SOLUTION MANUAL - Lake Forest College

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SOLUTION MANUAL
Introductory Mathematical Economics 2nd Edition
B. Wade Hands
Mathematical Economics Tutorial Students
Lake Forest College
Spring 2006
Last Updated: February 1, 2006
CHAPTER ONE
1.1
Q = BP–1 with B > 0. This can be rewritten as P = BQ–1.
(a) TR(Q) = PQ = BQ–1Q = B. Thus, MR(Q) = ∂TR / ∂Q = 0.
(b) Notice that demand is undefined when Q = 0. Thus, MR = 0 only applies for Q > 0.
1.2
Consider Q = aP + b where a, b > 0. Solving for P we have P 
Q b Q b
 
.
a a
a
2
bQ
TR (Q) 2Q b 2Q  b
Qb Q

 

(a) TR (Q)  PQ  Q  
, so MR (Q) 
.

a
a
Q
a
a
a
 a 
(b) Thus, graphically, the marginal revenue curve, which is twice as steep as the (inverse) demand
function, is:
MR = (2Q – b) / a
(slope = 2 / a)
Price (P)
P = (Q – b) / a
(slope = 1 / a)
b/a
b
-b / a
Quantity (Q)
1.3
Constant elasticity demand curve: Q = APa with a < 0 and A > 0. Thus, P = (Q/A)1/a.
Q 11 a  Q [ a1] a 
TR (Q)  1  a  Q 
, so MR(Q) 


 
Q
A1 a
A1 a
 a  A 
1/ a
(a) TR (Q)  PQ  Q(Q / A)1 / a 
.
(b) Notice that (Q/A)1/a > 0, so the sign of MR depends on the sign of (1+a) / a. It follows immediately
that MR = 0 if a = –1, MR > 0 if a < –1, and MR < 0 if –1 < a < 0.
1.4
Consider a supply curve Q = aP + b where a > 0 so that the supply curve is upward sloping. The inverse
supply curve is then P = (Q – b) / a. Thus, the supply curve cuts the price (vertical) axis if –b / a > 0 which
requires that b < 0. In contrast, the supply curve cuts the quantity (horizontal) axis if b > 0. Now,
ES 
dQS P
P
(Q  b) / a
b

 a  a
 1 .
dP QS
Q
Q
Q
It immediately follows that supply is elastic (that is, ES > 1) if b < 0 so that –b / Q > 0; that supply is
inelastic (that is, ES < 1) if b > 0 so that –b / Q < 0; and that supply is unit elastic (that is, ES = 1) if b = 0 so
that –b / Q = 0.
1.5
Demand for real money balances, m is m = e–απ where π is the expected rate of inflation and α > 0.
(a)
 m, 
dm 
  e  
  e   
  .
d m
m
e 
(b) Suppose the demand for real money balances is unit elastic. As there is a negative relationship
between money demand and inflation (i.e., consumers want to hold less money when inflation is
expected to increase), this supposition is the same as supposing that ε m,π = –1. In this case, the
supposition reduces to assuming that –1 = –απ, or that π = 1/α.
1.6
Given: q1 = f(P), q2 = g(P), and Q = q1 + q2 = f(P) + g(P). so now,
1,P 

dq1 P
dq P g   P
f  P
dQ P
P
fP
g P
,  2,P  2 
, and  Q,P 
.
 

  ( f   g )  

dP q1
q1
dP q2
q2
dP Q
Q q1  q2 q1  q2
Thus,  Q,P 


q fP
q g P q1
q
fP
g P
f ( P)
g ( P)

 1
 2
  1,P  2   2,P 
 1,P 
  2,P .
q1  q2 q1  q2 Q q1
Q q2
Q
Q
Q
Q
2
1.7
1.8
Given: x* = M / 2px =0.5M(px)–1. Thus,
 x , px 
p
px
dx * p x

 (1)0.5Mp x2  x  (1)0.5Mp x2 
 1.
dp x x *
x*
0.5Mp x1
 x ,M 
dx * M
M
M

 0.5 p x1 
 0.5 p x1 
 1.
dM x *
x*
0.5Mp x1
(a) TC(y) = y3 – 60y2 + 1210y + 600. Thus, MC(y) = ∂TC(y) / ∂y = 3y2 – 120y + 1210 and AVC(y) =
VC(y) / y = y2 – 60y + 1210. To find the minimums/maximums, consider the derivatives of MC and
AVC. MC΄ = 6y – 120. Set equal to zero, this implies that y = 20. Likewise, AVC΄ = 2y – 60. Set
equal to zero, this implies that y = 30. Further, as the coefficient on y2 is positive for both equations,
both are U-shaped parabolas (and not inverted U-shaped parabolas). Graphically, we have:
MC
$
AVC
1210
20
Quantity (y)
30
(b) TC(y) = y3 / 3 – 50y2 + 1500y + 50. Thus, MC(y) = ∂TC(y) / ∂y = y2 – 100y + 1500 and AVC(y) =
VC(y) / y = y2 / 3 – 50y + 1500. To find the minimums/maximums, consider the derivatives of MC and
AVC. MC΄ = 2y – 100. Set equal to zero, this implies that y = 50. Likewise, AVC΄ = 2y/3 – 50. Set
equal to zero, this implies that y = 75. Further, as the coefficient on y2 is positive for both equations,
both are U-shaped parabolas (and not inverted U-shaped parabolas). Graphically, we have:
MC
$
AVC
3500
50
Quantity (y)
75
3
1.9
Suppose TC(y) = ay3 + by2 + cy + d where a, b, c, and d are scalars. We are interested in finding the
necessary restrictions on a, b, c, and d so that MC and AVC are U-shaped and strictly positive for y > 0.
First, MC = 3ay2 + 2by + c and AVC = ay2 + by + c. Next, for both curves to be U-shaped, clearly a > 0 is
required. Otherwise the curves would be inverted U-shaped.
Second, in order for the curves to be U-shaped when y > 0, we need both curves to take a minimum when
y > 0. To insure this, set the derivative of both curves equal to zero; solve for y; and require this y to be
positive. In particular, MC΄ = 6ay + 2b = 0 requires y = –b / 3a. Requiring this to be positive, therefore,
requires b < 0 as we already have that a > 0. Likewise, AVC΄ = 2ay + b = 0 requires y = –b / 2a. Requiring
this to be positive, therefore, also requires b < 0 as we already have that a > 0.
Third, in order to insure that all values are positive when y > 0, it is enough to make sure that the values of
the functions are positive at these minimums. For MC, this requires that 3a(–b/3a)2 + 2b(–b/3a) + c > 0,
which requires c > b2 / 3a. For AVC, this requires that a(–b/2a)2 + b(–b/2a) + c > 0, which requires
c > b2 / 4a. As the first requirement is stronger, we need that c > b2 / 3a.
In summary, the requirements on the parameters are that a > 0, b < 0, and c > b2 / 3a. There are no
restrictions on d.
1.10
Let y = 15L2 – L3.
(a) MPL = dy / dL = 30L – 3L2.
(b) Graphically, this is an inverted U-shaped parabola:
Output (y)
75
MPL
5
10
4
Labor (L)
1.11
 m 
 1 
Equation (1.36) gave us C(Y) = mYd + C0, and solving the model gave us Y *  
T0  
A
1 m 
1 m 
where A = C0 + I0 + G0.
1.12
(a) The autonomous tax multiplier =
Y *
 m 
 
.
T0
1 m 
(b) The balanced budget multiplier =
Y *
 m   1  1 m 
 


  1.
T0 T G
1 m  1 m  1 m 

0
0
Mark-up = m = (P – MC) / P.
Recall that MR(Q) = P + (∂P/∂Q)∙Q and that ε = –(∂Q/∂P)∙(P/Q). Thus, MR = P∙( 1–1/ε). Moreover, the
profit maximizing condition is that MR = MC. Thus, we have that
P
1
 1
 P  MC , or, that m  .
MR  P1    MC which implies that


 
1.13
We have that TC(Q) = 200Q – 24Q2 + Q3. Thus, MC(Q) = 200 – 48Q + 3Q2. Under perfect competition,
we know that P = MC. Thus, P = 200 – 48Q + 3Q2. Further, in the long-run under perfect competition,
profit equals zero. So now we have:
Π = PQ – TC(Q)
Π = (200 – 48Q + 3Q2)Q – [200Q – 24Q2 + Q3]
Π = 200Q – 48Q2 + 3Q3 – 200Q + 24Q2 – Q3
Π = 2Q3 – 24Q2 = 0
Thus, Q = 0 or Q solves 2Q – 24 = 0, which gives us Q = 12. At this quantity, P = MC = 200 – 48(12) +
3(12)2 = $56.
1.14
For each of the following demand and cost curves, we are to find the profit maximizing quantity. For each,
we will find MR and MC, set these equal, and solve for Q*.
(a) P = 10 – 0.1Q implies that MR = 10 – 0.2Q.
TC(Q) = 2Q + 0.025Q2 implies that MC = 2 + 0.05Q.
Setting MC = MR and solving for Q yields Q* = 32,
5
(b) P = 22 – 2Q implies that MR = 22 – 4Q.
TC(Q) = Q3/3 – 10Q2 + 50Q + 45 implies that MC = Q2 – 20Q + 50.
Setting MC = MR and solving for Q yields Q2 – 16Q + 28 = 0. This factors to (Q–14)(Q–2) = 0. It is
easy to show that Q = 2 minimizes profits whereas Q = 14 maximizes profits. Therefore, Q* = 14.
(c) Q = –P + 25 implies that P = 25 – Q, which implies that MR = 25 – 2Q.
TC(Q) = Q3/12 – 2.5Q2 + 30Q + 100 implies that MC = 0.25Q2 – 5Q + 30.
Setting MC = MR and solving for Q yields Q2 – 12Q + 20 = 0. This factors to (Q–10)(Q–2) = 0. It is
easy to show that Q = 2 minimizes profits whereas Q = 10 maximizes profits. Therefore, Q* = 10.
(d) Q + 2P – 90 = 0 implies that P = 45 – 0.5Q, which implies that MR = 45 – Q.
ATC(Q) = Q2/2 – 20Q + 82.5 + 125/Q implies that VC = Q3/2 – 20Q2 + 82.5Q which implies that
MC = 1.5Q2 – 40Q + 82.5.
Setting MC = MR and solving for Q yields Q2 – 26Q + 25 = 0. This factors to (Q–25)(Q–1) = 0. It is
easy to show that Q = 1 minimizes profits whereas Q = 25 maximizes profits. Therefore, Q* = 25.
1.15
Demand of P = –bQ + c implies that MR = c – 2bQ. Moreover, AVC = a implies that VC = aQ and that
MC = a. Thus, setting MR = MC, we have that c – 2bQ = a so that Q* = (c – a) / 2b. This holds as long as
c > a, which makes sense as c is the maximum willingness to pay and a is the marginal cost of production.
Finally, P* = –b(c – a) / 2b + c = (c + a) / 2.
1.16
Suppose Q = L1/2 so that L = Q2. The cost of labor is fixed at $4, and the firm has a fixed cost of $100.
(a) Total cost = TC(Q) = wL + 100 = 4L + 100 = 4Q2 + 100.
(b) Average total cost = TC(Q) / Q = 4Q + 100/Q.
(c) Marginal cost = ∂TC(Q) / ∂Q = 8Q.
(d) Let P = –8Q + 96 so that MR = 96 – 16Q. Then setting MR = MC, we have that 96 – 16Q = 8Q or that
24Q = 96 so that Q* = 4. At this quantity, we know that P* = $64.
6
1.17
Suppose y = –L2 + 10L. A perfectly competitive firm faces a constant price of $10 and a constant wage rate
of $40. Thus, π(L) = py – wL = 10(10L – L2) – 40L = 60L – 10L2. Therefore,

 60  20 L  0. Solving
L
this gives us the optimal amount of labor to employ is L* = 3.
1.18
The supply of labor is given by w = L1/2. Production is characterized by y = 2L1/2. And the price of the
output is constant at $13.50 per unit.
Thus, π(L) = py – wL = 13.50(2L1/2) – L1/2(L) = 27L1/2 – L3/2. Therefore,

 13.5L1 / 2  1.5L1 / 2  0.
L
Solving this gives us the optimal amount of labor to employ is L* = 9. At this level of employment, the
firm faces a wage of 3.
1.19
Consider the standard consumer choice problem with U(x,y) = x + ln(y).
(a) The budget constraint is x∙px + y∙py = M. Thus, x = (M – y∙py) / px. Optimal choice further requires that
MRS 
Ux
p
p
1
 x . Thus, as Ux = 1 and Uy = 1/y, we have that
 x . Thus, y* = px / py and
Uy
py
1/ y p y
x* = (M – (px / py.)∙py) / px = (M – px) / px = (M/px) – 1.
(b) Given these optimal demand curves, we can find the standard six elasticities:
 x , px 
px
dx * p x
M
M

 (1) 2 
 (1) 2
dpx x *

px  M
px

 1
 px

 x, p y 
p
dx * p y

 0  x  0,
dp y x *
x*
 x ,M 
dx * M
1
M



dM x * p x  M  p x

 px
 y, py 
py
p
dy * p y

 (1) x2 
 1,
dp y y *
p y ( px / p y )
 y , px 
px
dy * p x
1



 1,
dpx y * p y ( p x / p y )
 y ,M 
dy * M
M

 0
 0.
dM y *
y*
 M
 
  M  px


7


,

 M
 
 M  px 
 M  px


 px 
px

,

 M
(c)  x, px  
 M  px

 so, if px = 0, then the own-price elasticity of the demand for good x would be unit

elastic.
1.20
There are two goods, x and y. There is a fixed amount of labor: Lx + Ly = 1. And the production functions
are such that x = 2Lx1/2 and y = 2Ly1/2 so that Lx = x2/4 and Ly= y2/4. The production possibilities curve,
which is always graphed in terms of the two goods, not labor, is found by simple substitution. In particular,
Lx + Ly = 1 implies that 1 


Now, to check for concavity, notice that y  (0.5) 4  x 2

y  ( x)(0.5) 4  x 2

3 / 2




1/ 2
x2 y2
x 2 
. Thus, y 2  41 
.
 4  x 2 so that y  4  x 2



4 
4
4


(2 x)  4  x 2

1 / 2

1 / 2

(2x)   x 4  x 2

1 / 2
and then
which is clearly negative. Thus, the production
possibilities curve is concave.
1.21
We have that C = mY + C0. Let 0 < θ < 1 and let y0 and y1 be real numbers. Now, consider
yˆ  y0  (1   ) y1 and notice that
C ( yˆ )   (my 0  C0 )  (1   )( my1  C0 )  my 0  (1   )my1  C0  mC ( y0 )  (1   )C ( y1 ).
Thus, C(y) is both concave and convex.
1.22
Let F be fixed costs. Then AFC = F / y. We aim to show that AFC is strictly convex in two different ways.
Second Derivative Test: AFC΄ = –F∙y–2 so that AFC΄΄ = 2F∙y–3 > 0, and thus AFC is strictly convex.
Definition: Consider any y0 and y1 both positive and y0 < y1. Suppose that AFC is not strictly convex. This
requires that there exist some y1 where y0 < y1 and AFC(y1) ≤ AFC(y0) + (y1–y0)AFC΄(y0). Thus, using that
AFC = F / y and AFC΄ = –F∙y-–2 we have that
 F 
F F

  y1  y0   2  ,
 y 
y1 y0
0 

which reduces to
y 0  y1 y 0  y1
1
1

. This further reduces to
as (y0–y1) < 0. But this equation

2
y1 y 0
y
y0
y0
1
implies that y0 ≥ y1, which is a contradiction.
8
1.23
Suppose the wage is fixed at w̃ so that the real wage is w = (1 – t)w̃. We also know that the labor supply
curve, S(w) is upward sloping so that ∂S(w)/∂w > 0. Finally, the elasticity of labor supply is:
S 
S ( w) w

.
w S ( w)
Now, as T = t∙S(w)∙w̃:
T
~tw
~ S ( w)  w  S ( w)  w
~tw
~ S ( w)  ( w
~)
 S ( w)  w
t
w
t
w
T
~ 1  tw
~ S ( w)  1   S ( w)  w
~ 1  t  S ( w)  w 
 S ( w)  w




t
w S ( w) 

 (1  t ) w S ( w) 
T
~ 1  t   
 S ( w)  w
S

t
 (1  t )

Finally, for the Laffer effect to hold, tax revenue must fall as the tax rate increases, i.e., ∂T/∂t < 0, it must
be that 1 
1.24
t
(1  t )
  S  0 which requires  S 
.
(1  t )
t
(a) Von Thünen’s definition of the real rate of interest, r, is
r
p  w National Income  Wage Income Profit  Rent


w
Wage Income
Wage Income
which is the classical definition of the rate of interest being the return on advanced labor costs.
(b) For von Thünen, savings equals wage income less consumption, which says that workers do all of the
savings.
pw  w 2  pa  wa
 p  w
 p  w  paw1  a .
(c) Maxw r(w)∙S(w) = r ( w)  S ( w)  
w  a  
w
 w 
Thus, the first order condition is: –1 + paw–2 = 0, which solves as w* = (pa)1/2 as required.
1.25
Let w = r = 1 and y = f(L,K) = LK / 100.
(a) MinL,K L + K such that y = 0.01LK, which requires K = 100y / L.
MinL L + 100y / L.
The first order condition is 1 – 100y / L2 = 0. The solution is that L* = 10y1/2 and K* = 10y1/2.
(b) LRTC(y) = wL* + rK* = 10y1/2 + 10y1/2 = 20y1/2.
9
(c) LRATC(y) = LRTC(y) / y = 20y–1/2. LRMC(y) = ∂LRTC(y) / ∂y = 10y–1/2. Graphically, these curves
look like the following:
Cost ($)
LRATC
LRMC
Output (y)
(d) Yes. The production function exhibits increasing returns to scale as y = LK (the exponents sum to
more than 1). Thus, as the firm employs more and more inputs, it makes output at an even faster rate
(which lowers the per-unit cost).
1.26
(a) Suppose (p1 + wt1)x1 + (p2 + wt2)x2 = wT and U(x1, x2) = x1x2. Let t2 = 0 and p2 = 1, then the budget line
requires that x2 = wT – (p1 + wt1)x1. The optimization problem then becomes:
Max x1 ∙ [wT – (p1 + wt1)x1]
The first-order condition is:
wT – 2p1x1 – 2wt1x1 = 0.
The solution is x1* 
(b)

 wT
wT
wT
 
and x2*  wT  ( p1  wt1 )
.
2
2 p1  2wt1
 2 p1  2wt1 
x1* 2 p1  2wt1 T  wT (2t1 ) 2Tp1  2wt1T  2wTt1
2Tp1



 0 , which makes sense. As one’s
2
w
sign()
sign()
(2 p1  2wt1 )
wage increases, one purchases more golf even though there is limited time.
10
CHAPTER TWO
2.1
2.2
2.3
Consider the utility function
U ( x1 , x2 )  x11/ 2 x12/ 2 and the inequality from (2.26) to verify diminishing
marginal rate of substitution
U 22U11  2U1U 2U12  U12U 22  0 .
By computing
U1 ,U 2 ,U11 ,U 22 ,U12 it will be possible to directly verify the inequality.
x12/ 2
U1  1/ 2
2x1
U11  
x12/ 2
4x13 / 2
x11 / 2
U 2  1/ 2
2x2
U 22  
x11 / 2
4x23 / 2
U 12 
1
1/ 2 1/ 2
1
2
4x
x
Plugging straight into the inequality (2.26) it can be shown that
U 22U 11  2U 1U 2U 12  U 12U 22  
2.4
1
1/ 2
1
4x
x12 / 2
0
Consider the additively separable utility function U ( x) 
n
U ( x ).
i
i 1
i
From (2.26),
U 22U11  2U1U 2U12  U12U 22  0
for the utility function U(x) to exhibit a diminishing marginal rate of substitution for any two goods. Now,
since U(x) is additively separable U12=0. So the above condition becomes
U 22U11  U12U 22  0.
Diminishing marginal utility for all goods implies
U11,U 22  0. Therefore, since U12 ,U 22  0, it follows
that the above condition will hold. Thus, diminishing marginal utility for all goods is sufficient for
diminishing marginal rates of substitution when the utility function is additively separable.
2.5
11
2.6
(a) Proof: First, recall
xi M
M x
 i,M 
n
px
Then, we know
i
i 1
n
i
dx dM

1
dM dM
  pi
xi M xi
1
M xi M
p
i 1
n
`
 M . Differentiating with respect to M, we get
i
i 1
p i xi xi M
1
M M xi
n

i 1
n
  bi  i , M  1
i 1
n
(b) Proof: Recall
p x
i 1
i
 M , bi 
i
x i p j
pi xi
, and  i , j 
.
p j x i
M
Differentiate the first equation with respect to p j ,
n
p
i 1
i
xi
0
p j
n
x j   pi
i 1
i j
n

i 1
M
pj
p i x i x i p j
M  x j
M p j x i
n
b 
i
i 1
n
b 
i 1
2.7
xi p j xi M
0
p j xi p j M
i
(a) Consider the function
i, j
i, j
  xi
pj
M
 b j
y  f ( L, K )  ALa K b with A  0, a  0 , and b  0 .
12
To check for the degree of homogeneity the function f ( rL , rK ) needs to be considered.
In this case
degree
f (rL, rK )  Ar a La r b K b  r ab ALa K b which implies the function y is homogeneous of
r  ab
(b) Consider the function
then
y  f ( L, K )  AL2 K 2  BL3 K 3 with A  0 and B  0 ,


f (rL, rK )  Ar 2 L2 r 2 K 2  Br 3 L3 r 3 K 3  r 2 AL2 K 2  Br 1 L3 r 1 K 3 ,
but no homogeneity can be detected.
(c ) Consider the function
AL1 a K 1b
y  f ( L, K ) 
BL  CK
0  b  1, then y  f (rL, rK ) 
A  0, B  0, C  0,0  a  1 and
Ar 1 a L1 a r 1b K 1b
AL1a K 1b
 r 1a b 
which implies the
BrL  CrK
BL  CK
function y is homogeneous of degree
r 1  a  b .
2.8
2.9
2.10
2.11
(a) Consider the Cobb-Douglas production function
y  f ( x1 , x2 )  x11/ 2 x12/ 2 .
From problem 2.10,

And recall that MRTS 
MRTS   x1 x2 
x1 x2  MRTS
 x2
f
 1.
x1
f2
f1 
Using this fact, we calculate
x12 / 2
2x11 / 2
f2 
x11 / 2
2x12 / 2
These yields:
MRTS 
x2
x1
and
From here we can calculate  :
13
MRTS
1
 x1 x2 
f1 and f 2 :

(b) Now consider
calculate
f1 
f1 x1
1
1
f 2 x2  MRTS 


 x1 x2  

y  f ( x1 , x2 )  ax1p  1  a x2p

1/ p
. Using the same concepts as in part (a) we
f1 and f 2 :

1
ax1p  1  a x2p
p

1 / p 1
apx1p 1 and f 2 

1
ax1p  1  a x2p
p

1 / p 1
1  a  px2p1
Thus yielding:
1 p
a  x2 
 
MRTS 
1  a  x1 
From here we can calculate  :




1 p


1
 a  x2   x1 
   1 




 
   


p
 1  a  x1   x2   a 
 x2     1  p 
 
1  p    
  1  a 
 x1   

2.12
Consider the production function y = f(x). Since f exhibits constant returns to scale, f is homogenous of
degree 1. From Theorem 2.2, fi is homogenous of degree 0 (r=0), for any good i. Applying this and
Theorem 2.1 to fi ,
n
rf i   f ij ( x) x j
j 0
n
0   fij ( x) x j
j 0
Furthermore, since f exhibits diminishing marginal productivity, fii < 0 for all i. In order for the above
condition to hold, there must exist a good j, such that fij >0. This implies that goods i and j are
complements in production.
2.13
2.14
2.15
14
2.16
2.17
2.18
2.19
2.20
2.21
We are given that individual A’s utility curve is defined by
U xB , y B   x1B/ 2 y1B/ 2 .
U x A , y A   x1A/ 2 y1A/ 2
The proof of this will be complete when it is shown that
and
MRS A  MRS B .
Therefore we calculate each of these and get
MRS A 
And using the fact that x B
MRS A  MRS B 
2.22
 x  xA
and
yA y  yA

xA x  xA
yA
xA
and
MRS B 
yB
xB
y B  y  y A we get:
which reduces to
 y
y A    x A .
 x
We are given that 0  y  y , and we know the ATC is subaddative if C ( y )  C ( y )  C ( y  y ) .
Let
y  y , then C ( y )  C (y ) . Also, C ( y  y )  C ( y  y ) , alternatively written
C ( y  y )  C ( y (1   )) . But C (y )  C ( y ) and C ((1   ) y )  (1   )C ( y ) .
Thus C (y )  C ((1   ) y )  C ( y )  (1   )C ( y ) .
Recall that
y  y , then this can be re-written
C ( y )  C ( y  y )  C ( y )  (1   )C ( y )  C ( y )  C ( y  y )  C ( y ) .
2.23
CHAPTER THREE
3.1
p* 
100  
100  
*
and q 
2
2
15
So
100  
100  
 0 and
0
2
2
Therefore, -100<α<100.
3.2
To start with consider the supply and demand functions for the model,
Q S  S P,   cP   . Therefore,
P *
1

.
 a  c
(a) Positive, because a  0 , c  0 and
a  c.
Q D  DP   aP  b and
(b) Negative.
(c) Negative.
(d) Positive.
3.3
3.4
Consider the two functions that characterize the short-run, perfectly competitive firm
y  f L   4L1/ 2
(3.36)
  py  wL  vK
(3.37)
The function (3.36) implies that
Therefore (3.37) becomes
L
y2
, which can be substituted into (3.37) for the variable L.
16
  py  w
y2
 vK and finding the profit maximizing condition yields
16
p
p2

2wy
*
*
 p
 0 . This yields y  8 and L  4 2 which equals the conditions (3.42) and
w
y
16
w
(3.43) and confirms the answer.
3.5
3.6
3.7
(a)

0
y
16

R c

0
y y

R c

y y
 2 R  2C

0
y 2 y 2
(b) We need to find
y *
*
*
. We know MR ( y ( A), A)  MC ( y ( A)) .
A

MR y * MR MC y *


A
y * A
y * A
y *  MR MC
MR

[

]
*
*
A y
A
y
MR
y
 ()
( )
A




 ()
 MR MC
A
( )
( )
[

]
y *
y *
*
3.8
Consider the function

y  L3 / 2 K 3 / 2 . The second order conditions imply that f LL  0 , f KK  0 , and
f LL f KK  f LK  0 .
2
But
f LL 
3K 3 / 2
3L3 / 2

0
f

 0 since labor and capital inputs cannot be negative. This
and
KK
4K 1/ 2
4 L1 / 2
implies that the second order conditions do not hold.
3.9
3.10
Consider an imperfectly completive firm with total revenue, TR = TR(y, R) = 2yR + 116R, and total cost,
TC = TC(y, R) = y2 + 4y + 5R2. Therefore, the profit function π = π(y,R) = TR – TC = 2yR + 116R – y2 –
4y -5R2.
(a) To solve the first-order conditions for profit maximization, we first taking the partial derivative of π
with respect to y and R, yielding πy = 2R – 2y – 4 and πR = 2y + 116 – 10R. Setting πy and πR equal to
0, yields the following profit maximizing output y and market research expenditure R, y* = R – 2 and
R* = (y+58)/2
17
(b) The second order conditions are: πyy < 0, πRR < 0, and πyyπRR – (πyR)2 > 0. Using the answer to (a) we
have, πyy=–2, πRR=–10, and πyyπRR – (πyR)2 = (–2) (–10) – 2 = 18. So the second order conditions hold.
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.22
3.23
3.24
(a) Using the equations given:


 E1 
 E1 
 E  aE 2  
 E1  E2   aE1  E2 2  E1 1  aE1  E2 
C1  
 E1  E2 
 E1  E2 

with

C2 found in the exact same manner.
(b) Using the profit function for firm 1,  1
 pE1 1  aE1  aE2   wE1 ,
18
 1
 p  2apE1  apE2  w
E1
Solving for
E1 we obtain the following function:
E1 
 p  aE2  w
2ap
and without loss of generality,
E2 
Now plugging in
 p  aE1  w
2ap
E2 into E1 yields:
E1* 
 p  w
3ap
E2 yields the same thing.
and plugging that into
Ei*  p  w

0
(c)
a
3a 2 p
Ei*
1

0
w 3ap
Ei*
1

0
p 3ap
3.25
Consider the Stackelberg leadership model where firm 1 is the leader. This model tells us that firm 1 will
maximize profits based on q2 = R2(q1) = (b – q1)/2. From (3.86) we have the following profit function for
firm 1:
1  (q1  R2 (q1 )  b)q1  FC
b-q
 b)q1  FC
2
1
1
1   q12  bq1  FC
2
2
1  (q1 
The optimization problem becomes:
Max   
1
1 2 1
q1  bq1  FC
2
2
The first order condition is:
 q1 
1
b  0.
2
19
The solution is q1 
*
b
bb/2 b
*
*
 .
and q2  R2 ( q1 ) 
2
2
4
3.26
3.27
3.28
CHAPTER FOUR
4.1
We know from (4.3) that TC ( y ) 
 MC ( y)dy.
(a) Since MC(y) = 50 + 20y – 9y2, we have TC(y) = 50y + 10y2 – 3y3 + C. Using the initial value
condition TC(5) = 700, we have TC(y) = 575 + 50y + 10y2 – 3y3.
(b) Since MC(y) = 75 - 36y + 12y2, we have TC(y) = 75y - 18y2 + 4y3 + C. Using the initial value
condition TC(10) = 3000, we have TC(y) = 50 + 75y - 18y2 + 4y3.
4.2
4.3
The choice problem is subject to px + y = M, which implies y = M – px. Using this along with the original
maximization problem, we have, Max{x} U(x) = v(x) + (M – px). Thus, the first order condition is: v’(x) –
p(x) = 0. This implies: v’(x) = p(x). So we have the following:
x0
CS ( x0 )   p ( x)dx  p ( x0 ) x0
0
x0
CS ( x0 )   v' ( x)dx  p ( x0 ) x0
0
CS ( x0 )  v( x0 )  p ( x0 ) x0
CS ( x0 )  U ( x0 )  M
And with a little algebra, this becomes U(x0) = M + CS(x0).
4.4
20
4.5
4.6
4.7
4.8
4.9
PV 
V [1  .08]T
 9,818.15
.08
The problem supplies all the crucial information in order to compute the present value.
(a) t  20 years , $1000coupon / year , FV  $10,000 ,
and i  6%
Using formula (4.42) the present value can be calculated which equals the maximum paid for the bond
19
PV  1000
1
1
10,000

 $14,276.16
t
(1  0.06) (1  0.06) 20
(b) same as above but
19
PV  1000
1
i  8%
1
10,000

 $11,651.82
t
(1  0.08) (1  0.08) 20
4.10
4.11
Machine A has a 10 year life span and generates profits of $10,000/year
Machine B has a 6 year life span and generates profits of $10,000/year and a scrap value.
Assume
i  6% .
In order to find the scrap value the difference between the machines’ present values needs to be calculated.
10
PVA  10,000
1
6
PVB  10,000
1
1
 $73,600
(1  0.06) t
1
 $49,175
(1  0.06) t
ScrapValue  PVA  PVB  73,600  49,175  24,427
FVofScrapvalue  24,427  (1.06) 6  34,650
4.12
4.13
4.14
21
4.15
4.16
4.17
4.18
First,
x(t * ) is the quantity of lumber available, and denote x 
We know that rx (t
*
x
x(t * )
and x  * .
t
t
)  x  0 and we differentiate with respect to r to obtain
x t * x t *
x(t )  r *

0
t r t * r
*
 x(t * ) 

t * x
x
[ * r *]
r t
t
t *
x(t * )

x
x
r
r *
*
t
t
x
t *
 f '[ x(t )]  r we know x  rx . Thus x  rx is negative. So,
Since
<0.
x
r
This means as interest rates rise, the optimal time to cut down the forest decreases.
4.19
Using discrete time analysis, we first form an equation for the present value of profit,
:
  R  e rT
For the firm to maximize its profit from this machine, it must maximize the amount of durability it is going
to obtain from the machine. Therefore, the cost of the marginal durability must equal that of the revenue
obtained per year of the machine:
 
R  e rT  C ' T *
4.20
4.21
4.22
4.23
(a) Plugging in
pn  1  e n into the given equation gives us NSB  B  Be n  nc .
22
Thus,
*
NSB
 Be n  c . Solving for n* then gives the following equation:
n
 c 
 ln 

B 

*
n 

(b) Using the equation from (a) yields 8 firms would be needed to maximize the net benefit to society.
CHAPTER FIVE
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
5.10
5.11
5.12
5.13
5.14
23
5.15
5.16
5.17
5.18
CHAPTER SIX
6.1
6.2
6.3
Consider the function y  f (x) with
x   n and constant returns to scale. This implies that the function
y  f (x) exhibits homogeneity of degree 1. By Theorem 2.2, it is known that f i exhibits homogeneity of
degree 0 and therefore ( Hf )( X )  0 and the determinant
Hf  0 implying that the Hessian is singular.
6.4
6.5
(a) Plugging in values given into the IS-LM on page 97 yields:
IS (Y , r )  (ar  I 0 )  (bY  cr )  D  0
MS
LM (Y , r )  fY  er 
0
p
Rearranging these statements yields:
ar  bY  cr  ( I 0  D)
fY  er 
Now, using Cramer’s Rule, we derive
MS
p
Y * and r * .
 I  D 
 b (a  c) Y *   0 S 
M
 
 f

e   r *  

 p 
First
Y*:
24
( I 0  D) (a  c)
Ms
e
p
*
Y 
 b (a  c)
f
Y* 
e
e( I 0  D )  M S ( a  c ) / p
be  f (a  c)
r * is obtained through a similar process:
bM s p  f ( I 0  D)
r 
be  f (a  c)
*
(b) Doing direct differentiation on the answers in (a) we obtain the following:
Y *
e

D b(e)  f (a  c)
Y *
(a  c)

S
p (be  f (a  c)
M
r *
f

D be  f (a  c)
r *
b

S
p (be  f (a  c)
M
(c) They are consistent.
6.6
6.7
Since gross substitution means z ij  0 , then all elements of
differentiable inverse functions by Theorem 6.1
6.8
6.9
6.10
25
Jf ( x )  0, x
0
0
  N . So there exist unique
6.11
(a) Recall that
 x1  2.77
 x    2.21

 2 
Therefore, using the labor coefficients given:
2.77
 2.21 (0.4,0.4)  1.992  2


(b) Now, recall that
1.94 0.83
( I  A) 1  
 and using equation 6.47:
0.55 1.66 
 p1* 
1.94 0.83
 *   2(0.4,0.4) 

0.55 1.66 
 p2 
p1*  2
p2*  2
(c) Repeating the process of part (b) with the new labor coefficients yield:
p1*  2.27
p2*  1.826
Thus, the price of the labor intensive good increased.
6.12
(a) From equation (6.30), x* = (I-A)-1d. So then we have,
 x1* 
.5 .2 1 10 22.86
 *   (I  
)    

.1 .4  2   7.14 
 x2 
(b) From equation (6.47), p* = waL(I-A)-1. So then we have,
p
*
1
6.13
1
 .5 .2 
p  5.2 .8 I  
   3.57 7.86
.
1
.
4

 
*
2

We are given that A is productive, i.e.
We will use Cramer’s Rule to find
x  xA .
x1* and show it is strictly greater than 0, and then find xi* to show it is
greater than or equal to 0.
(1  A) x *  d or written in matrix form,
26
1  a11  a12   a1n   x1*   d1 
 
 a
   x2*  d 2 
 21 1  a22
*

 

   

    

 1  ann   xn*  d n 
  an1
*
Now we use Cramer’s Rule to find x1 .
1  a12   a1n 
0 1  a
 
22




 


0  a n1  1  a nn 

x1 
1  a11  a12   a1n 
a
 
 21 1  a 22
 

 



 1  a nn 
  a n1
Here, we know the numerator is positive because the leading principle minor of n-1 degrees is positive.
The denominator is a P matrix by Theorem 6.3, and thus the determinant is positive. Therefore,
x1* is
strictly positive.
Now each xi  0 because we do not know a ij terms, thus if there are no 0 terms, the principle minors are
*
positive. If there are 0 terms, then xi  0 .
*
n
6.14
We first note that
V x
j
j 1
j
 Vx. Then using equations (6.43), V=p(I-A), and (6.30), x=(I-A)-1d:
n
n
j 1
i 1
V j x j  Vx  p( I  A) x  p( I  A)( I  A) 1 d  pd   pi di
n
So,
n
V x   p d .
j 1
j
j
i 1
i
i
This means that the total value added for each good is equal to the sum of the
price times the final consumption demand of each good. Or put another way, the total value added for each
good is equal to the final consumption revenue for each good.
6.15
27
6.16
Consider an additively separable utility function U ( x) 
n
U
i 1
i
( xi ) exhibiting diminishing marginal
 2U
 0 . By Theorem 6.5 the Hessian needs to be symmetric and be an NP
utility for all goods U ii 
2
xi
matrix in order to be negative definite. In the case of this utility function the matrix looks like

0

0

0
0 0 0
 0 0 
with n rows and columns. The diagonal is negative due to the property of exhibiting
0  0

0 0 
marginal utility and the off diagonal entries are all zero because of the utility function being additively
separable. This matrix is symmetric and also qualifies as a NP matrix because the determinant is simply
the product of the diagonal entries. By taking the principal minors the signs alternate and therefore the
Hessian is negative definite.
6.17
6.18
CHAPTER SEVEN
7.1
Consider the coefficient matrix
vector
0.2 0.6
A
 and d1  100 , d 2  200 which form the column
0.5 0.2
100 
 0.8  0.6
1
(
I

A
)

.
Using
(7.4)
we
determine

x

(
I

A
)

d
 0.5 0.8  and using
 200 




Cramer’s rule we can determine
 d1 a12 
d a 
22 
x1   2

det( I  A)
x .
100  0.6
200 0.8 

  588.24
.34
 0.8 100 
 0.5 200
  617.65
x2  
.34
28
7.2
7.3
7.4
7.5
7.6
n
7.7
We first note that
 *   * ( w, p)  pf ( x * ( w, p))   wi xi * ( w, p)
by equation (7.48). Now
i 1
taking the derivative of this with respect to p, we have,
n
 * n 
x 
x
   pf i i   f ( x * ( w, p))   wi i
p
p 
p
i 1 
i 1
 x 
 * n
   pf i  wi  i   f ( x * ( w, p))
p
i 1
 p 
And then by equation (7.37) we know
pfi  wi  0 , for all i. So we have,
7.8
7.9
By (7.53) it is known that
y* 

p
p
y *
  pwi
Therefore
wi
Also by (7.53) it is known that
 xi   wi
*
xi
  wi p
p
*
Therefore

The order of differentiation does not matter, and therefore
x
y *
  pwi   i validation the condition.
wi
p
*
29
 *
 f ( x * ( w, p))  y * .
p
7.10
7.11
7.12
7.13
7.14
n
7.15
(a)
e
i 1
 ( yi [a  bxi ]) 2
2
i
 yi2  2 yi [a  bxi ]  [a  bxi ]2
 yi2  2 yi a  2 xi yi b  a 2  2abxi b 2 xi2
Now differentiate with respect to a and b to obtain first-order conditions:
n
ei
   2 yi  2a  2bxi  0
a i 1
n
n
i 1
i 1
 yi   a  bxi
n
y
i 1
n
i
 na  b xi
i 1
n
ei
   2 xi yi  2axi  2bxi2  0
b i 1
n
n
 x y   ax
i 1
i
i
n
x y
i 1
(b)
i
i
i 1
n
i 1
i 1
 bxi2
n
n
i 1
i 1
 a  xi  b xi2
ei
 n
 2a
n
ei
   xi
ba
i 1
n
i
n
ei


xi2

2
b
i 1
n xi2   xi2  0
n
(c)
y
i 1
n
i
 na  b xi
i 1
30
n
n
i 1
i 1
na   yi  b xi
n
a *  y  bx
n
n
n
i 1
i 1
i 1
 xi yi  ( y  bx ) xi  b xi2
n
x y
i 1
i
i 1
i
b* 
n
i 1
i 1
i 1
n
n
n
i 1
i 1
i 1
i
 y  xi  b( xi2  x  xi )
n
n
i 1
n
i 1
n
 xi y i  y  xi
( xi2  x  xi )
i 1
7.16
n
 y  xi  bx  xi  b xi2
n
x y
n
i
i 1
Recall that:
 1 (q1 , q2 )  (q1  q2  b)q1  cq1  t1q1
 2 (q1 , q2 )  (q2  q1  b)q2  cq2  t 2 q2
Using the given equations for profit for each of the respective companies, start by finding
 1
 2q1  q2  b  c  t1
q1
 2
 2q2  q1  b  c  t 2
q2
Now, solve
 1
 2
for q1 and
for q 2 :
q1
q2
b  c  t1  q 2
2
b  c  t 2  q1
q2 
2
q1 
And then plug
q 2 into q1 and resolve for q1 :
31
 1
 2
and
:
q1
q2
q1* 
Here we can impose symmetry for
q 2* . Now, using the tax revenue function TR(t1 , t 2 )  t1q1*  t 2 q2* :
TR(t1 , t 2 )  t1 (
To find
t1* and t 2* , find
b  c  2t1  t 2
3
b  c  2t1  t 2
b  c  2t 2  t1
)  t2 (
)
3
3
TR
since we can impose symmetry:
t
TR
2 bct t
 t
 0
t
3
3
3
Solve for t :
bc
2
t
7.17
7.18
CHAPTER EIGHT
8.1
8.2
8.3
8.4
Consider the function U ( x1 , x2 )  a1 ln x1  a2 ln x2
Subject to
M  p1 x1  p2 x2
(a) The Lagangrian is
L( x1 , x2 ,  )  a1 ln x1  a2 ln x2   (M  p1 x1  p2 x2 )
And the corresponding first-order conditions are
Lx1 
a1
 p1  0
x1
L x2 
32
a2
 p 2  0
x2
L  M  p1 x1  p2 x2  0
Solving for
x1* and x 2* yields x1* 
Ma1
Ma 2
*
and x 2 
p2
p1
(b) The answer is the same as (8.2) just with added a parameters that unequally distribute the products
whereas in (8.2) only price determines the distribution of the two goods.
8.5
8.6
8.7
8.8
8.9
Consider an economy that produces only two goods, x and y. These goods are produced using labor L
input according to the production functions x  2( Lx )
1/ 2
and
y  2( Ly )1/ 2 .
(a) Differentiating these equations with respect to their labor gives us the following marginal return
functions: MR x 



2( Lx )1 / 2   ( Lx ) 1 / 2 and MR y 
2( L y )1/ 2  ( L y ) 1/ 2 .
Lx
L y

Differentiating these again with respect to there labor gives,
MR y
L y
( L )    12 ( L )
1 / 2
y
y
3 / 2

MR x
1
( Lx ) 1 / 2   ( Lx ) 3 / 2 and
Lx
2


. Since labor L must be nonnegative, these are always negative
and thus exhibit diminishing marginal returns.
(b) Solving the production functions for labor, we have L x 
the constraining labor condition (Lx + Ly = 1), we have
1 2
1
x and L y  y 2 . Putting these into
4
4
1 2 1 2
x  y  1 . Which simplifies to
4
4
x2  y2  4 .
(c) This is simply the equation for a circle of radius 2 centered at the origin. When we restrict our
ourselves to x,y>0 we have a quarter-circle that has the standard concave shape.
33
(d) We want to maximize xy, subject to
x 2  y 2  4 . Finding the Lagrangian, we have
L( x, y,  )  xy   ( x 2  y 2  4) . This yields the following first order condition:
Lx  y  2x  0
L y  x  2y  0
L  x 2  y 2  4  0
Solving this system for x* and y* yields, x*  y* 
2.
8.10
8.11
(a)
C1  r (e0  e1 )  e1
C1  re0  re0  e1
(b)
L(c0 , c1 ,  )  U (c0 , c1 )  [c1  rc0  re0  e1 ]
Lc0 
U
 r  0
c0
U 0  U1r  0
Lc1 
U
 0
c1
 U 1  *
r
U0
U1
L  c1  rc0  re0  e1  0
(c) Differentiate each first order condition with respect to r:
Lc0
r
Lc1
r
 U 00
c0
c

 U 01 1    r
0
r
r
r
 U10
c0
c 
 U 11 1 
0
r
r r
c
 c0

r  1  c0  e0  0
r
r
r
(d)

c0  [U1  (e0  c0 )(U 01  U11r )]

r
D

 U 00
 U1
0
D
34
c0
c

 U 01 1  r

r
r
r
c0
c
r  1  e0  c0
r
r
c1 U10 r

0
r
D
In interest rises, you consume less in the present day and consume more later.
8.12
8.13
8.14
8.15
(a) Using the specific utility function set up in the text; the dual problem is set up as follows:
Min
p1 x1  p2 x2
Subject to: U  x1 x 2
From this, the Lagrangian is:
L( x1 , x2 ,  )  p1 x1  p2 x2   ( x1 x2  U )
The first order conditions are then:
(1) Lx1  p1  x2  0
(2) Lx2  p2  x1  0
L  x1 x2  U  0
(3)
Solving (1) for
Using

yields
* 
 p1
px
, which then can be used in equation (2) to get x2  1 1 .
x2
p2
x2 in equation (3) gives us an equation for x1 :
x1 (
p1 x1
) U  0
p2
p1 2
x1  U
p2
xU1 
p2
U
p1
Taking this result and plugging it back into our equation for
xU2 
p1
p2
p2
U 
p1
(b) Yes they are symmetric, as it has already been shown.
35
p1
U
p2
x2 yields the following:
(c) To find the expenditure function, we utilize the following formula:
E U ( p,U )  p1 x1U ( p,U )  p2 x2U ( p,U )  p1
p2
U  p2
p1
p1
U
p2
Simplifying this yields:
E U ( p,U )  2 p1 p2 U
(d) Recall from the text that x 1 
*
M
M
U
*
and x 2 
. Using E ( p,U )  2 p1 p 2 U for M gives us
2 p1
2 p2
the following two equations:
x1* 
2 p1 p 2 U
x 2* 
2 p1 p 2 U
2 p1
2 p2

p2
U  x1U
p1

p1
U  x 2U
p2
Thus, xi  xi for i  1,2 .
*
U
(e) Recall, to calculate the cross effects for this problem, we need to utilize the following expression:
x1* x1U
x *

 x2* 1
p2 p2
M
Now find the specific derivatives required:
x1U
1

p 2
2
U
p1 p 2
x1*
1

M
2 p1
To solve, plug these back into the original expression to obtain:
x1U 1

p 2 2
U
M
1

*
0
p1 p 2 2 p 2 2 p1
Now to calculate the own effects, we utilize the following equation:
x1* x1U
x

 x1* 1
p1 p1
M
After finding the appropriate derivatives and plugging them into the above expression, we obtain the
following:
x1*  U p 2 2 p1 p 2 U


p1
2 p13 2
4 p12
36
This holds, thus implying the Slutsky equations for this demand function holds for good 1.
8.16
(a) Consider the function U (C1 , C2 )  C1C2
Subject to the
C1  C2  1
The Langragian is
L(C1 , C2 ,  )  C1C2   (1  C1  C2 )
and the corresponding fist-order conditions are
LC1  C2    0
LC2  C1    0
L  1  C1  C2  0
Solving for
C1* and C2* yields C1* 
1
 C 2*
2
(b) Consider the function U (C1 , C2 )  C1C2
Subject to the
C1  C2  1  n(1  C1 )
The Langragian is
L(C1 , C2 ,  )  C1C2   (1  n(1  C1 )  C1  C2 )
and the corresponding fist-order conditions are
LC1  C 2  n    0
LC2  C1    0
L  1  n(1  C1 )  C1  C2  0
Solving for
C1** and C2** yields C1** 
(c) U (C 1 , C 2 ) 
*
*
1
n 1
**
and C 2 
2
2
1
4
1 n 1
U (C 1** , C *2 * )  
2
2
As long as
n  0 individuals will be better of because U (C1** , C *2* )  U (C1* , C *2 ) or in other words as
long as the population is growing, individuals will be better off with the social security system.
(d) Yes, the government will be able to finance this “pay as you go” social security system, because
Borrowed  S  NY and Paidout  (1  n)S  N 0
37
We also know that
N Y  (1  n) N 0 and therefore Borrowed=Paidout
8.17
8.18
(a) L( K , L,  )  vK  wL   ( y  f ( K , L))
f
0
K
f
LL  w  
0
L
L  y  f ( K , L)  0
LK  v  
2 f
LKK  
K 2
2 f
LLL   2
L
2 f
LLK  
LK
To satisfy all second order conditions, all leading principle minors must be negative:
H1  0
f
K  0
H2 
f
2 f


K
K 2
f
 f   2 f
 f  2 f
 f
 f   2 f
 f   2 f
H 3  0  [ ][(
)

(

)]

[
][{

}

{

}]  0
K
K L2
L LK
L
K LK
L K 2
0

(b) Prove MC increasing when w increasing, i.e.
LK
K 
 f KK

fK  0
w
w w
LL
L 
 1  f LL

fL  0
w
w w
MC

0
0
w
w
L
K
L
  fK
 fL
0
w
w
w
38
 f KK
 0

  f K
K

w
8.19
 K 
 f K   w  0
 L 
 f L     1
 w 
0     0
 w 
0
 f LL
 fL
 f KK
0
0
0
 fK
 f LL
 fL
1
0
H3

 f KK (  f L )  f KK f L
()

 () since f KK  0 
()
( )
( )
Consider a consumer with fixed money income M, who purchases goods x1 and x2 so as to maximize the
utility function
U ( x1 , x2 )  b  a ln( x11/ 2 x12/ 2 ) , where a and b are positive constants.
(a) Finding the Lagrangian, we have
L( x1 , x2 ,  )  b  a ln( x11/ 2 x12/ 2 )   ( M  p1 x1  p2 x2 ) . This
yields the following first order condition:
Lx1 
a
 p1  0
2 x1
L x2 
a
 p 2  0
2 x21
L  M  p1 x1  p2 x2  0
Solving this system yields: x1 * 
M
a
M
, x2 * 
, and  * 
.
2 p1
2 p2
2 x1 p1
(b) Finding the bordered Hessian,
0
H   p1
 p2
 p1
a
2 x12
 p2
0
a
2 x22
0
Then we have the following principle minors,
H1  0
H 2   p12  0
 pa
  p2 a 
0
H 3  0  ( p1 ) 1 2   ( p2 )
2 
2
x
2
x
1 
 2

39
So the utility function satisfies the second order conditions.
(c) The marginal utility of money income is given by
*.
So differentiating with respect to pi and M
yields,
 *
0
pi
 *  a

M M 2
So the marginal utility of money income is constant with respect to prices but not money income.
8.20
8.21
8.22
8.23
8.24
8.25
8.26
In this problem we want to show that
V
V
 x *j
 0 . For this, we are simply going to prove for the
p j
M
case of j=1,2 since the steps would be similar to proving the identity in general. Now, recall that
V ( p, M )  U ( x1* ( p1 , M ), x2* ( p2 , M )) . From this expression we know the following:
(1)
V
 * ( p, M )
M
(2)
x
x
V U x1 U x2


 * p1 1  * 2  * x2*
p 2 x1 p 2 x2 p 2
p 2
p 2
To prove that (2) is true, consider the following claim:
n
p
i 1
i
xi
  x *j
p j
Since we are only considering this for n=2 cases, the proof looks as follows:
40
( M  p1 x1  p2 x2 )  0
 p1
x1
x
 p 2 2  x2  0
p2
p2
x2   p1
x1
x
 p2 2
p2
p2
 x2  p1
x1
x
 p2 2
p2
p2
Therefore, returning to expression (2), we see this result proves the equality. Using the results of (1) and
(2) we see that Roy’s Identity holds.
8.27
8.28
8.29
8.30
8.31
8.32
CHAPTER NINE
9.1
9.2
9.3
Consider the two-good consumer choice problem in (9.24). Assuming that U1 = U2 where U1,U2>0 and p2
> p1. There are the following four cases:
Case: x1*>0 and x2*>0. From (9.26b) we have
implies
* 
U1
U
and from (9.26d) we have  *  2 . But this
p1
p2
U1
p
p
 1 . Since U1 = U2, 1  1 but this implies p1 = p2. A contradiction.
U 2 p2
p2
41
Case: x1*=0 and x2*=0. From (9.26f),
 *M  0.
*  0 into (9.26a) implies U1  0 .
*  0 .
Substituting
A contradiction.
Case: x1*=0 and x2*>0. From (9.26d), we have
* 
But since M>0, this implies
* 
U2
U
and from (9.26a) *  1 . These imply
p2
p1
U 2 U1
p
U
p

and thus 1  1 . But since U1 = U2, we have 1  1 but this implies
p2 U 2
p2
p1
p2
p1  p2 . A contradiction.
Case: x1*>0 and x2*=0. From (9.26b), we have
* 
* 
U1
U
and from (9.26c)  *  2 . These imply
p1
p2
U1 U 2
U
p
p

and thus 1  1 . Since U1 = U2, we have 1  1 and thus p2  p1 , which is
U 2 p2
p1 p 2
p2
true.
Therefore it follows that x1*>0 and x2*=0.
9.4
9.5
R( L, K )  8L1 / 2 K 1 / 2  LK
RL  4K 1 / 2 L1/ 2  K
RK  4 L1 / 2 K 1 / 2  L
From (9.37b),
From (9.37a),
*  1 because s  v  1.
4
K 1/ 2
 K 1
L1 / 2
K 1/ 2
4 1/ 2  K  1
L
L* 
From (9.37d),
2K * 
16 K
(1  K ) 2
16 K *
16 K * 1 / 2 1 / 2
16 K *

8
(
)
K

K*  0
* 2
* 2
* 2
(1  K )
(1  K )
(1  K )
2K * (1  K * ) 2  16K *  32K * (1  K * )  16K *  0
2
2K * (1  2K *  K * )  16K *  32K *  32K *  16K *  0
2
2
42
2
2K *  4K *  2K *  16K *  16K *  0
2
3
2
 14 K *  12 K *  2K *  0
2
3
2K * ( K *  6K *  7)  0
2
Since K  0 ,
*
So, L 
*
Then y
*
( K *  1)( K *  7)  0 , and since K *  0 , K *  7 .
16
* 7  1.75
(1  7) 2
 Q( L* , K * )  71 / 21.751 / 2  3.5 .
We know in this production function,
L*  K *  3.5 . Hence
L*  K * , thus without the constraint,
7 3 .5

.
1 .5 3 .5
9.6
9.7
9.8
9.9
9.10
9.11
Consider the problem faced by a macroeconomic policy maker who can control the level of real national
income Y and the rate of inflation

via various policy instruments.
(a) The short-run Phillips curve shows the trade off between unemployment and inflation. If one were to
graph it in (Y ,  ) space, it would be downward slopping curve that bows inward towards the origin.
(b) Given our maximization problem, we have the following Lagrangian and Kuhn-Tucker Conditions:
43
L(Y ,  ,  )  U (Y ,  )   ( (Y  Y )   e   )
LY  U Y (Y ,  )    0
YLY  Y U Y (Y ,  )     0
L  U  (Y ,  )    0
L   U  (Y ,  )     0
L   (Y  Y )   e    0


L    (Y  Y )   e    0
(c) Given
U (Y ,  )  Y   2 , we have the following Lagrangian and Kuhn-Tucker Conditions:
L(Y ,  ,  )  Y   2   ( (Y  Y )   e   )
LY  1    0
YLY  Y 1     0
L  2    0
L    2     0
L   (Y  Y )   e    0


L    (Y  Y )   e    0
Given
 *  0 , we have  2 * *  0 and thus *  2 * .
1  (2 *)  0 and thus  * 
And given
Y *  0 , we have
1
e
. And then since  *   (Y * Y )   we have
2
1
1  2 e
e
.
  (Y * Y )   and thus Y *  Y 
2
2 2
(d)
Differentiating our results from (c) and given that
comparative statics terms:
 e  0 and   0 , we have the following signed
 *
Y * 1
Y *  e 2  
Y *
 0,
,


0
 1,

 0,
e
4
 e




Y
 * 1  *
 ,
 0.

2 Y
(e)
All of the comparative statics terms in (d) make economic sense. Their interpretations are as follows:
Y *
e
- as  increases, Y* decreases or as the excepted rate of inflation increases, output decreases,
 e
Y *
- as  increases, Y* increases or as the responsiveness to inflation increases, output increases,

Y *
- as Y increases, Y* increases equal proportionally or as the full-employment level increases,
Y
44
output increases equal proportionally,
 *
e
- there is no relationship between  and  * or there
e

is no relationship between the expected rate of inflation and the actual rate of inflation,
 *
- as 

increases,
 * increases or as the responsiveness to inflation increases, the actual rate of inflation
increases,
 *
- there is no relationship between Y and  * or there is no relationship between the
Y
full-employment level and the actual rate of inflation.
9.12
9.13
9.14
9.15
Consider the GDP maximization problem from (9.47).
(9.56) Gives the formula to set up the corresponding dual.
Min NI i  i Li
i
i a xi  p x ,
Subject to: i a yi  p y ,
i  0,
From (9.47) we know that
 aTx

 a Lx
a Kx

aTy   1 2

a Ly   3 / 2 2 ,Total GDP=76, and T
a Ky   2 1 


L K  100 120 140
In this case the subscript i’s in (9.56) correspond to the 3 input T,L,K and by knowing total GDP the
inequalities can be changed to equalities. Therefore (9.56) becomes
Min 1001  1202  1403  76
1 ,2 ,3
3
11  2  23  1,
2
Subject to: 21  22  13  1,
1 , 2 , 3  0
45
Solving for
1 , 2 , 3 yields 1  0 , 2 
2
1
, and 3  .
5
5
9.16
9.17
In this problem we want to maximize the following utility function with the usual budget constraint:
U ( x1 , x2 )  ( x1 1)( x2  1)
Subject to:
M  p1 x1  p2 x2
The Lagrangian is then set up as follows with its first order conditions:
L( x1 , x2 ,  )  ( x1  1)( x2  1)   (M  p1 x1  p2 x2 )
(a)
(1)
Lx1  x*2  1  * p1  0
(2)
x1* Lx1  x1* ( x*2  1  * p1 )  0
(3)
Lx2  x1*  1  * p2  0
(4)
x*2 Lx2  x*2 ( x1*  1  * p2 )  0
(5)
L  M  p1 x1*  p2 x*2  0
(6)
* L   (M  p1 x*  p2 x* )  0
1
2
x1*  0 and x2*  0
Under these conditions, using equation (2) and
that
x1*  0 , we only have to solve for * using the fact
x2  1  * p1  0 :
* 
From this we get our first restriction:
1
p1
p1  0 . Using this we can then say that *  0 , which then
allows us to examine equation (5) as an equality:
M  p1 x1*  p2 x *2  0
M  p1 x1*  0
x1* 
(b)
M
p1
x1*  0 and x2*  0
Utilizing the same process in (a) here, but instead using equation (4), we see that equation (3) becomes
46
* :
an equality statement, which can be solved for
1
p2
* 
From this we then get our next restriction:
p2  0 . This then allows us to say that *  0 , which
turns equation (5) into an equality that can be solved for
M
p2
x *2 
(c)
x 2* :
x1*  0 and x2*  0
Under these new conditions, equations (1), (3) and (5) for the first order conditions of the Lagrangian
can be made equalities. Solving equation (1) for
 
*
Using this, solve equation (3) for
* yields:
x *2  1
p1
x1* :
x 
*
1
p 2 ( x *2  1)
p1
Now take this result and solve equation (5) for
M  p1 (
p 2 ( x *2  1)
p1
1
x 2* :
 1)  p 2 x *2  0
M  p 2 ( x *2  1)  p 2  p 2 x *2  0
 2 p 2 x2*   M
x2* 
M
2 p2
Here we can impose symmetry and say that x1 
*
9.18
47
M
.
2 p1
48
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