Because the problems in this chapter do not involve optimization (cost minimization principles
are not presented until Chapter 10) they tend to have a rather uninteresting focus on functional
form. Computation of marginal and average productivity functions is stressed along with a few
applications of Euler’s theorem. Instructors may want to assign one or two of these problems for
practice with specific functions, but the focus for Part (4) problems should probably be on those
in Chapters 10 and 11.
Comments on Problems
cases of a number of fixed proportions technologies.
9.2 This problem provides some practice with graphing isoquants and marginal productivity
relationships.
9.3 This problem explores a specific Cobb–Douglas case and begins to introduce some ideas
about cost minimization and its relationship to marginal productivities.
9.4 This problem involves production in two locations and develops the equal marginal
products rule.
9.5 This problem is a thorough examination of most of the properties of the general two-input
Cobb–Douglas production function.
9.6 This problem is an examination of the marginal productivity relations for the CES
production function.
9.7 This problem illustrates a generalized Leontief production function and provides a two-
input illustration of the general case, which is treated in the extensions.
9.8 Application of Euler’s theorem to analyze what are sometimes termed the “stages” of the
average–marginal productivity relationship. The terms “extensive” and “intensive”
margin of production might also be introduced here, although that usage appears to be
archaic.
Analytical Problems
CHAPTER 9:
Production Functions
9.9 Local returns to scale. This problem introduces the local returns– to– scale concept and
presents an example of a function with variable returns to scale.
9.10 Returns to scale and substitution. This problem shows how returns to scale can be
incorporated into a production function while retaining its input substitution features.
9.11 More on Euler’s theorem. This problem shows how Euler’s theorem can be used to
study the likely signs of cross-productivity effects.
Solutions
9.1 a, b.
With the small-mower technology,
1k
and
1l
are needed to mow 5,000 sq.
ft. To mow
40,000 8 5,000,
need to scale inputs up by 8, that is,
8k
and
8.l
Similar calculations show that with the large-mower technology, use
10k
and
5.l
c. To mow half of the total 40,000 with each technology, use half of the inputs from
part (b), so allocate
4k
and
4l
to the small-mower technology and
5k
and
2.5l
to the large-mower technology, for a total of
9k
and
6.5.l
To mow only a quarter of 40,000 with the small-mower technology,
allocate a quarter of the inputs from part (b) (
2k
and
2l
) to production using
the small-mower technology and
34
of the inputs from part (b) (
7.5k
and
3.75l
) to the large-mower technology for a total of
9.5k
and
5.75.l
We can interpret fractions of
k
and
l
as use of the input for only part of
k per
period
8l per
period
5
8
10
Small-mower
technology
Large-mower
technology
an hour.
d. We know from part (c) that the combinations
( 9, 6.5)kl
and
( 9.5, 5.75)kl
can be used to mow at least 40,000 sq. ft. Let’s take for
1
1
2
2
2,
2,
2 2 ,
.
k l k
l l k
k l k
l k l
Substituting these solutions into Equations 5 and 6,
11
22
min , 2 ,
min , .
2
k l l k
kl k l
One can rearrange this equation into Equation 1.
9.2 Given production function
22
0.8 0.2 .q kl k l
a. When k = 10, total labor productivity is
2
10 0.2 80,
l
TP l l
To find where
l
AP
reaches a maximum, take the first-order condition:
80 0.2 0.
l
dAP =
dl l
The maximum is at
= 20.l
When
20,l
40.q
The graph is provided after
b. Marginal labor productivity is
10 0.4 .
l
dq
MP = l
dl
To find where this is 0, set
10 0.4 0,
l
MP l
implying
25.l
c. If
20,k
2
20 0.2 320 ,
320
20 0.4 .
l
l
TP l l q
MP l
l
AP
reaches a maximum at
= 40,l
= 160.q
At
50,l
20 0.4 0.
l
MP l
d. Doubling of
k
and
l
here multiplies output by 4 (compare parts (a) and (c)).
Hence, the function exhibits increasing returns to scale.
0.2 0.8
a. Given Sam spends $10,000 in total and equal amounts on both inputs, he spends
$5,000 on each. At the $50 per hour, he uses inputs
100,k
100,l
and
produces output
10.q
Total cost is 10,000 (by design).
b. We have
0.8
0.02 ,
k
ql
MP kk
0.2
0.08 .
l
k
MP l
Solving,
33k
and
132.l
Total cost is 8,250.
10,000 10 12.12,
8,250
input for Cheers. If she does, she can either resist increasing the number of stools
or can allow an increase in stools in exchange for a higher salary.
12
12
0.5 0.5
12
12
0.5 0.5
12
1
2
(10 ) (50 )
5 25
1.
25
qq
ll
ll
ll
ll
l
l
b. In addition to the previous equation, we have
12 .l l l
Solving these two
equations for
1
l
and
2
l
as functions of
l
yields
12
25
,.
26 26
l
l l l
Substituting into the production function and then summing over the two
locations, total output is
0.5 0.5
1 2 1 2
10 50 10 26 .q q q l l l
10,
b.
1
,,
qk
q k k
e Ak l
k q q
1
q l k
c.
( , ) ,f tk tl t Ak l
1
,11
lim lim( ) .
qt tt
q t t
e t q
t q q
2 2 2 2 2 2 2 2 2 2 2 2 2
2 2 2 2 2
( 1) ( 1)
(1 ).
kk ll kl
f f f A k l A k l
A k l
This expression is positive (and the function is concave) only if
1.