Solution Manual for Microeconomic Theory Basic Principles and Extensions, 12th Edition by Walter Nicholson, Christopher M. Snyder

Microeconomic Theory:
Basic Principles and Extensions
12th Edition




Solutions Manual


Walter Nicholson & Christopher Snyder




Preface
© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or
posted to a publicly accessible website, in whole or in part.

, CHAPTER 2:
Mathematics for Microeconomics
The problems in this chapter are primarily mathematical. They are intended to give students
some practice with the concepts introduced in Chapter 2, but the problems in themselves offer
few economic insights. Consequently, no commentary is provided. Results from some of the
analytical problems are used in later chapters, however, and in those cases the student will be
directed back to this chapter.




Solutions



2.1 f ( x, y) = 4 x 2 + 3 y 2 .



a. f x = 8 x, f y = 6 y.



b. Constraining f ( x, y ) = 16 creates an implicit function between the variables. The
dy f −8 x
slope of this function is given by =− x = for combinations of x and y
dx fy 6y
that satisfy the constraint.




© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or
posted to a publicly accessible website, in whole or in part.

, dy 8 1 2
c. Since f (1, 2) = 16 , we know that at this point =− =− .
dx 62 3



d. The f ( x, y ) = 16 contour line is an ellipse centered at the origin. The slope of the
line at any point is given by dy dx = − 8x 6 y . Notice that this slope becomes
more negative as x increases and y decreases.




2.2 a. Profits are given by  = R − C = −2q 2 + 40q − 100. The maximum value is found
by setting the derivative equal to 0:

d
= − 4q + 40 = 0 ,
dq

implies q* = 10 and  * = 100.



b. Since d 2 dq 2 = − 4  0, this is a global maximum.



c. MR = dR dq = 70 − 2q. MC = dC dq = 2q + 30. So, q* = 10 obeys
MR = MC = 50.

2.3 First, use the substitution method. Substituting y = 1 − x yields
f ( x) = f ( x,1 − x) = x(1 − x) = x − x 2 . Taking the first-order condition, f  ( x) = 1 − 2 x = 0,
and solving yields x* = 0.5, y* = 0.5 , and f ( x* ) = f ( x* , y* ) = 0.25. Since f ( x* ) = −2  0,
this is a local and global maximum.

Next, use the Lagrange method. The Lagrangian is L = xy +  (1 − x − y ). The first-
order conditions are


© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or
posted to a publicly accessible website, in whole or in part.

, Lx = y −  = 0,
L y = x −  = 0,
L = 1 − x − y = 0.

Solving simultaneously, x = y. Using the constraint gives x* = y* = 0.5,  = 0.5, and
x* y* = 0.25.




2.4 Setting up the Lagrangian, L = x + y +  (0.25 − xy ). The first-order conditions are

Lx = 1 −  y ,
L y = 1 −  x,
L = 0.25 − xy = 0.

So x = y. Using the constraint ( xy = x 2 = 0.25) gives x* = y* = 0.5 and  = 2. Note that
the solution is the same here as in Problem 2.3, but here the value for the Lagrangian
multiplier is the reciprocal of the value in Problem 2.3.




2.5 a. The height of the ball is given by f (t ) = −0.5 gt 2 + 40t. The value of t for which
height is maximized is found by using the first-order condition: df dt = − gt + 40 = 0, implying
t * = 40 g .



b. Substituting for t * ,

2
 40   40  800
f (t ) = −0.5 g   + 40   =
*
.
 g   g  g

Hence,

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or
posted to a publicly accessible website, in whole or in part.