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ANSWERS TO EVEN NUMBERED EXERCISES FOR CHAPTER 1 Section 1.1
Ex er cise cise 1.1-2: 1.1-2: Supporti ng hyperpl hyperpl ane for CES preferences preferences 1−
(a) For σ > 1 show that the CES utility function U ( x) = ( x1
1 σ
1−
1
+ x1 ) σ
1 1
1−
σ
is strictly quasi-
concave. 1−
Hint: Sh Show that the utility function u( x) = f (U ( x)) is concave, where f (U ) = U (b) Show that this is true for
σ
1 σ
.
<1 . 1−
HINT: Consider the mapping f (U ) = −U
1 σ
.
(b) For any x > 0 obtain an expression for the supporting hyperplane of the upper contour set through x .
ANSWER
(a) Suppose that both both (− z 0 , q 0 ) and (− z 0 , q 0 ) ∈Y . Then q 0 ≤ F ( z 0 ) and q1 ≤ F ( z1 ) . We need to show that every convex combinati ation ( − z λ , qλ ) ∈ Y . Since F ( z ) is concave, F ( z λ ) ≥ (1 (1 − λ ) F ( z 0 ) + λ F ( z1 ) .
Appealing to the previous inequalities it follows that 1
F ( z λ ) ≥ (1 − λ ) q 0 + λ q
=q . λ
Hence the convex combination is indeed in the production set. (b (b)) Sinc Sincee F (1,1) ,1) = 1 , the the pr prod oduc ucti tion on plan plan ( −1, −1,1) ,1) is on the boundary of
Y
(c) To support the production production plan, it must be profit profit maximizing. That is, (− z* , q* ) = arg Max{ p3q − p1z1 − p2 z2 | q ≤ z11/ 3 z2 / 3} . z , q
Since profit is increasing in q the constraint must be binding at the maximum. Therefore z *
= arg Max{Π ( z) = z
p3 z11/ 3 z 2/ 3 − p1 z1 − p2 z2} .
First Order Conditions page
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∂Π ∂Π = p3 13 z1−2/ 3 z2 2/3 − p1 = 0 and = p3 23 z11/3 z2−1/ 3 − p2 = 0 . ∂ z1 ∂ z2 Substitu Subs tituting ting for z = (1,1) ,
∂Π 1 ∂Π 2 = 3 p3 − p1 = 0 and = p − p = 0 ∂ z1 ∂ z2 3 3 2 Hence p = (1, (1, 2,3) 2, 3) is a supporting supporting price vector.
Ex er cise 1.1-4: Robi nson Cru soe Economy Econ omy
Robinson Crusoe Crusoe lives alone on an island off the coast of New Zealand. He has a production set
Y
= {(− z1 , y2 ) | y2 ≤ 16 z11/3 ,
z1 ≥ 0} and an endowment vector
= = (32,0) .
ω
His preferences are represented by the utility function U ( x) = ln x1 + ln x2 . (a) Solve for his optimal choice of input input and hence his optimal production production plan and consumption plan x* . (b) Depict the production production set and the set
Y + ω in
a neat figure and indicate the optimal
production and consumption consumption plans. Explain what it means for the optimal optimal production plan to be supported supported by a price vector vector p = ( p1 , p2 ) . (c) Solve for the price vector that supports supports the optimal production production plan. (d) Depict this supporting supporting price line, Crusoe’s budget budget set and indifference curve though though * x .
(e) Hence explain why the supporting supporting price vector is a WE price vector if Robinson Crusoe is a price-taker. ANSWER
(a) If Robinson Robinson chooses input input level z1 , his consumption of commodity 1 is x1
= ω 1 − z1 = 32 − z1 . Output of commodity 2 is y2 = 16 z11/3 . Since his endowment of
commodity 2 is zero, his consumption of commodity 2 is x2 = y2 . His utility is therefore U
= ln x1 + ln x2 = ln(32 − z1) + ln 16 z11/3 = ln(32 − z1) + ln 16 + 13 ln z1 .
FOC
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dU dz1
21 September 2011
=
1 32 − z1
−
1 3
z1
= 0.
Solving, z1* = 8 . Then y2* = 32 and x* = (24,32)
(b) A price vector supports a production production plan if, at those price, the plan is profit maximizing. The more heavily shaded region is the production production set, bounded bounded by the production function y2 = 16 z11/ 3 = 16( − y1)1/ 3 . Adding the endowment vector
= ω =
(32,0)
shifts the set to the right by 32.
p ⋅ y = p ⋅ y*
y*
x2 , y2
x
32) = (−8, 32 Y
*
= (24,32)
Y + ω
ω
(c)
Π = p ⋅ y =
p2 y2
x1 , y1
− p1 z1 = p216 z11/3 − p1 z1 .
FOC
∂Π 16 = p2 z1−2/ 3 − p1 = 0 . ∂ z1 3 Since z1* = 8 ,
∂Π 4 = p2 − p1 = 0 . ∂ z1 3 page
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Thus the price vector p = (4,3) (4, 3) is a supporting supporting price vector. vector. At these prices, maximized profit is 64. 64. The iso-profit iso-profit line though the the profit-maximizing profit-maximizing production production plan is depicted depicted in the figure. (d) In the next figure figure the budget line line for Crusoe has been been added. Since profit is 64, the budget set is B = {x | x ≥ 0, p ⋅ x ≤ 64 + p ⋅ ω = 192} .
Note that the right-hand right-hand side is the the dividend (the (the profit) plus plus the value of Crusoe’s Crusoe’s endowment. (e) If you solve for the optimal optimal consumption vector in this this budget set you will find that that is it x* = (24,32) . Therefore the prices that support the optimal production plan are also market clearing prices.
p ⋅ y = p ⋅ y
*
x2 , y2 p ⋅ x
y
*
x
= (−8, 32 32) Y
= p ⋅ y + p ⋅ ω
*
= (24,32)
Y + ω
ω
x1 , y1
Ex er cise 1.1-6: Robin son Cru soe Economy Econ omy with wi th 4 commoditi es
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Robinson Crusoe has a production set endowment vector function U ( x) =
Y
= {(− z, y3 ) | y3 ≤ z11/ 3 z2 2 / 3 ,
z ≥ 0} and an
(32,160,, 0,16) . His preferences are represented by the utility = = (32,160
ω
4
∑
ln x j .
j =1
(a) Solve for the optimal production production plan and hence the optimal consumption consumption vector x* . (b) Obtain a price vector vector ( p1 , p2 , p3 ) , where p2 = 1 , that supports the optimal production plan. (c) What must be the price of of commodity 4 if x* is Crusoe’s optimal consumption bundle in his his budget set? HINT: There is only one consumer so so Crusoe must consume consume his endowment. ANSWER
(a) Utility is strictly increasing. Therefore x = (ω1 − z1 , ω2 − z2 , z11/ 3 z2 2 / 3 , ω 4 ) . Hence
= ln(32 − z1) + ln(160 − z2 ) + ln( z11/ 3 z2 2 / 3) + ln ω 4
U
= ln(32 − z1 ) + ln(160 − z2 ) + 13 ln z1 + 23 ln z2 + 13 ln 2 + ln ω 4 . FOC 1 2 ∂U ∂U 1 1 3 = − =0, = − 3 = 0 . Solving, z* = (8,64) . ∂ z1 32 − z1 z1 ∂ z2 160 − z2 z2
Then the optimal production plan is ( z * , q* ) = (8, 64, 32) and so x* = (24,96,32,16) . (b) The supporting price price vector must be profit maximizing. maximizing. Therefore z *
= arg Max{Π = z
p3 z11/ 3 z2 2/ 3 − p1 z1 − p2 z2 }
FOC
∂Π 1 = 3 p3 z1−2/ 3 z2 2/3 − p1 = 0 . ∂ z1 ∂Π 2 = 3 p3 z11/ 3 z2−1/3 − p2 = 0 ∂ z2 Substituting for z * , ( p1 , p2 , p3 ) = (4,1, 3) (c) The consumer’s optimization optimization problem problem can written as follows. follows. Max{U ( x) | p ⋅ x ≤ p ⋅ x*} . page
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FOC
∂U * ( x ) ∂ x j p j
=
1 *
pj xj
= λ , j=1,…,4.
Substituting for x* . 1 1 1 1 . = = = p1 24 p2 96 p3 32 p416 Hence p = (4,1,3,6) is a supporting supporting price vector.
Section 1.2: CONSTRAINED OPTIMIZATION
E xerci xerci se 1.2-2: Con sumer Ch oice
A consumer with income I has has utility function U ( x) = x1α 1 ... xnα n ,
α
> 0 and faces the
price vector p. (a) Find an increasing function function f such that u ( x) = f (U ( x)) is concave and hence quasiconcave. (b) By Proposition 1.2-4 1.2-4 the FOC for the modified problem problem are both necessary and sufficient. Hence solve for the consumer’s optimal choice. ANSWER
(a) Facing a price vector vector p the cost of a bundle x is p ⋅ x thus the consumer c onsumer must satisfy the budget constraint p ⋅ x ≤ I . The function u = ln U is increasing and ln U =
n
c oncave functions and is therefore concave and so quasi∑ α ln x is the sum of concave i
i
i =1
concave. Then we consider consider the concave maximization maximization problem, problem, Max{u = x
n
∑ α ln x | x ≥ 0,0, i
i
p ⋅ x ≤ I } .
i =1
For such a problem we need only check for a solution to the necessary conditions since they are also sufficient. (b) Writing the problem problem in standard form, the consumer consumer seeks to solve page
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Max{u ( x) | h( x) = I − p ⋅ x ≥ 0, x ≥ 0} . x
Note that the gradient vector n
L( x, λ )
=∑
α j
∂h ≠ 0 thus the KT constraint qualifications are satisfied. ∂ x
ln x j + λ ( I −
j =1
n
∑=
pj xj )
j 1
First order conditions
∂L α j = − λ p j ≤ 0, with equality if x j* > 0 . ∂ x j x j Since the first term increases without bound as x j goes to zero, this constraint cannot be satisfied if x j = 0 . It follows follows that x j* > 0, j = 1, ..., n . Then, from the first order conditions, α ∂L α j = − λ p j = 0 and so x j* = j . λ p j ∂ x j x j
Substituting into the budget constraint, n
n
n α j
∑= p x = ∑= λ = I . Hence λ = j
j
j 1
Then x j* =
j 1
α j λ p j
=
α j
I
n
pj
∑=
α i
∑= α
j
j 1
I
.
i 1
Ex er cise cise 1.2-4: 1.2-4: Satisf Satisf ying th e constrai constrai nt qual if ication with a sin sin gle cons constr tr ain t
There is a single constraint g ( x) ≥ g ( x ) and x ⋅
∂g (x) < 0 . ∂ x
(a) Show that the linearized set at x has a non-empty interior. interior. (b) Hence show that the constraint qualifications hold at x . ANSWER
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(a) Define h( x) = g ( x) − g ( x ) so that the constraint can be written in the standard form h( x) ≥ 0 . The gradient of h at x is
∂g ( x ) so the linearized constraint is ∂ x
∂g ( x ) ⋅ ( x − x ) ≥ 0 . ∂ x ∂g By hypothesis x ⋅ ( x ) < 0 thus the constraint is satisfied strictly at x = 0 . Then it is ∂ x also satisfied at interior points of the linearized feasible set in some neighborhood neighborhood of x = 0 . Hence the interior of the linearized feasible set is non-empty.
(b) Since x ⋅
∂g ∂h ∂g ( x ) < 0 , it follows that ( x ) = ( x ) ≠ 0 . Thus the constraint ∂ x ∂ x ∂x
qualifications are both satisfied at x .
1.3 THE ENVELOPE THEOREM
Ex er cise cise 1.3-2. 1.3-2. I ll ustratin g the Envelope Th eorem
To produce q units of output, a profit-maximizing profit-maximizing firm requires 12 q 2 units of the single input. The price of the the output is p and the price of the input is r . (a) Write rite do down wn an exp express ession ion for prof rofit π ( q, p, r ) . (b) Solve for the profit maximizing output q* and hence show that maximized profit is
Π ( p , r ) = π ( q ( p , r), p, r ) = *
(c) Confirm that
1 2
p2 r
.
∂Π ∂π * ∂Π ∂π * = (q , p, r ) and = (q , p, r ) . ∂ p ∂p ∂r ∂r
ANSWER
To produce q units of output, a profit-maximizing profit-maximizing firm requires 12 q 2 units of the single input. The price of the the output is p and the price of the input is r . (c) Write rite do down wn an exp express ession ion for prof rofit π ( q, p, r ) . page
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(d) Solve for the profit maximizing output q* and hence show that maximized profit is
Π ( p , r ) = π ( q ( p , r), p, r ) = *
(c) Confirm that
(a)
π
p
1 2
2
r
.
∂Π ∂π * ∂Π ∂π * = (q , p, r ) and = (q , p, r ) . ∂ p ∂p ∂r ∂r
= pq − 12 rq 2 .
(b) This is a strictly concave function so the the first order condition condition is both necessary and sufficient.
∂π = p − rq = 0 at q(r, p) = p / r . ∂q Substituting into the expression for profit, maximized profit is
Π (r , p ) =
pq( r, p) − 12 rq( r, p)
2
= 12
2
p / r .
(c) Hence
∂Π ∂Π = p / r and = − 12 ( p / r ) 2 . ∂ p ∂r Since q (r , p) = p / r we can rewrite these expressions as follows.
∂Π ∂Π = q(r, p) and = − 12 q(r , p) 2 .. ∂ p ∂r Since
π
= pq − 12 rq 2 ,
∂π ∂π = q and = −q 2 ∂ p ∂r
1.3-4: 1.3-4: En velope velope Th eorem with a corner solu solu tion
A consumer has a utility function U ( x) = ln(α + x1 ) + ln x2 , income I and and faces a price vector p = ( p1 , p2 ) >> 0 . (a) Show that if she consumes both commodities, commodities, her optimal consumption consumption bundle is
⎛ I x( p, I , α ) = ⎜
− p1α
⎝ 2 p1
,
I + p1α ⎞
⎟.
2 p2 ⎠
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(b) Hen Hence write down an expression for maximized utility F (α ) = U ( x( p, I, α )) and show that dF dα
=
2 p1 I
.
+ α p1
(c) Explain why, for large differentiable for all
α ,
F (α )
= ln α + ln( I /
p2 ) and confirm that F is is continuously
≥ ≥ 0 .
α
ANSWER
(a) If x* >> 0 the marginal utility per dollar is the same for both commodities, that is 1 ∂U
∂xi
pi
=
1 p1 (α
=
+ x1 )
1 p2 x2
=
2 p1α
+ p⋅ x
=
2 p1α + I
.
Note that we have appealed to to the ratio rule for for the third equality. equality. Since utility is strictly strictly monotonic, total expenditure is equal to income, hence the fourth equality. Solving, x*
I
=(
− p1α 2 p1
,
I
+ p1α 2 p2
)
(b) Substituting into the utility utility function, F (α ) = U (α , x* (α )) = ln(
p1α
+I
2 p1
) + ln ln(
p1α + I
2 p2
)
= − ln 2 p1 − ln 2 p2 + 2 ln( p1α + I ) Differentiating by dF dα
=
α ,
2 p1 p1α + I
(c) From the answer to (a) we note note that x1* cannot be strictly positive if
> > I / p1 . For a
α
corner solution with x1* = 0 , the marginal utility per dollar must be lower for commodity 1. That is 1 ∂U p1 ∂x1
(0, I / p2 ) <
1 ∂U p2 ∂x2
(0, I / p2 ) .
Substituting, page
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1
1
p1α
< . I
Then F (α ) = U (0, I / p2 ) = ln α + ln I / p2 .
= = I / p1 .
It is then easy to check that the derivative is continuous at
α
This exercise illustrates three three general propositions. propositions. First, given the strict quasi-
Remark:
concavity of the utility function and the convexity of the feasible set, the solution
(
)
x p, I , α is a continuous function function in all its arguments. Second, given the continuity of
the solution, F (α ) is continuou c ontinuously sly differentiable for all
≥ ≥ 0 . Finally, by the envelope
α
theorem, dF α d α
=
1 ∂U x ( p, I , α )) = . ( x ( p, I , α ) + α ∂α
Section 1.4 FOUNDATIONS OF CONSTRAINED OPTIMIZATION
Ex er cise cise 1.4-2: 1.4-2: Generali Generali zation of the Ku hn -Tu cker cker conditi ons
(a) Suppose that x
= arg Max{a0 ⋅ x | Ax ≤ z , x
x j ∈ + , j ∈ I , x j ∈ , j ∈ J , x j ∈ − , j ∈ K }
exists and that the interior of the feasible set is non-empty. Modify the proof or Proposition Proposition 1.1-5 to obtain new necessary conditions. conditions. (b) Suppose that x
= arg Max{ f ( x) | hi ( x) ≥ 0, i = 1, ..., m, x
x j ∈ + , j ∈ I , x j ∈ j ∈ J , x j ∈ − , j ∈ K }
Explain how Proposition 1.4-3 must be restated. ANSWER
(a) For the linear model, it follows from the the supporting hyperplane hyperplane theorem that
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q − q − λ ′( z − z ) ≤ 0 fo for all ( − z, q) ∈ Y .
(1)
Consider the activity vector x = x + δ . The associated change change in input input requirements is z − z
= Aδ and the change in output is q − q = a0′ ( x − x ) = a0′δ .
Substituting into (1), a0′δ
− λ ′Aδ = (a0′ − λ ′ A) ⋅δ ≤ 0 .
Next set all the changes in in x except the jth component equal to zero. This inequality then becomes (a0 j −
m
∑ λ a )δ ≤ 0 . i
ij
(2)
j
i =1
For i ∈ I the argument is an in Chapter 1. For i ∈ J both x j + α and x j − α are feasible as long as
α is
(a0 j −
sufficiently small. Therefore
m
m
∑ λ a )α ≤ 0 and (a − ∑ λ a )(−α ) ≤ 0 i
0 j
ij
i =1
i
ij
i =1
and so m
a0 j
−∑
=0.
λ i aij
i =1
For i ∈ K , there are two cases. If x j < 0 we can argue as above that a0 j − If x j = 0 , x j − α is feasible if (a0 j −
m
∑ λ a )δ i
m
a0 j
ij
j
−∑
λ i aij
∑ λ a i
ij
=0.
i =1
> > 0 . Therefore, appealing to (2),
α
m
i =1
m
= ( a0 j − ∑
λi aij )( −α )
≤ 0 and so
i =1
≥0
i =1
(b) The FOC for the non-linear optimization optimization problem are the FOC for the linearized optimization problem. problem. Thus the FOC become
∂L ≤ 0, i ∈ I , with equality if xi > 0 ∂ xi page
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∂L = 0, j ∈ J ∂ x j
∂L ≥ 0, k ∈ K , with equality if xk > 0 ∂ xk
SECTION 1.5
Exercise Exercise 1.51.5-2:M 2:M ul ti-product fi rm with Joint Costs Costs
A firm produces two products. The production function function for product j is F j ( K j , L j ) =
K j L j . Capital equipment can be be used in the production production of either either or both
products. Thus if K 0 units of capital are rented, the capital use constraints are K j
≤ K0 ,
j = 1, 2 . The unit cost of labor labor is 2 and the unit unit cost of capital is 10.
(a) Explain why the Lagrangian for minimizing the total cost of producing (q1 , q2 ) can be written as follows L( q, K )
=−
2q12 K1
−
2q2 2 K 2
− 10 K0 + λ1( K0 − K1 ) + λ 2 ( K 0 − K2 ) .
(b) Hence, or otherwise show that all the capital equipment will be used for the production of both products. products. (c) Solve for the minimized total cost. (d) Fix q2 and depict the marginal cost of producing commodity commodity 1. (a) From the production function qt2 = K t Lt . Thus, given that K t units of capital have been rented, labor demand demand is Lt = qt / K t . Hence total cost is is 2
2q12 K1
+
2q22 K 2
+ 10K 0 .
(b) FOC
∂L = −10 + λ1 + λ 2 = 0 (since ∂K 0
qt
> 0 ⇒ K t > 0 )
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∂L 2qt 2 = − λ = 0 since ∂Kt K t 2 t
Since
2qt 2 2
K t
K t > 0, t = 1, 2 .
> 0 th the second FOC cannot be satisfied unless λ t > 0, t = 1,2 . Hence the
second constraint must hold with equality and so K t = K 0 , t = 1, 2
(a) From the FOC
2q12 K 02
+
2q22 K 02
= 10 , hence
K0
=
1 1 2 (q1 + q22 ) 2 . Substituting, minimized 5
1
total cost is C (q) = 4 5 ( q12 + q22 ) 2 .
(b) Differentiating by q1 ,
∂C 4 5q1 = = ∂q1 (q 2 + q 2 ) 12 1
2
4 5 (1 +
q22
1
)2 2
q1
Thus, as q1 increas increases, es, marg margina inall co cost st rise risess fr from om zero zero to 4 5 .
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