Microeconomic Theory:
Basic Principles and Extensions
12th Edition
Solutions Manual
By
Walter Nicholson & Christopher Snyder
, 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.
dy 8 1 2
c. Since f (1, 2) = 16 , we know that at this point =− =− .
dx 62 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 = − 8 x 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.
Since d 2 dq = − 4 0, this is a global maximum.
2
b.
c. MR = dR dq = 70 − 2q. MC = dC dq = 2q + 30. So, q* = 10 obeys
MR = MC = 50.
,2 Chapter 2: Mathematics for Microeconomics
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
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,
df (t * ) 800
=− 2 .
dg g
c. Differentiation of the original function at its optimal value yields
df (t * )
= −0.5(t * ) 2 .
dg
Because the optimal value of t depends on g ,
, Chapter 2: Mathematics for Microeconomics 3
2
df (t * ) 40 −800
= − 0.5(t * )2 = −0.5 = 2 ,
dg g g
as was also shown in part (c).
d. If g = 32, t * = 5 4. Maximum height is 800 32 = 25. If g = 32.1, maximum
height is 800 32.1 = 24.92, a reduction of 0.08. This could have been predicted
from the envelope theorem, since
−800 −25
df (t * ) = 2 dg = (0.01) −0.08.
32 32
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. The first-order condition for maximum volume is given by
V
= 3x 2 − 16 xt + 12t 2 = 0.
t
Applying the quadratic formula to this expression yields
16 x 256 x 2 − 144 x 2 16 x 10.6 x
t= = = 0.225 x.
24 24
The second value given by the quadratic (1.11x) is obviously extraneous.
c. If t = 0.225 x, V 0.67 x 3 − 0.04 x 3 + 0.05 x 3 0.68 x 3 .
So volume increases without limit.
d. This would require a solution using the Lagrangian method. The optimal solution
requires solving three nonlinear simultaneous equations, a 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 the Lagrangian: L = x1 + 5ln x2 + (k − x1 − x2 ). The first-
order conditions are
Lx1 = 1 − = 0,
5
Lx2 = − = 0,
x2
L = k − x1 − x2 = 0.
Hence, = 1 = 5 x2 . With k = 10, the optimal solution is x1* = x2* = 5.
b. With k = 4, solving the first-order conditions yields x1* = −1 and x2* = 5.
