Econ 310 Microeconomic Theory II
Solutions to Assignment 3
Solution 1 (a) The …rm’s problem’s cost-minimization problem is:
s.t.
min w1 x1 + w2 x2
x1 ;x2 0
The Lagrangian is:
p
p
x1 + 2 x2
p
p
( x1 + 2 x2
L = w1 x1 + w2 x2
y:
y) :
The …rst-order conditions are:
@L
1 1=2
=0
= w1
x
@x1
2 1
@L
1=2
= w2
x2
=0
@x2
p
p
@L
=
x1 + 2 x2 y = 0:
@
1=2
w1
1
Equations (1) and (2) yield w
= 12 xx21
. Thus, x2 = 2 w
w2
2
and solve for the conditional input demands:
x1 (w1 ; w2 ; y) =
w22 y 2
;
(4w1 + w2 )2
x2 (w1 ; w2 ; y) =
(1)
(2)
(3)
2
x1 . Substitute it into (3)
4w12 y 2
:
(4w1 + w2 )2
Thus the cost function is:
c (w1 ; w2 ; y) = w1
w1 w2 y 2
w22 y 2
4w12 y 2
+
w
=
:
2
4w1 + w2
(4w1 + w2 )2
(4w1 + w2 )2
There are two ways to solve for the output supply function in this question. A student
gets full credit for either method.
Method 1: Consider the pro…t-maximization problem:
max p y
y
c(w1 ; w2 ; y):
The …rst-order condition is:
p
2w1 w2 y
= 0;
4w1 + w2
which yields
y(w1 ; w2 ; p) =
1
p (4w1 + w2 )
:
2w1 w2
Method 2: One can also set up the pro…t-maximization problem directly:
max
y;x1 ;x2 0
py
w 1 x1
s.t.
w 2 x2
p
p
x1 + 2 x2
y:
Since the objective function is strictly increasing in y, the constraint must hold with equality.
Thus the above problem simpli…es to
p
p
max p ( x1 + 2 x2 )
x1 ;x2 0
w1 x1
w 2 x2 :
The …rst-order conditions are:
@L
1 1=2
= p x1
w1 = 0
@x1
2
@L
1=2
= px2
w2 = 0:
@x2
(4)
(5)
Then one can directly solve for the input demand functions from the above two equations:
x1 =
p2
p2
,
and
x
=
:
2
4w12
w22
Thus output supply function is
y =
p
p
p
p (4w1 + w2 )
p
+2
=
:
x1 + 2 x2 =
2w1
w2
2w1 w2
[Note: Obviously, method 1 is much more convenient given that we’ve already got the cost
function.]
p
p
(b) Given f (x1 ; x2 ) = min x1 ; 2 x2 , the …rm’s cost-minimization problem:
min w1 x1 + w2 x2
x1 ;x2
p
p
s.t. min [ x1 ; 2 x2 ]
y:
We analyze each of the two possible cases:
p
p
(i) Suppose x1 2 x2 at the optimum. Then the above problem becomes
p
s:t: 2 x2
min w1 x1 + w2 x2
x1 ;x2
y:
Since the constraint does not concern x1 , it is optimal to set x1 as low as possible. Therefore,
p
p
in this case, it is optimal to have x1 = 2 x2 .
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(ii) Suppose
p
x1
p
2 x2 at the optimum. Then the problem becomes
min w1 x1 + w2 x2
s:t:
x1 ;x2
p
x1
y:
p
p
Similarly, it is optimal to have x1 = 2 x2 .
p
p
Therefore, both cases indicate that x1 = 2 x2 at the optimum. Moreover, the conp
p
straint holds with equality at the optimum, which implies x1 = 2 x2 = y. Thus, the cost
function is given by
c (w1 ; w2 ; y) = w1 x1 + w2 x2 = w1 y 2 + w2 y 2 =4:
Again, there are two ways to solve for the output supply function. For method 1, use
the cost function to derive it. The pro…t-maximization problem is
max py
y
w1 y 2
w2 y 2 =4:
The …rst-order condition is
1
w2 y = 0
2
Solving this problem yields the output supply function
p
2w1 y
y(w1 ; w2 ; p) =
2p
:
4w1 + w2
For method 2, one can set up the pro…t-maximization problem and solve for the output
supply function. The pro…t-max problem is:
p
p
max p (min [ x1 ; 2 x2 ])
x1 ;x2 0
w 1 x1
w2 x2 ;
where I have used the result that the constraint holds with equality and substituted the
constraint into the objective function to eliminate y. Again, consider two cases:
p
p
(i) Suppose x1 2 x2 at the optimum. Then the problem becomes
p
max p2 x2
w1 x1
x1 ;x2 0
w 2 x2 :
p
Obviously in this case, it is optimal to have the lowest possible value of x1 . Thus, x1 =
p
2 x2 .
p
p
(ii) Suppose x1
2 x2 at the optimum. Similarly, it is optimal to have the lowest
p
p
possible value of x2 . Thus, x1 = 2 x2 .
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p
p
Both of the above cases imply x1 = 2 x2 , that is, x1 = 4x2 . Thus the pro…t-max
problem becomes
p
max p2 x2 w1 4x2 w2 x2 :
x2 0
The …rst-order condition is
Thus, x2 =
p
4w1 +w2
p
p
x2
4w1
w2 = 0:
2
and thus the output supply function is
p
y = 2 x2 =
2p
:
4w1 + w2
[Again, obviously it is much more convenient to use method 1 to derive the output supply
if we already have the cost function.]
Solution 2 (a) The …rm’s short-run cost minimization is given by:
min L + 2E + 4K
L;E
s:t: K
p
L+
p
E
y:
The constraint holds with equality at the optimum. Rearranging the constraint, the above
is equivalent to:
p
p
y
L+ E = :
min L + 2E + 4K
s.t.
L;E
K
The Lagrangian is:
L = L + 2E + 4K
The …rst-order condition is:
hp
L+
p
E
1 1=2
@L
= 1
L
=0
@L
2
1 1=2
@L
= 2
E
=0
@E
2
p
p
@L
y
=
L+ E
= 0:
@
K
yi
:
K
(6)
(7)
(8)
From (6) and (7), L = 4E. Substitute it into (8) and solve for the short-run conditional
2
2
input demands E = 3yK and L = 4 3yK . Thus the short-run cost function is:
c w; y; K = 6
y
3K
2
+ 4K:
(b) There are two ways to solve this problem. A student gets full credit for presenting
either of the two methods.
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Method 1: In the long run, the …rm also chooses the level of capital to minimize cost:
y
3K
min c (w; y; K) = 6
K
2
+ 4K:
The …rst-order condition is
4y 2
+ 4 = 0:
3K 3
Thus the long-run input demands are:
y 2=3
; E (w; y) =
K (w; y) = p
3
3
y
3K (w; y)
2
=
1
4
(3y)2=3 ; L (w; y) = 4E (w; y) = (3y)2=3 :
9
9
(9)
And the cost function is
c (w; y) = 6
y
3K (w; y)
2
+ 4K (w; y) =
4 2=3
2
(3y)2=3 + p
y :
3
3
3
(10)
Method 2: Set up the long-run cost-minimization problem directly:
min L + 2E + 4K
L;E;K
s:t: K
p
p
L + E = y:
The Lagrangian is:
L = L + 2E + 4K
h
K
p
p
L+ E
i
y :
The …rst-order condition is:
@L
@L
@L
@E
@L
@K
@L
@
1
K L 1=2 = 0
2
1
= 2
K E 1=2 = 0
2
p
p
L+ E =0
= 4
p
p
= K
L+ E
y = 0:
= 1
(11)
(12)
(13)
(14)
Then one can solve the above system of equations and show that indeed the input demand
functions are given by (9) and the long run cost function by (10).
Solution 3 (a) A consumer maximizes the given utility function subject to a budget constraint. Notice that the utility function is a Cobb-Douglas function. Apply the formula to
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get the Marshallian demand for good 1:
x1 (p1 ; p2 ; y) =
y
y
= :
2p1
2a
(Of course, you can also derive the above by setting up the Lagrangian and solving the
…rst-order conditions.) Adding up over all the consumers gives the aggregate demand for
good 1:
100
50
200
2250
20
+ 10
+ 10
=
:
2a
2a
2a
a
Each …rm chooses q to maximize its pro…t:
max aq
q
q2:
The …rst-order condition is p 2q = 0. Thus, the …rm’s the supply function is q = a=2.
Adding up over all the …rms gives the aggregate supply for good 1: np=2. Equating aggregate
demand with aggregate supply yields:
2250
na
=
:
a
2
q
. The pro…t for an individual …rm is a q
q 2 = 4500
.
Therefore, a = 4500
n
4n
(b) In the long run, …rms enter the market such that pro…t is equal to F:
4500
= F:
4n
Therefore, the long-run number of …rms in the market is n = 4500
. The equilibrium number
4F
of …rms decreases with entry cost F. Intuitively, the higher the cost, the fewer …rms enter
the market.
Solution 4 (a) If there is at least one …rm charging a price lower than the marginal cost
c, then it must be true that pmin < c. Then the …rm who is charging the lowest price pmin
is making a negative pro…t. This …rm will be better o¤ charging any price c, which gives
it non-negative pro…ts. Therefore, there are pro…table deviations for the …rm(s) charging
pmin . Then this price vector cannot be a NE.
(b) [There are many ways to prove this case. A student gets full credit for identifying
ANY of the following pro…table deviations.]
Consider a …rm j which is not charging pmin . It is making zero pro…t because its price
is not the lowest and it gets zero demand. However, if …rm j charges any price pb such that
c < pb < pmin , it will make a positive pro…t. This is because now …rm j becomes the one
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who charges the lowest price. Thus it gets the entire market demand. Its pro…t is given
by = pbQ cQ > 0. Therefore, there are pro…table deviations for …rm(s) not charging
pmin . [One may realize that there may not exist …rms who are not charging pmin . That is,
it could be the case that all …rms are charging the same price: pmin . In such a case, a …rm
can pro…tably deviate by slightly decreasing its price. This way, the …rm captures the entire
market demand and at the same time increases its pro…t because the quantity has increased
by a great deal while the price only has decreased a little bit.]
Or, consider a …rm i which is charging pmin , where pmin > c:
(i) Suppose …rm i is the only …rm who is charging pmin . Then it is earning pro…t
= pmin Q cQ > 0 because pmin > c. Denote the second lowest price as psec . Firm i
can make strictly higher pro…ts by charging any price pb such that pmin < pb < psec . Because
pb < psec , is still charging the lowest price and getting the entire market demand. Under pb,
the pro…t is given by 0 = psec Q cQ > pmin Q cQ = .
(ii) Now suppose …rm i is not the only …rm who is charging pmin . In fact, there are
k 2 …rms charging pmin . Then …rm i is earning pro…t = pmin Qk c Qk = (pmin c) Qk .
But it can earn a even higher pro…t by charging a price slightly lower than pmin . Say the
new price is pmin ", where " is a very small positive number. By doing so, …rm i becomes
the …rm who is charging the lowest price. Therefore, it gets the entire market demand. Now
the pro…t becomes 0 = (pmin ") Q cQ pmin Q cQ = (pmin c) Q > (pmin c) Qk = .
The approximation is because " is a very small positive number.
In sum, we can …nd a pro…table deviation for any possible situation there is for this
case. Therefore, pmin > c cannot be a NE outcome.
[The results obtained here, together with Theorem 16 and the follow-up question we
discussed in class, we have proven that the Nash equilibrium in Theorem 16 is the only
possible NE for the Bertrand model.]
Solution 5 (a) (i) The aggregate demand is given by rearranging the market demand function: Qd = 50 2p t . (ii) The …rm’s pro…t-maximization problem is given by:
max pq
q
q2:
From the …rst-order condition, we get q = p=2. Therefore, the aggregate supply is Qs =
nq = np=2. (iii) In the equilibrium, Qd = Qs :
50
p
2
t
=
np
2
)
p =
50 t
n+1
)
Q =
(iv) Draw a graph with the market demand curve p + t = 50
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np
n (50 t)
=
:
2
2 (n + 1)
2Q and the marginal cost
curve mc (q) = 2q. The total surplus is given by
1
S2 = Q (50
2
t) =
(50 t)2 n
502 n
<
= S1 :
4 (n + 1)
4 (n + 1)
Therefore, the per-unit tax is distortionary in that it reduces the total surplus.
)p
(b) Given the market demand function, we have Qd = 50 (1+
. The aggregate supply
2
s
is the same as before: Q = np=2. In equilibrium,the market clears:
50
np
(1 + ) p
=
2
2
)
p =
50
n+1+
Draw a graph with the market demand curve p =
mc (q) = 2q. The total surplus is given by
1
50
S3 = Q
2 1+
=
)
50 2Q
1+
Q =
50n
:
2 (n + 1 + )
and the marginal cost curve
502 n
502 n
<
= S1 :
4 (n + 1 + ) (1 + )
4 (n + 1)
Therefore, the proportional tax is also distortionary in that it reduces the total surplus.
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