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Exam (elaborations) TEST BANK FOR Microeconomic Analysis 3rd Edition B
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, Answers toExercises
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 = {(y, −x) : y ≤ x2 } which is certainly not convex, but the input re-
√
quirement set is V (y) = {x : x ≥ y} 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 X ∂f(x)
n
= xi ,
dt ∂xi
i=1
so that
1 X ∂f(x)
n
1 dy
= xi .
y dt f(x) ∂xi
i=1
1.5 Substitute txi for i = 1, 2 to get
1 1
f(tx1 , tx2 ) = [(tx1 )ρ + (tx2 )ρ ] ρ = t[xρ1 + xρ2 ] ρ = 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 ho-
mogeneous of degree 1. Hence the TRS of a homothetic function has the
, 2 ANSWERS
form
g0 (h(x)) ∂h ∂h
∂x1 ∂x1
= .
g0 (h(x)) ∂h ∂h
∂x2 ∂x2
That is, the TRS of a homothetic function is just the TRS of the un-
derlying homogeneous function. But we already know that the TRS of a
homogeneous function has the required property.
1.9 Note that we can write
1ρ
1 a1 a2
(a1 + a2 ) ρ xρ + xρ .
a1 + a2 1 a1 + a2 2
1
Now simply define b = a1 /(a1 + a2 ) and A = (a1 + a2 ) ρ .
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
sufficient (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 semidefinite. The first principal minor of
the Hessian must have a negative determinant, and the second principal
minor must have a nonnegative determinant.
∂ 2 f(x) 1 −3 1 ∂ 2 f(x) 1 −1 − 1
2 = − x1 2 x22 = x12 x2 2
∂x1 4 ∂x1 ∂x2 4
∂ 2 f(x) 1 1 −3
2 = − x12 x22
∂x2 4
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