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Nicholson Microeconomics-Solutions-Manual

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CHAPTER 2

THE MATHEMATICS OF OPTIMIZATION

The problems in this chapter are primarily mathematical. They are intended to give students some practice
with taking derivatives and using the Lagrangian techniques, but the problems in themselves offer few
economic insights. Consequently, no commentary is provided. All of the problems are relatively simple
and instructors might choose from among them on the basis of how they wish to approach the teaching of
the optimization methods in class.

Solutions

2.1
2 2( , ) 4 3\uf03d \uf02bU x y x y

a.
8 6
\uf0b6 \uf0b6
\uf0b6 \uf0b6
U U
= x , = y
x y

b. 8, 12
c.
8 6
\uf0b6 \uf0b6
\uf03d
\uf0b6 \uf0b6
U U
dU dx + dy = x dx + y dy
x y

d.
for 0 8 6 0\uf03d \uf02b \uf03d
dy
dU x dx y dy
dx

8 4
6 3
\uf02d \uf02ddy x x
= =
dx y y

e.
1, 2 4 1 3 4 16\uf03d \uf03d \uf03d \uf0d7 \uf02b \uf0d7 \uf03dx y U

f.
4(1)
2 / 3
3(2)
\uf02d
\uf03d \uf03d \uf02d
dy

dx

g. U = 16 contour line is an ellipse centered at the origin. With equation
2 24 3 16\uf02b \uf03dx y
, slope of the line at (x, y) is
4
3
\uf03d \uf02d
dy x

dx y
.

2.2 a. Profits are given by
22 40 100\uf070 \uf03d \uf02d \uf03d \uf02d \uf02b \uf02dR C q q
*
4 40 10
\uf070
\uf03d \uf02d \uf02b \uf03d
d
q q
dq

2* 2(10 40(10) 100 100)\uf070 \uf03d \uf02d \uf02b \uf02d \uf03d
b. 2
2
4
\uf070
\uf03d \uf02d
d

dq
so profits are maximized
c.
70 2\uf03d \uf03d \uf02d
dR
MR q
dq

2 \uf076 Solutions Manual
2 30\uf03d \uf03d \uf02b
dC
MC q
dq

so q* = 10 obeys MR = MC = 50.

2.3 Substitution:
21 so \uf03d \uf02d \uf03d \uf03d \uf02dy x f xy x x

1 2 0
\uf0b6
\uf03d \uf02d \uf03d
\uf0b6
f
x
x

0.5 0.5, 0.25x = , y = f =

Note:
2 0\uf0a2\uf0a2 \uf03d \uf02d \uf03cf
. This is a local and global maximum.
Lagrangian Method:
? 1 )\uf06c\uf03d \uf02b \uf02d \uf02dxy x y

£
\uf06c
\uf0b6
\uf02d
\uf0b6

= y = 0
x

£
\uf06c
\uf0b6
\uf02d
\uf0b6

= x = 0
y

so, x = y.
using the constraint gives
0.5, 0.25\uf03d \uf03d \uf03dx y xy

2.4 Setting up the Lagrangian:
? 0.25 )\uf06c\uf03d \uf02b \uf02b \uf02dx y xy
.

£
1
£
1
\uf06c
\uf06c
\uf0b6
\uf03d \uf02d
\uf0b6
\uf0b6
\uf03d \uf02d
\uf0b6
y
x
x
y

So, x = y. Using the constraint gives
2 0.25, 0.5\uf03d \uf03d \uf03d \uf03dxy x x y
.

2.5 a.
2( ) 0.5 40\uf03d \uf02d \uf02bf t gt t

* 4040 0,\uf03d \uf02d \uf02b \uf03d \uf03d
df
g t t
dt g
.
b. Substituting for t*,
* 2( ) 0.5 (40 ) 40(40 ) 800\uf03d \uf02d \uf02b \uf03df t g g g g
.
*
2( ) 800
\uf0b6
\uf03d \uf02d
\uf0b6
f t
g
g
.
c.
2*1 ( )
2
\uf0b6
\uf03d \uf02d
\uf0b6
f
t
g
depends on g because t
*
depends on g.
so
* 2 2
2
40 800
0.5( ) 0.5( )
\uf0b6 \uf02d
\uf03d \uf02d \uf03d \uf02d \uf03d
\uf0b6
f
t
g g g
.
Chapter 2/The Mathematics of Optimization \uf076 3
d.
800 32 25, 800 32.1 24.92\uf03d \uf03d
, a reduction of .08. Notice that
2 2800 800 32 0.8\uf02d \uf03d \uf0bb \uf02dg
so a 0.1 increase in g could be predicted to reduce
height by 0.08 from the envelope theorem.

2.6 a. This is the volume of a rectangular solid made from a piece of metal which is x by 3x
with the defined corner squares removed.
b.
2 23 16 12 0
\uf0b6
\uf03d \uf02d \uf02b \uf03d
\uf0b6
V
x xt t
t
. Applying the quadratic formula to this expression yields
2 216 256 144 16 10.6
0.225 , 1.11
24 24
\uf0b1 \uf02d \uf0b1
\uf03d \uf03d \uf03d
x x x x x
t x x
. To determine true
maximum must look at second derivative -- 2
2
16 24
\uf0b6
\uf03d \uf02d \uf02b
\uf0b6
V
x t
t
which is negative only
for the first solution.
c. If
3 3 3 30.225 , 0.67 .04 .05 0.68\uf03d \uf0bb \uf02d \uf02b \uf0bbt x V x x x x
so V increases without limit.
d. This would require a solution using the Lagrangian method. The optimal solution
requires solving three non-linear simultaneous equations\u2014a task not undertaken here.
But it seems clear that the solution would involve a different relationship between t and
x than in parts a-c.

2.7 a. Set up Lagrangian
1 2 1 2? ln ( )\uf06c\uf03d \uf02b \uf02b \uf02d \uf02dx x k x x
yields the first order conditions:
1
2 2
1 2
£
1 0
?
0
£
0
\uf06c
\uf06c
\uf06c
\uf0b6
\uf03d \uf02d \uf03d
\uf0b6
\uf0b6
\uf03d \uf02d \uf03d
\uf0b6
\uf0b6
\uf03d \uf02d \uf02d \uf03d
\uf0b6
x
x x
k x x

Hence,
2 21 5 or 5\uf06c \uf03d \uf03d \uf03dx x
. With k = 10, optimal solution is
1 2 5.\uf03d \uf03dx x

b. With k = 4, solving the first order conditions yields
2 15, 1.\uf03d \uf03d \uf02dx x

c. Optimal solution is
1 20, 4, 5ln 4.\uf03d \uf03d \uf03dx x y
Any positive value for x1 reduces y.
d. If k = 20, optimal solution is
1 215, 5.\uf03d \uf03dx x
Because x2 provides a diminishing
marginal increment to y whereas x1 does not, all optimal solutions require that, once x2
reaches 5, any extra amounts be devoted entirely to x1.

2.8 The proof is most easily accomplished through the use of the matrix algebra of quadratic
forms. See, for example, Mas Colell et al., pp. 937\u2013939. Intuitively, because concave
functions lie below any tangent plane, their level curves must also be convex. But the
converse is not true. Quasi-concave functions may exhibit \u2015increasing returns to scale\u2016;
even though their level curves are convex, they may rise above the tangent plane when all
variables are increased together.

4 \uf076 Solutions Manual
2.9 a.
1
1 21 0.
\uf062\uf061\uf061 \uf02d\uf03d \uf03e f x x

1
1 22 0
\uf062\uf061\uf062 \uf02d\uf03d \uf03e f .x x

2
1 111 ( 1) 0.
\uf062\uf061\uf061 \uf061 \uf02d\uf03d \uf02d \uf03c f x x

2
1 222 ( 1) 0.
\uf062\uf061\uf062 \uf062 \uf02d\uf03d \uf02d \uf03c f x x

11
1 212 21 0.
\uf062\uf061\uf061 \uf062 \uf02d\uf02d\uf03d \uf03d \uf03e f f x x

Clearly, all the terms in Equation 2.114 are negative.
b. If
1 2
\uf062\uf061\uf03d \uf03dy c x x

/1/
2 1
\uf061 \uf062\uf062 \uf02d\uf03d x c x
since \u3b1, \u3b2 > 0, x2 is a convex function of x1 .
c. Using equation 2.98,
2 22 2 2 22 2 2 2 2
1 12 211 1222
( 1) ( ) ( 1) \uf062 \uf062\uf061 \uf061\uf061 \uf061 \uf062 \uf062 \uf062\uf061\uf02d \uf02d\uf02d \uf02d\uf02d \uf03d \uf02d \uf02d \uf02d f f f x x x x
=
2 22 2
1 2(1 )
\uf062\uf061\uf061 \uf062 \uf062 \uf061 \uf02d\uf02d\uf02d \uf02d x x which is negative for \u3b1 + \u3b2 > 1.

2.10 a. Since
0, 0\uf0a2 \uf0a2\uf0a2\uf03e \uf03cy y
, the function is concave.
b. Because
11 22, 0\uf03cf f
, and
12 21 0\uf03d \uf03df f
, Equation 2.98 is satisfied and the function
is concave.
c. y is quasi-concave as is
\uf067
y
. But
\uf067
y
is not concave for \u3b3 > 1. All of these results
can be shown by applying the various definitions to the partial derivatives of y.

5
CHAPTER 3

PREFERENCES AND UTILITY

These problems provide some practice in examining utility functions by looking at indifference
curve maps. The primary focus is on illustrating the notion of a diminishing MRS in various
contexts. The concepts of the budget constraint and utility maximization are not used until the next
chapter.

3.1 This problem requires students to graph indifference curves for a variety of functions,
some of which do not exhibit a diminishing MRS.
3.2 Introduces the formal definition of quasi-concavity (from Chapter 2) to be applied to the
functions in Problem 3.1.
3.3 This problem shows that diminishing marginal utility is not required to obtain a
diminishing MRS. All of the functions are monotonic transformations of one another, so
this problem illustrates that diminishing MRS is preserved by monotonic transformations,
but diminishing marginal utility is not.
3.4 This problem focuses on whether some simple utility functions exhibit convex
indifference curves.
3.5 This problem is an exploration of the fixed-proportions utility function. The problem also
shows how such problems can be treated as a composite commodity.
3.6 In this problem students are asked to provide a formal, utility-based explanation for a
variety of advertising slogans. The purpose is to get students to think mathematically
3.7 This problem shows how initial endowments can be incorporated into utility theory.
3.8 This problem offers a further exploration of the Cobb-Douglas function. Part c provides
an introduction to the linear expenditure system. This application is treated in more detail
in the Extensions to Chapter 4.
3.9 This problem shows that independent marginal utilities illustrate one situation in which
diminishing marginal utility ensures a diminishing MRS.
3.10 This problem explores various features of the CES function with weighting on the two
goods.

6 \uf076 Solutions Manual
Solutions

3.1 Here we calculate the MRS for each of these functions:
a.
3 1\uf03d \uf03dx yMRS f f
\u2014 MRS is constant.
b. 0.5
0.5
0.5( )
0.5( )\uf02d
\uf03d \uf03d \uf03dx y
y x
MRS f f y x
y x
\u2014 MRS is diminishing.
c.
0.50.5 1\uf02d\uf03d \uf03dx yMRS f f x
\u2014 MRS is diminishing
d.
2 2 0.5 2 2 0.50.5( ) 2 0.5(```
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