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Solution Manual for Microeconomic Theory Basic Principles
and Extensions 10th Edition Chapters 2 19 by Nicholson
Complete downloadable file at:
https://testbanku.eu/Solution-Manual-for-Microeconomic-Theory-
Basic-Principles-and-Extensions-10th-Edition-Chapters-2-19-by-
Nicholson
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 to here.
Solutions
2.1 U (x, y) = 4x
2
2
a.
=6y
y
U
=8x ,
x
U
b. 8, 12
c. dU =
dy=8x dx+6y dy
y
U
dx +
x
U
d.
fordU= 08 xdx+ 6 ydy= 0
dx
dy
3y
4x
=
6y
8x
=
dx
dy
e.
x = 1, y = 2 U = 4 1 + 3 4 = 16
f.
= 2/
3(2)
4(1)
=
dx
dy
g. U = 16 contour line is an ellipse centered at the origin. With equation
2 2
4 x 3 y 16 , slope of the line at ( x, y ) is
3y
4x
=
dx
dy
.
2.2 a. Profits are given by
2
R C 2 q 40 q 100
= 4q+ 40 q= 10
dq
d
+40(10) 100 = 100 )
= 2(
2
b.
= 4
dq
d
2
2
so profits are maximized
c.
dR
MR = = 70 2q
dq
dC
MC = = 2q + 30
dq
so q * = 10 obeys MR = MC = 50.
2.3 Substitution:
2
y 1 x so f xy x x
= 1 2x= 0
x
f
x = 0.5 , y = 0.5, f = 0.
Note:
f 2 0
. This is a local and global maximum.
Lagrangian Method:
£ xy (1 x y )
= y = 0
x
= x = 0
y
so, x = y.
using the constraint gives
x y 0.5, xy 0.
2.4 Setting up the Lagrangian:
£ x y (0.25 xy ) .
y
x
x
y
So, x = y. Using the constraint gives
2
xy x 0.25, x y 0..
2.5 a.
2
f t ( ) 0.5 gt 40 t
df
= g t + 40 = 0 t
dt g
b. Substituting for t* ,
f t ( ) 0.5 (40 g g ) 40(40 g ) 800 g .
2
f t
g
g
c.
2
f 1
t
g 2
depends on g because t
depends on g.
so
2
f 800
= t
g g g
d.
a reduction of .08. Notice that
2 2
800 g 800 32 0.8so a 0.1 increase in g could be predicted to
reduce height by 0.08 from the envelope theorem.
98 Fixed Costs
98 0
14 ) 0
2
14
( ) 15 * 14 (
2
C
C
pq TC q C
c. If p=20, q = 19. Following the same steps as in b., and using C=98, we get
19 98 ) 82. 5
2
19
( ) 20 * 19 (
2
pq TCq
So, profits increase by 82.
d. Assuming profit maximization, we have p = MC(q)
98
2
( 1 )
( 1 ) 98
2
( 1 )
98 ) ( 1 )
2
(
1 1
2 2 2
p
p
p
q p p
q
pq
p q q p
e. i. Using the above equation, π(20) - π(15) = 82.
ii.
- 5
2
( 20 ) ( 15 ) ( 1 )
20
15
2
20
15
p
p
p dp
The 2 approaches above demonstrate the envelope theorem. In the first case,
we optimize q first and then substitute it into the profit function. In the second
case, we directly vary the parameter (i.e., p ) and essentially move along the
firm’s supply curve.
Analytical Problems
2.9 Concave and Quasiconcave Functions
The proof is most easily accomplished through the use of the matrix algebra of
quadratic forms. See, for example, Mas Colell et al., pp. 937-939. 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 “increasing returns to scale”; even though their level curves are convex,
they may rise above the tangent plane when all variables are increased together.
A counter example would be the Cobb-Douglas function which is always quasi-
concave, but convex when α+β > 1.
2.10 The Cobb-Douglas Function
a.
1
1
1 2
f > 0. x x
2
1
1 2
f > 0. x x
11
2
1 1
f ( 1) < 0. x x
22
2
1 2
f ( 1) < 0.
x x
12 21
1 1
1 2
f f > 0.
x x
Clearly, all the terms in Equation 2.114 are negative.
b. If x x
y =c=
2
1
c x
x
/
1
1/
2
since α, β > 0, x 2
is a convex function of x 1
c. Using equation 2.98,
x x x x
f f f = ( 1)( )( 1)
2 2
2
2 2
1
2 2 2 2
2
2 2
1
2
22 12
11
x x
2 2
2
2 2
1
which is negative for α + β > 1.
2.11 The Power Function
a. Since
y 0, y 0
, the function is concave.
b. Because
11 22
f , f 0 , and
12 21
f f 0 , Equation 2.98 is satisfied and the
function is concave. Because
1 2
f , f 0 Equation 2.114 is also satisfied so
the function is quasi-concave.
c. y is quasi-concave as is
y
. But
y
is not concave for
. This can
be shown most easily by
1 2 1 2 1 2
f (2 x , 2 x ) [(2 x ) (2 x ) ] 2 f ( x , x )
2.12 Taylor Approximations
a. From Equation 2.85, a function in one variable is concave if
f ' '( x ) 0
Using the quadratic Taylor to approximate
f ( x )
near a point a:
2
f ( x ) f ( a ) f '( a )( x a ) 0. 5 f ''( a )( x a )
f ( a ) f '( a )( x a )
(because
( ) 0 ( ) 0
2
f a and x a )
The RHS above is the equation of the line tangent to the point a and so,
we have shown that any concave function must lie on or below the tangent
to the function at that point.
b. From Equation 2.98, a function in 2 variables is concave if
2
11 22 12
f f f
and we also know that due to the concavity of the function,
2
12 1 2 22 2
2
11 1
2
d y f dx f dxdx f dx
So, 0.5(
2
12 1 2 22 2
2
11 1
f dx 2 f dxdx f dx ) 0.
This is the third term of the quadratic Taylor expansion where
dx x a , dy y b
.
Thus, we have
( , ) ( , ) ( , )( ) ( , )( )
1 2
f x y f ab f ab x a f ab y b
t
E x
t
Since x
t t
E x
t
P x t F t t
1 ( )
1 ,
( ) 1
1
( ) 1 ( )
2
2
2
Thus, Markov’s inequality holds.
f. 1. 1
9
1
3
( )
2
1
3
2
1
2
dx x
x
f x dx
4
5
12
16 1
12
1
3
( ) ( )
2
1
4
2
1
3
dx x
x
Ex xf x dx
9
1
9
1
3
( 1 0 )
0
1
3
0
1
2
dx x
x
P x
2
2
x
x
x
P A
P xand A
f x A
2
3
32
3
8
3
( ) ( )
2
0
4
2
0
3
dx x
x
E xA xf xA dx
- Eliminating the lowest values for x should increase the expected
value of the remaining values.
2.14 More on Variances and Covariances
a.
2 2
2 2 2 2
2 2 2
( ) ( ( ))
( ) 2 ( ( )) (( ( )) ) ( ) 2 ( ( ) ( )) ( ( ))
( ) [( ( )) ] ( 2 ( ) ( ( )) )
Ex E x
Ex ExE x E Ex Ex E xEx E x
Var x E x Ex Ex xE x E x
b.
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
[ ( ) ( ) ( ) ( )]
( , ) [( ( ))( ( ))]
E xy E x E y
E xy E x E y E y E x E xE y
E xy xE y yE x E xE y
Cov x y E x E x y E y
c.
2 2
Var ( ax by ) E [( ax by ) ]( E ( ax by )) (From part a.)
( ) ( ) 2 ( , )
( ) 2 ( ) ( ) ( ( )) 2 ( ) ( ) ( ( ))
( 2 ) ( ( ) ( ))
2 2
2 2 2 2 2 2 2 2
2 2 2 2 2
aVar x bVar y abCov x y
a E x abExy b Ey a Ex abEx E y b E y
Ea x axby b y aEx bE y
(From results of parts a. and b.)
d.
E (. 5 x . 5 y ). 5 E ( x ). 5 E ( y ) E ( x )
Remember that if 2 random variables x and y are independent, then
Cov(x, y) = 0
. 5 ( )
(. 5. 5 ). 25 ( ). 25 ( ) 0
Var x
Var x y Var x Var y
If x and y characterize 2 different assets with the properties
E ( x ) E ( y ),var( x )var( y ) we have shown that the variance of a
diversified portfolio is only half as large as for a portfolio invested in only
one of the assets.
