Microeconomia, solucion de varian

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Microeconomic Analysis
Third Edition

Hal R. Varian
University of California at Berkeley

W. W. Norton & Company • New York • London

Copyright c 1992, 1984, 1978 by W. W. Norton & Company, Inc.

All rights reserved Printed in the United States of America

THIRD EDITION

0-393-96282-2

W. W. Norton & Company, Inc., 500 Fifth Avenue, New York, N.Y. 10110 W.W. Norton Ltd., 10 Coptic Street, London WC1A 1PU

234567890

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 = dt so that 1 dy 1 = y dt f(x) 1.5 Substitute txi for i = 1, 2 to get f(tx1 , tx2 ) = [(tx1 )ρ + (tx2 )ρ ] ρ = t[xρ + xρ ] ρ = tf(x1 , x2 ). 1 2 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 g (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 g (0) = 0. 1.7 Let f(x) = g(h(x)) and suppose that g(h(x)) = g(h(x )). Since g is monotonic, it follows that h(x) = h(x ). Now g(h(tx)) = g(th(x)) and g(h(tx )) = g(th(x )) which gives us the required result. 1.8 A homothetic function can be written as g(h(x)) where h(x) is homogeneous ofdegree 1. Hence the TRS of a homothetic function has the
1 1

n

i=1

∂f(x) xi , ∂xi
n

i=1

∂f(x) xi . ∂xi

form

∂h g (h(x)) ∂h ∂x1 ∂x1 = . ∂h g (h(x)) ∂h ∂x2 ∂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 wecan write (a1 + a2 )
1 ρ

a2 a1 xρ + xρ a1 + a2 1 a1 + a2 2

1 ρ

.
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 y in Y and 0 ≤ t ≤ 1, we must have ty + (1 − t)y in Y . But divisibility implies that ty and (1 − t)y are in Y , and additivity implies that their sum is in Y . To show constant returns to scale, we mustshow 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 oflog 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 theproduction 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 = − x1 2 x2 2 4 ∂x1 ∂ 2 f(x) 1 −1 − 1 = x12 x2 2 ∂x1 ∂x2 4

∂ 2 f(x) 1 1 −3 2 = − x1 x22 24 ∂x2

Ch. 2 PROFIT MAXIMIZATION

3

Hessian =

− 1 x1 x2 4 1 −1/2 −1/2 x1 x2 4

−3/2 1/2

1 −1/2 −1/2 x2 4 x1 1 1/2 −3/2 − 4 x1 x2

1 −3/2 1/2 D1 = − x1 x2 < 0 4 1 −1 −1 1 D2 = x x − x−1 x−1 = 0. 16 1 2 16 1 2 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...