Microeconomic theory basic principles and extensions 11th edition nicholson solutions manual

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CHAPTER 9: Production Functions 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 9.1

This problem illustrates the isoquant map for fixed proportions production functions.

Parts (c) and (d) show how variable proportions situations might be viewed as limiting cases of a number of fixed proportions technologies.

9.2

This 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 is a thorough examination of most of the properties of the general two-input CobbDouglas production function.

9.6

This problem is an examination of the marginal productivity relations for the CES production function.

9.7

This illustrates a generalized Leontief production function. 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.

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Chapter 9: Production Functions

Analytical Problems 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. 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. Shows how Euler’s Theorem can be used to study the likely signs of cross productivity effects.

Solutions 9.1

a., b. k per period

Large-mower technology

10 Small-mower technology

8

5

8

l per period

With the small-mower technology, k = 1 and l = 1 are needed to mow 5,000 sq. ft.. To mow 40,000 = 8  5,000, need to scale inputs up by 8, that is, k = 8 and l = 8. Similar calculations show that with the large-mower technology, use k = 10 and l = 5. c.

To mow half of the total 40,000 with each technology, use half of the inputs from part b, so allocate k = 4 and l = 4 to the small-mower technology and k = 5 and l = 2.5 to the large-mower technology, for a total of k = 9 and l = 6.5.

To mow only a quarter of 40,000 with the small-mower technology, allocate a quarter of the inputs from part b ( k = 2 and l = 2 ) to production using the small-mower technology and 3 4 of the inputs from part b ( k = 7.5 and

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Chapter 9: Production Functions

89

l = 3.75 ) to the large-mower technology for a total of k = 9.5 and l = 5.75.

We can interpret fractions of k and l as use of the input for only part of an hour. d.

We know from part c that the combinations (k = 9, l = 6.5) and (k = 9.5, l = 5.75) can be used to mow at least 40,000 sq. ft. Let’s take for granted for now that 40,000 is in fact the most that can be mowed with these combinations. Then both these combinations are on the q = 40, 000 isoquant, as pictured in this graph:

Assuming that the isoquant is linear, one can use the point-slope form of a line to show that the equation for the line between the two combinations is 40 2l (1) k = − .

3 3 We made some assumptions to enable us to graph this isoquant, namely that 40,000 was the most that could be produced by the input combinations and that the isoquant was a line between the two points, not some other curve. The formal proof that Equation (1) is the q = 40, 000 isoquant is hard. Let (k1, l1 ) be the inputs used in the small-mower technology and (k2 , l2 ) in the large-mower technology. Letting (k , l ) be the total input combination, we have (2) k = k1 + k2 (3) l = l1 + l2 .

Total output of 40,000 is the sum of the output from technologies 1 and 2, both Leontieff: k  (4) 40, 000 = 5, 000 min ( k1 , l1 ) + 8, 000 min  2 , l2  .  2  For efficient production, there should be no wastage, so (5) min ( k1 , l1 ) = k1 = l1

k  k (6) min  2 , l2  = 2 = l2 . 2  2 Equations (2), (3), (5), and (6) are four equations which can be used to solve for the four unknowns k1 , l1 , k2 , l2 in terms of k and l . Sparing the details, the © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

90

Chapter 9: Production Functions

solution is k1 = 2l − k

l1 = 2l − k k2 = 2l − 2k l2 = k − l.

Substituting these solutions into Equations (5) and (6), min ( k1 , l1 ) = 2l − k

k  min  2 , l2  = k − l. 2  Substituting these expressions back into Equation (4), 40,000 = 5,000 ( 2l − k ) + 8,000 ( k − l ).

One can rearrange this equation into Equation (1).

9.2

Given production function q = kl − 0.8k 2 − 0.2l 2 . a.

When k = 10, total labor productivity is TPl = 10l − 0.2l 2 − 80, and average labor productivity is q 80 AP l = = 10 − 0.2l − . l l To find where APl reaches a maximum, take the first-order condition: dAP l 80 = − 0.2 = 0. dl l The maximum is at l = 20. When l = 20, q = 40. The graph is provided after part b.

b.

Marginal labor productivity is dq MPl = = 10 − 0.4l. dl To find where this is 0, set MPl = 10 − .4l = 0, implying l = 25.

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Chapter 9: Production Functions

c.

91

If k = 20, TPl = 20l − .2l 2 − 320 = q AP l = 20 − .2l −

320 l

MPl = 20 − .4l.

APl reaches a maximum at l = 40, q = 160. At l = 50, MPl = 20 − .4l = 0.

d.

9.3

Doubling of k and l here multiplies output by 4 (compare parts a and c). Hence, the function exhibits increasing returns to scale.

Given production function q = 0.1k 0.2l 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 k = 100, l = 100, and produces output q = 10. Total cost is 10,000 (by design).

b.

We have

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92

Chapter 9: Production Functions

MPk =

q l = .02   k k

0.8

0.2

k MPl = .08   . l Setting these equal yields l k = 4. Substituting into the production function,

q = 10 = 0.1k 0.2 (4k )0.8 = 0.303k.

Solving, k = 33 and l = 132. Total cost is 8,250.

9.4

c.

The cost savings in part b is 1,750. This can buy 35 more bar stools, for a total of k = 68. Substituting k = 68 and l = 132 into the production function gives q = 11.6, an increase in output from the 10 units produced in part a.

d.

Carla’s ability to influence the decision depends on whether she provides a unique 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.

a.

The firm should allocated labor such that MPl1 =MPl2 . Thus,

q1 q2 = l1 l2 

 (10l10.5 )  (50l20.5 ) = l1 l2

 5l1−0.5 = 25l2−0.5 l 1  1 = . l2 25 b.

9.5

In addition to the previous equation, we have l1 + l2 = l. Solving these two equations for l1 and l2 as functions of l yields l 25 l1 = , l2 = l.

26 26 Substituting into the production function summing the two locations, total output is q = q1 + q2 = 10l10.5 + 50l20.5 = 10 26l .

Given production function q = Ak  l  . a.

f k =  Ak  −1l   0.

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Chapter 9: Production Functions

93

f l =  Ak  l  −1  0. f kk =  ( − 1) Ak  − 2l   0. f ll =  (  − 1) Ak  l  − 2  0. f kl =  Ak  −1l  −1  0.

b.

q k k  =  Ak  −1l   =  . k q q q l k eq ,l =  =  Ak  l  −1  =  . l q q eq ,k =

c.

f (tk , tl ) = t  + Ak  l  . q t t eq ,t = lim  = lim( +  )t  +  −1q  =  +  . t →1 t q t →1 q

d.

Quasiconcavity follows from the signs in part a.

e.

Concavity looks at: f kk fll − f kl2 =  ( − 1)  (  − 1) A2 k 2 − 2l 2  −2 −  2  2 A2 k 2 −2l 2  −2 = A2 k 2 − 2l 2  − 2 (1 −  −  ).

This expression is positive (and the function is concave) only if  +   1.

9.6

a.

We have MPk =

q 1   1− =  k + l    k  −1 k 

= q 1−   k  − 1 1− 

q =  . k Similar manipulations yield

q MPl =   l

b.

1− 

.

Using the results from part a, 1−  MPk  l  RTS = =  .

MPl  k  Inverting, 1

l = RTS 1−  , k © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

94

Chapter 9: Production Functions

implying 1

k = RTS  −1 , l in turn implying 1 k ln   = ln RTS .  l   −1 From Equation (9.34), d ln ( k l ) 1 = = . d ln RTS  − 1 c.

Computing elasticities, −

q k  q  1 eq ,k =  =  = k q  k  1+ (l k )  q e q, l =   l

−

=

1 1+ ( k l )



=

1 1+ ( l k )

−

.

Putting these over a common denominator yields eq ,k + eq ,l = 1, which shows constant returns to scale. d.

9.7

The result follows directly from part a since 1 = .

1− 

Given production function f (k , l ) =  0 +  1 kl +  2 k +  3 l. a.

For constant returns to scale, f (tk , tl ) = tf (k , l ). But

f (tk , tl ) =  0 +  1 tk  tl +  2tk +  3tl

(

)

=  0 + t  1 kl +  2k +  3 tl , while

(

)

tf (tk , tl ) = t  0 + t  1 kl +  2k +  3 tl .

For these two equations to be equal, 0 = 0. b.

Assume 0 = 0 to ensure constant returns to scale. Then k MPl = 0.51   l

0.5

l MPk = 0.51   k

0.5

+ 3 + 2 .

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Chapter 9: Production Functions

95

Both are homogeneous of degree zero with respect to (k , l ) and exhibit diminishing marginal productivities. c

Footnote 6 provides the key formula in the special case of constant returns to scale: f f = l k f  f kl

( 1 2 )( k l )1 2 + 3  ( 1 2 )( l k )1 2 +  2   . = 12 −1 2  1 ( kl ) +  2 k + 3l  ( 1 4 )( kl )     For  = 0, one of the factors in the numerator has to be 0. For this to be true for all (k , l ) , either 1 = 3 = 0 or 1 = 2 = 0. In other words, 1 = 0 and either 3 = 0 or 2 = 0.

For  = 1, the numerator has to equal the denominator. Expanding out the numerator gives

12

+

13

+

13

(l k )

+

1 2

(l k )

+

1 2

12

4 2 and the denominator gives

12

12

2

(k l )

12

+  2 3

(k l ) .

12

4 4 4 For these two expressions to be equal for all (k , l ) requires 13 = 12 = 2 3 = 0.

This condition holds if any two of the three parameters 1 , 2 , 3 are 0.

For  = , the denominator must be 0. This only holds for all (k , l ) if the second factor is 0, i.e., −1 2 ( 1 4 )( kl ) = 0.

But for this condition holds for all (k , l ) only if 1 = 0.

9.8

Because q = f (k , l ) exhibits constant returns to scale, it is homogeneous of degree 1. In this special case, Euler’s theorem (Equation 2.123) states q = f (k , l ) = f k k + f l l .

Rearranging, q k = fl + f k   . l l If fl  q l , then f k  0. But no firm would ever produce in such a range.

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96

Chapter 9: Production Functions

Analytical Problems 9.9

Local returns to scale a.

If f (tk , tl ) = tf (k , l ), then f (tk , tl ) t eq ,t = lim  t →1 t f (tk , tl ) f (k , l ) = lim t →1 f ( k , l ) = 1.

b.

eq ,t = lim t →1

f (tk , tl ) t  t f (tk , tl )

f  t  f = lim   k +  l   t →1 k l  f  = eq ,k + eq ,l .

c.

We have

  (1 + t −2 k −1l −1 ) t  eq ,t = lim    t →1 t q   t = lim  q 2 2t −3 k −1l −1   t →1 q  = 2qk −1l −1 1  = 2q  − 1 q  = 2 − 2q.

Hence, eq ,t  1 for q  0.5, and eq ,t  1 for q  0.5. d.

9.10

The intuitive reason for the changing scale elasticity is that this function has an upper bound of q = 1 and gains from increasing the inputs decline as q approaches this bound.

Returns to scale and substitution a.

Let  denote the elasticity of substitution and RTS the marginal rate of technical substitution associated with F . Then F  f  −1 fl f RTS = l = = l = RTS .  −1 Fk  f f k fk

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Chapter 9: Production Functions

97

Because f is homogeneous of degree 1, it is homothetic, and so RTS is a function only of the input ratio: i.e., RTS = B(k l ). Hence, k l = b( RTS ), where b = B−1. We have d ( k l ) RTS = d RTS k l RTS = b( RTS ) k l

= b( RTS ) =

RTS k l

d ( k l ) RTS

d RTS k l = . where the last set of equalities follow from RTS = RTS . . Hence, the elasticity of substitution is the same for both functions.

In view of the formula for  in the statement of the question, we can also write f f  = = k l . f f kl b.

A general Cobb-Douglas for arbitrary returns to scale can be written F (k , l ) = k al b  a b  =  k a +b l a +b    = ( k  l1− )

a +b

a +b

= f (k , l ) a +b ,

where a . a+b The formula f f (1)  = k l f f kl can be applied to show the elasticity of substitution for f equals 1. Since F is just f scaled by the exponent a + b, the preceding results imply that the elasticity of substitution for F also equals 1.

A general CES function can be written

=

F (k , l ) = ( k  + l  )

 

= f ( k , l ) , © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

98

Chapter 9: Production Functions

where

.

f (k , l ) = ( k  + l  ) .

1

The formula in Equation (1) above can be used to show that the elasticity of substitution for this f is 1 = .

1−  Hence the elasticity of substitution for general CES function F is the same as this

9.11

More on Euler’s Theorem

a.

By Euler’s Theorem:  xi fi =kf . i

Differentiating with respect to x j :

f j +  xi fij = kf j . i

Multiplying by x j :

x j f j + x j  xi fij = kx j f j , i

implying

 ( x f + x  x f ) =k  x f . j

j

j

j

i ij

j

i

j

j

  xi x j fij =(k − 1) x j f j . i

j

j

  xi x j fij =(k − 1)k f . i

b.

j

Using Young’s Theorem, we have f12 = f 21. For n = 2 and k = 1, the above expression becomes x12 f11 + 2 x1 x2 f12 + x22 f 22 = 0.

If we assume diminishing marginal productivity, then f11 , f 22  0. Since x1 and x2 denote quantities, they are always positive numbers. Thus, in order for the above relation to hold, f12  0. Therefore, with two inputs and constant returns to scale, increasing one input must increase the marginal product of the other input.

If k  1, we have increasing returns to scale and x12 f11 + 2 x1 x2 f12 + x22 f 22 = k (k − 1) f .

Since k  1, k (k − 1) f  0. As described above, under the assumption of diminishing marginal productivity, f11 , f 22  0. Thus, f12  0.

If k  1, we have decreasing returns to scale and

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Chapter 9: Production Functions

99

x12 f11 + 2 x1 x2 f12 + x22 f 22 = k (k − 1) f .

Since k  1, k (k − 1) f  0. Under the assumption of diminishing marginal productivity, f11 , f 22  0. Therefore, the left-hand side of the expression must be

negative, and this could happen if f12  0 or f12  0. Thus, in this case, the sign of f12 cannot be determined without knowing the form of the production function. c.

Under the assumption of diminishing marginal productivity, fii  0 for i = 1,..., n.

Thus, we cannot infer the sign of any one of the cross-partial derivatives. But, for k  1, we know that  xi x j fij  0 i

j

for the expression in part a to hold. Therefore, at least one cross-partial derivative must be positive. For the case of k  1, like in the case of only two inputs, the sign for the sum of the cross-partial derivatives is ambiguous. d.

Given n

f ( x1 , x2 ,..., xn ) =  xii . i =1

we have n

f (tx1 , tx2 ,..., txn ) = t 1 + 2 +...+ n  xii . i =1

Hence, returns to scale will be determined by k = 1 +  2 + ... +  n : fi =  i xii −1  x j j 

j i

fii =  i ( i − 1) xii − 2  x j j 

j i

 −1

fij =  i j xii −1 x j j

 x . k

k  j i

k

Using the results from part a yields  xi x j fi j = i j f − i f = k (k − 1) f . i

j

i

j

i

So, the sign of

  −  i

i

j

j

i

i

is given by the sign of k .

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