condition for a maximum is not satisfied. In fact∗ = 1 is a minimum;

the functionx2-2x + 1 has no maximum; they should fill their house with puppies.

2.(a)

maxu = 50p - p2−28h

p s

1
= 50p - p2-28
2
36p - p2

∂u s
36 - 2p = 0
∂p
pM= 18

Therefore, h = 1812 = 9.

2.(b)u S36·18−182= 324

uG=·18142+ 3·18 = 135
Solutions to Exercises 54

2.(c)
maxu=us +ug
p

= 36p - p2
1 2
+ p + 3p
4
3
39p - p2
4

∂u 3
= 39−p= 0
∂p 2
p∗ = 26

3.(a)
3
maxu = 8m - 2m2− x
m i 10

∂u I
= 8−4m= 0
∂m
m M= 2

Therefore, music is blasting 2.11 = 22 hours per day.

3
3.(b)u i= 8·2−2·22− 10 (2·10) = 2

3.(c)
3
maxu = 8m - 2m2-(10m)
m 10
= 5m - 2m2

∂u
= 5−4m= 0
∂m
5
m∗ =
4
Solutions to Exercises 55

4.(a)
2
1
maxπf = 50f - 5f - b
f 3

∂π f = 50 - 10f - b = 0 1
∂f 3
f=b+5
1
(10)
3
2
1
maxπb= 100b - 10b - f
b 2

∂π b 1
= 100−20b−f= 0
∂b 2
b=f+5
1
(11)
2

Solving equations (8) and (9) simultaneously, we find that fM= 8 andbM= 9.
1 2
πM
f = 50·8−5 8 − · 9 = 275
3
2
πbM= 100·9−10 9 − · 8 = 650
1
2

4.(b)
maxπ=πf +πb
f,b

2 2
1 1
= 50f - 5f - b + 100b - 10b - f
3 2

∂π 50 - 10f - b + 10b - f =1 0 1
∂f 3 2
10 8
+b (12)
3 9

∂π 10 f - b +1100 - 20b - f = 0 1
=
∂b 3 3 2
b=f+
12 90
13
19 19

Solving equations (10) and (11) simultaneously, we find that f∗ = 17.2 andb∗ 15.6.
2
π∗ 1
f = 50·17.2−5 17.2− · 15.6 =3 140
2
πb∗ 100·15.6−10 15.6− · 17.2 = 1,070
1
2
Solutions to Exercises 56

4.(c) Beatrice transfers between $135 and $420 to Flo.

5.(a)
pc=MCc

700 = 10c + 100

cM= 60

pb =MCb

10 = b - 140 + c

bM= 90

5.(b)
maxπ=πc +πb
c,b

1
= 700c - 5c2+ 100c + 10b - 2
2 b −140b+bc

∂π = 700 - 10c - 100 - b = 0

∂c
b = 600 - 10c (14)

∂π = 10 - b + 140 - c = 0
∂b
b = 150 - c (15)

Solving equations (12) and (13) simultaneously, we find that c∗ = 50 andb∗ = 100.

5.(c) Competitive market outcome: (c M, bM (60,90)

π M=πc+πb

1
= 700 · 60 - 5 · 602+ 100·60 + 10·90− 2
2 90 −140·90 + 90·60
Solutions to Exercises 57

18,000 + 4,050

22,050

Joint profit maximization outcome: (c∗ , b∗(50,100)

π ∗ =πc+πb

1
= 700 · 50 - 5 · 502+ 100·50 + 10·100− 2
2 100 -14000 + 5000
17,500 + 5,000

22,500

Total profits from joint profit maximization, π∗ , are greater than total profits from individ-
total profit maximization, πMIf Bonnie transfers a minimum of $500 and a maximum of

$950 to Clyde, then πc∗ ≥πcMandπb∗ ≥πbM. Thus (c∗ , b∗(50,100) is Pareto optimal.

5.(d) Since this is a negative externality, the creator of the externality, Clyde, pays the tax.
The negative externality imposed on Bonnie by Clyde is now internalized by Clyde.

5.(e) If Clyde and Bonnie are able to negotiate a mutually beneficial contract where Bonnie
Clyde is paid to reduce his output fromMtoc∗ , then they will end up at the Pareto optimal
outcome, (c∗ , b∗ ).

6. Please note two corrections of the problem in the text:

(1) After the second sentence, insert the following sentence: “The total pollution level had
been at 60.
(2) In the next to last sentence, change “pollution abated” to “pollution abated.”

Because the government is issuing a total of 30 pollution permits, and because the total
pollution level had been 60, there must be 60−30 = 30 units of abatement. Therefore

x1+x2= 30.
Solutions to Exercises 58

Factory 1's marginal cost of pollution abatement is MC1= 120x1 It could get rid of a
marginal unit of pollution by spending MC1 or, alternatively, it could buy an additional
pollution permit, which would allow it to produce that marginal unit of pollution. Let p be
the market price of a pollution permit. In equilibrium we should have MC1=p. Similarly,
in equilibrium we should have MC2=p. It follows that MC1=MC2 , or

120 times1= 3x22.

Solving the two equations for x1andx2simultaneously, we find x1= 10 andx2= 20.
Solutions to Exercises 59

Chapter 18 Solutions

1.(a) Correction. The textbook should read: 'What condition or conditions must hold for the
TV to be considered a public good?” The answer is: Nonexclusivity in use. Once the
When the television is bought, neither Fabio nor Paolo can be prevented from watching it.

1.(b) Each of them has a dominant strategy of not paying. The Nash equilibrium is (Don’t
Pay, Don’t Pay) with payoffs of (0, 0). Note that they would both have been better off
with (Pay, Pay) and payoffs of (50, 50). This game resembles the Prisoner’s Dilemma.

Paul
Pay Don’t Pay
Pay 50, 50 -100, 200
Fabio
Don’t Pay 200, -100 0, 0

1.(c) The Nash equilibria are (Pay, Don’t Pay) with payoffs of (100, 400) and (Don’t Pay, Pay)
with payoffs of (400, 100). This is a game of Chicken.

Paul
Pay Don't Pay
Pay 250, 250 100, 400
Fabio
Don't Pay 400, 100 0, 0

2.Correction There are two errors in the textbook. The original utility function is shown as
ui=xi y√i The correct utility function isi=xI+yi The new√ utility function where
movie streaming is a public good is shown asui=xyi The √correct utility function is
√
ui=x+y i. Delete part (d). Here is the solution for the corrected problem:

2.(a) Since MRSx,y = 1 = 1 for both Fabio and Paolo, their optimal allocation are (xf, andf) =
2√ x
√
(0.25,9.75) and (xp, andp) = (0.25,19.75). Their utility levels areuf0.25 + 9.75 = 10.25
andup0.25 √ + 19.75 = 20.25.
Solutions to Exercises 60

2.(b)u f= √
0.5 + 9.75≈10.46>10.25 andp= 0.5 √+ 19.75≈20.46>20.25.

2.(c) According to the Samuelson optimality condition, the sum of the marginal benefits from
the public good must equal the marginal cost of the public good:

1 1 1
M B f(x) +MBp (x) = √ + √ = √ =1
2x2x x
1
Therefore the Pareto optimal quantity of the public good is x = 1 >. The allocation from
2
part (a) is not Pareto optimal.

3.(a) Each pig would be willing to contribute a maximum of h itowards buying a stone house
as long as the utility with a house is at least as great as the utility without a house. Each
Pig faces a budget constraint of y.i+hi=Mi . With a house,yi=MiInvalid inputi , and without a
house,yI=Mi .
Pig 1: (3 + 1)y1>=(3 + 0)y1⇔ 4(M-h
1 1 )≥3M1⇔ h≤
1
1M
1 4
Pig 2: (2 + 1)y2≥(2 + 0)y2⇔ 3(M22 )≥2M2⇔ h≤
2
1M
2 3
Pig 3: (1 + 1)y3≥(1 + 0)y3⇔ 2(M−h
3 3 )≥1M3⇔ h≤
3
1M
3 2

3.(b) The maximum amount each pig is willing to contribute is as follows: 1=·2,800 =14 700,
1
h2=·1,800 = 600, andh31,00012 = 500. Since 700 + 600 + 500 = 1,800, the pigs
3
buy the house.

4.(a)MRSx,y = 1
i 10 √ x

4.(b)
M B i (x) = 1
i

1
1,000 =1
10x√
x∗ = 10,000
Solutions to Exercises 61

5. If agent 1 has to pay the full cost of all units of the public good, he maximizes u.1 (x, y1) =
√ x+y 1subject tox+y 1=M1 Equating marginal benefit with marginal cost yields

x = 1/4. Note that agent 1's marginal benefit from another unit of the public good, net of
Its full cost is positive if x < 1/4, and negative if x > 1/4.
Similarly, if agent 2 has to pay the full cost of all units of the public good, he maximizes
√ 2subject tox+y 2=M2 , which leads to x = 1. Agent 2’s marginal
u2 (x, y2 ) = 2 x+y
The benefit from another unit of the public good, net of its full cost, is positive if x < 1, and

negative ifx >1.

Now suppose agent 2 were to contribute 0. Then, agent 1 would contribute 1/4, but then
agent 2 would not like to contribute 0. Therefore, agent 2 contributing 0 will never happen
in equilibrium. That is, if one chooses x = 1/4, there is 1/4 unit of the public good which
is available for agent 2’s consumption. Agent 2 wants to maximize u2 (x, y2 ) = 2 x+y 2 , √
but the first 1/4 unit of the public good is already there at no cost to him. His constrained
Utility maximum is still at x = 1, and so he buys an additional 3/4 unit of the public good.
Therefore, agent 2 does not benefit from agent 1's previous purchase of the public good.

6.(a)Ti=C(x∗ )−vj (x∗ )

j=i
T11,000 - (200 + 300 + 400) = 100
T21,000 - (100 + 300 + 400) = 200
T3= 1,000−(100 + 200 + 400) = 300
T4= 1,000 - (100 + 200 + 300) = 400

6.(b)MB i (1) = 100 + 200 + 300 + 500 = 1,100 > 1,000 = MC(1)
I
Yesonce marginal benefit exceeds marginal cost, the park gets built. Family 4 pays a tax of
T4= 1,000−(100 + 200 + 300) = 400. Since T4is unchanged, family 4’s net benefit is also
unchanged.

6.(c)MB i (1) = 100 + 200 + 300 + 300 = 900 < 1,000 =MC(1)
i
Since marginal cost exceeds marginal benefit, the park does not get built. Family 4 does
not pay any taxes;4= 0. Family 4's net benefit is zero, just as in part (a).
Solutions to Exercises 62

6.(d)
M B 1 (1) +MB2 (1) +MB3 (1) +MB4 (1) =MC(1)

100 + 200 + 300 + 400 = 1,000

Solutions to Exercises 63

Chapter 19 Solutions

1. Currently, the risk-averse consumers' and the risk-neutral consumers' utilities are as
follows:
√ + (10 25)
1 50)
ui (L) = (10 √
1 ≈60.36
2 2
uk (P) = 0
If the consumer pays consumer $13 to bear the risk, both of them would be better off.
utilities are now as follows:
√
ui (P) = 10 37≈60.83
1 = 0.50
uk L = 13 + 0 - 25 1
2 2

2.(a)E(L) = (5) 13+ (50) +13(500) = 185

1
3

1
2.(b)uug (5) + ug (50) 1
+ ug (500) = 1 1 1 1 1
3 3 3 3 2 ·52+·5022+·500242,087.5
2

2.(c) His certainty equivalent of the lottery is the dollar amount P such that he is indifferent.
between accepting the certain payment and playing the lottery.

ug (P) =ug (L)

1 2
P = 42,087.5
2
P≈290.13

2.(d) George is risk loving; his certainty equivalent of the lottery is greater than the expected
value of the lottery.

3.(a)E(L) = (6) 14+ (2) + (2)

1 + (0)1= 2.5
4 4
1
4

3.(b)u a ua (6) + ua14(2) + ua (2)14+ ua (0) = 1

4
1
4
1
4
62+ 22+ 22+ 02= 11
ui ui (6) + u14i (2) + uI (2)14 + ui (0) =14(2(6) + 2(2)
1 + 2(2) +1 2(0)) = 5
4 4
Solutions to Exercises 64

3.(c) An individual with a certainty equivalent less than 3 prefers to draw water at the foot of
the hill, while an individual with a certainty equivalent greater than 3 prefers to hike up
the hill.
Jack's and Jill's certainty equivalent are derived as follows:
ua(P) =ua (L)⇔P2= 11 ⇔Pa3.3
ui (P) = ui (L) ⇔ 2P = 5 ⇔ Pi= 2.5
Therefore, Jack prefers to hike up the hill, and Jill prefers to draw water at the foot of the
hill.

3.(d) Jack is risk loving. aE(L). Jill is risk neutral;PI=E(L).

4.(a)E(L) = ·8 +12 ·0 = 4 1
2

4.(b)u a ua (8) + ua12(0) = 1 1 √38 +30 √ = 1

2 2
um (L) = um12 (8) + um (0)12 = (8 + 0) =12 4
1 + u (0) =
us (L) = u(8) 1 1 83+ 03= 256
2 s 2 2

4.(c) Compare their utility with protection to their utility without protection.

3 24−6
ua (P) = ≈1.82>1 =ua (L)
3
24−6
um (P) = = 6 >4 =um (L)
3
3
24−6
us (P) = = 216 <256 =us (L)
3
Adam and Michael will buy protection. Stella will not buy protection.

4.(d) The neighborhood thug would have to set a price low enough such that Stella will buy
24−P 3
protection:us (P)≥us (L)⇔ 3 ≥256 ⇔P≤4.95

1
5.uk (L) = uk (10) + uk (-5)26 + uk (-1) = (10)
3 + (-5) + 16(-1) = -2 26 3 1
6 6 6
Since Ko's utility from the lottery is negative, he does not accept it.
Solutions to Exercises 65

1 ) + E(L )1+ E(L ) =

6.E(L) = E(L 1 1 30 1 40 1 50 17
31 2 3 3 3 3 100 -20 + 3 100 ·15 + 3 100 -10 = 3
Since Will paid $6 for the lottery ticket, the net expected value of his lottery ticket is− 3. 1
Solutions to Exercises 66

Chapter 20 Solutions

1
1.(a)E(L) = (3,000) + (2,000)13 + (1,000) = 132,000
3
Harry is willing to pay $2,000, and he ends up buying type B and type C cars.

1
1.(b)E(L) = (2,000) + (1,000)12 = 1,500
2
Now Harry is willing to pay $1,500. However, this results in his getting type C cars only.
Eventualmente Harry fracasa como empresario.

The market price for used cars would be $1,500.

The market price for uninspected used cars would be $X/2.

2.(c)X−300 = X
2 X = 600

2.(d) Any car worth less than $600 will not get inspected, and there are 600/3 = 200 such cars.
Each uninspected car will be sold for 600/2 = $300.

3.(a)Ep (Loss) = 0.6(1,000,000) = 600,000

Ej 200,000
El (Loss) = 0.1(1,000,000) = 100,000

3.(b) Placido is risk loving since WTP p= 500,000<600,000 =Ep(Loss).

Jose is risk loving since WTPj= 175,000<200,000 =Ej (Loss).
Luciano is risk averse since WTPl= 125,000>100,000 =El Loss.

3.(c)E(Payout) = 1 6 + 1 2 + 1 1 1,000,000 = 300,000

3 10 3 10 3 10

3.(d) If the price of insurance equals the expected payout, only Placido will buy insurance since
W T P p500,000 < 300,000 = Price. The insurance company will make a loss, because
if Placido is the only buyer, E(Payout) = 6 1,000,000 = 600,000 > 300,000 = Price.
10
Solutions to Exercises 67

4.(a) If Kevin does not buy insurance, his expected utility from locking the door 100 percent of the time.

the time is H−250, and his expected utility from locking the door 80 percent of the time
isH−750 + 100 =H−650. Therefore Kevin will lock his door 100 percent of the time.

An insurance policy costs 0.02·$5,000 = $100.

4.(c) An insurance policy costs 0.06·$5,000 = $300.

4.(d) Insurance premiums range between $100 and $300. First, suppose an insurance policy
costs $100. Kevin’s net benefit from locking the door 100 percent of the time is H - 100.
His net benefit from locking the door 80 percent of the time is H - 100 + 100 = H. Therefore

Kevin will lock his door 80 percent of the time. Clearly, if homeowners lock their doors 80
percent of the time when the premium is $100, insurance companies will eventually revise
premiums upward to $300. Kevin’s net benefit from locking the door 100 percent of the
time is now H−300, while his net benefit from locking the door 80 percent of the time is
nowH−300 + 100 =H−200. Therefore Kevin will still lock his door 80 percent of the
time.
In equilibrium, insurance policies cost $300. Kevin locks his door 80 percent of the time,
and his net benefit is H−200.

4.(e) If homeowners lock their doors 100 percent of the time, policies cost $100, and each
homeowner’s net benefit is H−100 > H−200.

5.(a)
E(u, e= 1) = E(u, e= 0)

0.1 √* 10c + 0.9√* 5c - 1 = 0

c≈0.1844

E(π, e= 1) = 0.1(10−10c) + 0.9(5−5c)≈4.49

5.(b)
E(u, e= 2) = E(u, e= 0)
Solutions to Exercises 68

0.9 √10c + 0.1 5c√ - 2 = 0

c≈0.4245

E(π, e= 1) = 0.9(10−10c) + 0.1(5−5c)≈5.47

5.(c)
E(u, e= 2) > E(u, e= 1)

0.9 √* 10c + 0.1√* 5c - 2 > 0.1 *√ 10c + 0.9 *√ 5c - 1

c > 1.82

The agent would have to be paid greater than 182 percent of the output in order to guarantee
that he works at high effort. Clearly, this is not a sustainable mechanism.

Wage = (900) + 23(3,000) = 131,600.

6.(b)X <600 and Y < 600.

6.(c) X > 400 and Y < 400.