Problem Set #10
Suggested Answers to Problems 11.1, 11.3, 11.5, 11.6 and 11.7
11.1 Digging clams by hand in Sunset Bay requires only labor input. The total number of
clams obtained per hour (q) is given by
q = 100L
where L is labor input per hour.
a. Graph the relationship between q and L.
300
250
b. What is the average
productivity of labor in
Sunset Bay? Graph this
relationship and show
that APL diminishes for
increases in labor input.
TPL
200
150
100
APL = 100/L. APL moves
inversely with L See graph
below.
50
0
0
c.
2
4
6
8
Show that the marginal productivity of labor in Sunset Bay is given by
MPL = 50/L
dq/dL = .5(100)L-.5
Graph this
relationship and show
that MPL< APL for all
values of L. Explain
why this is so.
See graph on the left.
The Marginal
“drives” the average.
When the marginal is
below the
average it pulls it
down.
120
100
80
60
APL
40
20
MP
L
0
0
2
1
4
6
8
11.3. Power Goat Lawn Company uses two sizes of mowers to cut lawns. The smaller
mowers have a 24-inch blade and are used on lawns with many trees and obstacles.
The larger mowers are exactly twice as big as the smaller mowers and are used on
open lawns where maneuverability is not so difficult. The two production
functions available to Power Goat are:
Output per hour
(Square Feet)
8000
5000
Large Mowers
Small Mowers
Capital Input
(# of 24” mowers)
2
1
Labor Input
1
1
a. Graph the q= 40,000 square feet isoquant for the first production function. How
much K and L would be
used if these factors were
K
16
combined without waste?
14
Q=40,000 with 5*2 = 10 units of
12
capital and 5*1 =5 units of
10
labor. However, the production
q=40,000K
technology allows no
8
substitutability. Thus, the
6
production function is
4
lexographical.
2
0
0
2
4
6
8
10
L
b. Answer part (a) for the second function.
For q=40,000,
K =8 and L =
8.
See graph on
right.
K
16
14
12
10
q=40,000K
8
6
4
2
0
0
2
4
6
8
10
12
14
16
L
2
c. How much K and L would be used without waste if half of the 40,000-square-foot
lawn were cut by the method of the first production function and half by the
method of the second? How much K and L would be used if three-fourths of the
lawn were cut by the first method and one-fourth by the second? What does it
mean to speak of fractions of K and L?
40,000 sq. ft.
LM
SM
LM/SM mix
L
K
L
K
0/100$%
0
0
8
8
25%/75%
1.25
2.5
6
6
50%/50%
2.5
5
4
4
75%/25%
3.75
7.5
2
2
100%/0
5
10
0
0
Fractions of K and L are fractions of an hour.
d. On the basis of your observations in part (c), draw a q=40,000 isoquant for the
combined production functions.
40,000 sq. ft.
LM/SM mix
0/100$%
25%/75%
50%/50%
75%/25%
100%/0
L
8
7.25
6.5
5.75
5
K
8
8.5
9
9.5
10
13
12
11
10
q=40,000
9
8
7
6
5
0
5
10
15
Observe that even when two technologies use fixed inputs, if combinations of those
technologies can be made, substitutability between the primary inputs be made.
3
11.5 Suppose that
q = LK 0<<1, 0<<1, +=1.
a. Show that eq,L = , eq,K = 
eq,L
= (q/L)(L/q)
= (L-1K)(L/ L-1K)
= 
eq,K
= (q/K)(K/q)
= (LK-1)(K/ LK-1)
= 
b. Show that MPL>0, MPK>0, 2q/L2 <0 , 2q/K2 <0
MPL = (q/L) = L-1K For L, K >0 this is positive.
MPK = (q/K) = LK-1 For L, K >0 this is positive.
2q/L2 = (MPL/L) = (-1)L-2K Since <1, this is negative
2q/K2 =(MPK/K) = (-1)LK-2 Since <1, this is negative
c. Show that the RTS depends only on K/L, but not on the scale of production, and that
the RTS (L for K) diminishes as L/K increases
RTS (L for K) = MPL/MPK
=
(L-1K )/LK-1
=
(K)/(L)
This is a CRS function. That is, any increase in all inputs a will increase output by a
(aL)(aK) = a+LK = a LK
Increasing the scale of production will not affect the RTS, because the scale term “a”
cancels out in the RTS expression.
Further L/K enters the RTS inversely, so L/K increases implies RTS falls
4
11.6. Show that for the constant returns-to-scale CES production function
q = [K+ L] 1/
a. MPK = (q/K) 1- and that MPL = (q/L) 1-
MPL = (q/L) =(/)[K+ L] (1-)/ L-1
=
(q/L) (1-)
MPK = (q/K) =(/)[K+ L] (1-)/ K-1
=
(q/K) (1-)
b. RTS(L for K) = (K/L) 1-. Use this to show that  = 1/(1-)
RTS = MPL/MPK
=
=
[(q/L) (1-)]/[(q/K) (1-)]
[K/L] (1-)
Observe that ln(RTS) = ln [K/L]
d (lnRTS)/ d( ln K/L) =
(1-)
If the implicit function theorem holds (and the relationship is invertible,
d( ln K/L) /d (lnRTS) =
1/(1-) =

c. Determine the output elasticities for K and L. Show that their sum equals 1.
eq,L
=
(q/L)[L/q] =
=
=
MPL[L/q] =(q/L) (1-)[L/q]
L/q
L /[K+ L]
(q/K)[K/q] =
=
=
MPK[K/q] =(q/K) (1-)[K/q]
K/q
K /[K+ L]
Similarly
eq,K
=
d. Prove that (q/L) = (q/L) Hence show ln (q/L) =  ln (q/L)
q/L =
(q/L) (1-)
=
(L/q) 
Since  = 1/(1-)
Taking logarithms
ln(q/L) =  ln (q/L)
5
11.7 Consider a production function of the form
q = o + 1 (KL)½ + 2 K+ 3 L
where
0< i<1 i = 0, …, 3.
a. If this function is to exhibit constant returns to scale, what restrictions should be placed
on the parameters o, … 3?
CRS implies that this function is homogeneous of degree 1. Thus, we need for
mq = o + 1 (mKmL)½ + 2 mK+ 3 mL
This can only hold if o =0. No other restrictions are necessary.
b. Show that in the constant-returns-to-scale case this function exhibits diminishing
marginal productivities and that the marginal productivity functions are homogeneous of
degree zero.
q/L = ½ 1 (K½ /L½) +3>0
MPL/L= - ¼ 1((K½ /L3/2 )<0
Similarly
q/K = ½ 1 (L½ /K½) +2 >0
MPK/K= - ¼ 1((K½ /L3/2 ) <0
The degree-zero homogeneity of the marginal productivity functions is obvious, since K
and L are reciprocally related in each case.
c. Calculate  in this case. Is  constant?

=
[d(K/L)/dRTS][RTS/(K/L)}
Observe that the RTS is
RTS = MPL/MPK = [½ 1 (K½ /L½) +3]/[ ½ 1 (L½ /K½) +2]
Sorry, but this isn’t pretty! It is obvious from the RTS that we cannot develop an
expression for RTS that will allow for application of the implicit function theorem. By
brute force, I managed to get an expression for  that is
=
¼ 1 (K/L)1/2(3-2)
¼ (K/L) + ½ 1(K/L)( 2+3) +23
If correct, this is clearly not a constant. This is what you would expect, given the linearly
additive terms in the RTS.
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