Exam (elaborations) TEST BANK FOR Microeconomic Analysis 3rd Edition By Hal R. Varian (Solution Manual)
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Exercises
Microeconomic
Analysis
Third Edition
Hal R. Varian
University of California at Berkeley
W. W. Norton & Company New York London
ANSWERS
Chapter 1. Technology
1.1 False. There are many counterexamples. Consider the technology
generated by a production function f(x) = x2. The production set is
Y = f(y;−x) : y x2g which is certainly not convex, but the input requirement
set is V (y) = fx : x
p
yg which is a convex set.
1.2 It doesn't change.
1.3 1 = a and 2 = b.
1.4 Let y(t) = f(tx). Then
dy
dt
=
Xn
i=1
@f(x)
@xi
xi;
so that
1
y
dy
dt
=
1
f(x)
Xn
i=1
@f(x)
@xi
xi:
1.5 Substitute txi for i = 1;2 to get
f(tx1; tx2) = [(tx1) + (tx2)] 1
= t[x
1 + x
2] 1
= tf(x1; x2):
This implies that the CES function exhibits constant returns to scale and
hence has an elasticity of scale of 1.
1.6 This is half true: if g0(x) > 0, then the function must be strictly
increasing, but the converse is not true. Consider, for example, the function
g(x) = x3. This is strictly increasing, but g0(0) = 0.
1.7 Let f(x) = g(h(x)) and suppose that g(h(x)) = g(h(x0)). Since g is
monotonic, it follows that h(x) = h(x0). Now g(h(tx)) = g(th(x)) and
g(h(tx0)) = g(th(x0)) which gives us the required result.
1.8 A homothetic function can be written as g(h(x)) where h(x) is homogeneous
of degree 1. Hence the TRS of a homothetic function has the
2 ANSWERS
form
g0(h(x)) @h
@x1
g0(h(x)) @h
@x2
=
@h
@x1
@h
@x2
:
That is, the TRS of a homothetic function is just the TRS of the underlying
homogeneous function. But we already know that the TRS of a
homogeneous function has the required property.
1.9 Note that we can write
(a1 + a2) 1
a1
a1 + a2
x
1 + a2
a1 + a2
x
2
1
:
Now simply dene b = a1=(a1 + a2) and A = (a1 +a2)1
.
1.10 To prove convexity, we must show that for all y and y0 in Y and
0 t 1, we must have ty + (1−t)y0 in Y . But divisibility implies that
ty and (1 − t)y0 are in Y , and additivity implies that their sum is in Y .
To show constant returns to scale, we must show that if y is in Y , and
s > 0, we must have sy in Y. Given any s > 0, let n be a nonnegative
integer such that n s n − 1. By additivity, ny is in Y ; since s=n 1,
divisibility implies (s=n)ny = sy is in Y .
1.11.a This is closed and nonempty for all y > 0 (if we allow inputs to be
negative). The isoquants look just like the Leontief technology except we
are measuring output in units of log y rather than y. Hence, the shape of
the isoquants will be the same. It follows that the technology is monotonic
and convex.
1.11.b This is nonempty but not closed. It is monotonic and convex.
1.11.c This is regular. The derivatives of f(x1; x2) are both positive so the
technology is monotonic. For the isoquant to be convex to the origin, it is
sucient (but not necessary) that the production function is concave. To
check this, form a matrix using the second derivatives of the production
function, and see if it is negative semidenite. The rst principal minor of
the Hessian must have a negative determinant, and the second principal
minor must have a nonnegative determinant.
@2f(x)
@x21
= −1
4x
−32
1 x
12
2
@2f(x)
@x1@x2
=
1
4x
−1
2
1 x
−12
2
@2f(x)
@x22
= −1
4x
12
1 x
−3
2
2
Ch. 2 PROFIT MAXIMIZATION 3
Hessian =
"
−1
4x
−3=2
1 x1=2
2
1
4x
−1=2
1 x
−1=2
2
1
4x
−1=2
1 x
−1=2
2
−1
4x1=2
1 x
−3=2
2
#
D1 = −1
4x
−3=2
1 x1=2
2 < 0
D2 =
1
16x
−1
1 x
−1
2
− 1
16x
−1
1 x
−1
2 = 0:
So the input requirement set is convex.
1.11.d This is regular, monotonic, and convex.
1.11.e This is nonempty, but there is no way to produce any y > 1. It is
monotonic and weakly convex.
1.11.f This is regular. To check monotonicity, write down the production
function f(x) = ax1 −
p
x1x2 + bx2 and compute
@f(x)
@x1
= a − 1
2x
−1=2
1 x1=2
2 :
This is positive only if a > 1
2
q
x2
x1
, thus the input requirement set is not
always monotonic.
Looking at the Hessian of f, its determinant is zero, and the determinant
of the rst principal minor is positive. Therefore f is not concave. This
alone is not sucient to show that the input requirement sets are not
convex. But we can say even more: f is convex; therefore, all sets of the
form
fx1; x2: ax1 −
p
x1x2 + bx2 yg for all choices of y
are convex. Except for the border points this is just the complement of
the input requirement sets we are interested in (the inequality sign goes in
the wrong direction). As complements of convex sets (such that the border
line is not a straight line) our input requirement sets can therefore not be
themselves convex.
1.11.g This function is the successive application of a linear and a Leontief
function, so it has all of the properties possessed by these two types of
functions, including being regular, monotonic, and convex.
Chapter 2. Profit Maximization
4 ANSWERS
2.1 For prot maximization, the Kuhn-Tucker theorem requires the following
three inequalities to hold
p
@f(x)
@xj
−wj
x
j = 0;
p
@f(x)
@xj
− wj 0;
x
j
0:
Note that if x
j > 0, then we must have wj=p = @f(x)=@xj.
2.2 Suppose that x0 is a prot-maximizing bundle with positive prots
(x0) > 0. Since
f(tx0) > tf(x0);
for t > 1, we have
(tx0) = pf(tx0) − twx0
> t(pf(x0) − wx0) > t(x0) > (x0):
Therefore, x0 could not possibly be a prot-maximizing bundle.
2.3 In the text the supply function and the factor demands were computed
for this technology. Using those results, the prot function is given by
(p;w) = p
w
ap
a
a−1
− w
w
ap
1
a−1
:
To prove homogeneity, note that
(tp; tw) = tp
w
ap
a
a−1
− tw
w
ap
1
a−1
= t(p;w);
which implies that (p;w) is a homogeneous function of degree 1.
Before computing the Hessian matrix, factor the prot function in the
following way:
(p;w) = p
1
1−aw
a
a−1
a
a
1−a −a
1
1−a
= p
1
1−aw
a
a−1(a);
where (a) is strictly positive for 0 < a< 1.
The Hessian matrix can now be written as
D2(p; !) =
@2(p;w)
@p2
@2(p;w)
@p@w
@2(p;w)
@w@p
@2(p;w)
@w2
!
=
0
BBB@
a
(1−a)2 p
2a−1
1−a w
a
a−1 − a
(1−a)2 p
a
1−a w
1
a−1
− a
(1−a)2 p
a
1−aw
1
a−1 a
(1−a)2 p
1
1−a w
2−a
a−1
1
CCCA
(a):
Ch. 2 PROFIT MAXIMIZATION 5
The principal minors of this matrix are
a
(1 − a)2 p
2a−1
1−a w
a
a−1 (a) > 0
and 0. Therefore, the Hessian is a positive semidenite matrix, which
implies that (p;w) is convex in (p;w).
2.4 By prot maximization, we have
jTRSj =
@f
@x1
@f
@x2
= w1
w2
:
Now, note that
ln(w2x2=w1x1) = −(ln(w1=w2) + ln(x1=x2)):
Therefore,
d ln(w2x2=w1x1)
d ln(x1=x2)
= d ln(w1=w2)
d ln(x2=x1)
− 1 = dln jTRSj
d ln(x2=x1)
− 1 = 1= − 1:
2.5 From the previous exercise, we know that
ln(w2x2=w1x1) = ln(w2=w1) + ln(x2=x1);
Dierentiating, we get
d ln(w2x2=w1x1)
d ln(w2=w1)
= 1− dln(x2=x1)
d ln jTRSj = 1−:
2.6 We know from the text that Y O Y Y I. Hence for any p, the
maximum of py over YO must be larger than the maximum over Y , and
this in turn must be larger than the maximum over Y I.
2.7.a We want to maximize 20x − x2 − wx. The rst-order condition is
20 − 2x − w = 0.
2.7.b For the optimal x to be zero, the derivative of prot with respect to
x must be nonpositive at x = 0: 20−2x−w < 0 when x = 0, or w 20.
2.7.c The optimal x will be 10 when w = 0.
2.7.d The factor demand function [Show Less]