10. A firm has two plants that produce outputs of three goods. Its total labor f

Question

10. A firm has two plants that produce outputs of three goods. Its total labor force is fixed. When a fraction of its labor force is allocated to its first plant and a fraction 1 - A to its second (with 0 s of the three goods are given by the vector 1), the total out 8 2 4 10 a. Is it possible for the firm to produce either of the two output vectors /5 7 ^ if outputs cannot be thrown away? b. How do your answers to part (a) change if outputs can be thrown away? c. How will the revenue-maximizing choice of the fraction ? depend upon the selling prices (Pi, p2, p3) of the three goods? What condition must be satisfied by these prices if both plants are to remain in use?

Explanation / Answer

I’ll use x instead of lambda.

Let three good be G1,G2 and G3.

a) To check whether a particular output vector is producible, we need to find the value of x that produces output of any one of the goods, and use that value of x to check whether the output of other goods is produced or not.

The output of each good is given in its corresponding row.

Output of G1 : 8x + 2(1-x) = 2+6x

Output of G2 : 4x + 6(1-x) = 6 -2x

Output of G3 : 4x + 10(1-x) = 10 -6x

i) equating output of G1: 2+6x = 5

   6x = 3

   x = 0.5

we need check whether this values of x ( = 0.5) produces the desired output

G2 and G3.

If we substitute x = 0.5 , output of G2 = 6-2(0.5) = 5

output of G3 = 10-5(0.5) = 7

Hence when the labor is divided equally between plants this output is

produced.

ii) equating output of G1: 2+6x = 7

6x = 5

x = 5/6

we need check whether this values of x ( = 5/6) produces the desired output

G2 and G3.

If we substitute x = 0.5 , output of G2 = 6-2(5/6) = 13/3

output of G3 = 10-5(5/6) = 35/6

  

Clearly the output of G2 and G3 is not what we wanted. Hence this output

be produced.

b) If outputs can be thrown away.

The question asks:

“If the division of labor produced an output, say O, can you throw some of the

goods from O, to get the desired output, say O’ ? ”

  

Note that to throw away some goods to get the desired output, O’, the actual

output,O, for each good has to be greater than or equal to that of each good in

output O’ .

i) The first output vector can be produced without even throwing away any

goods, hence it can be produced.

(Just don’t throw any goods away)

ii) We have verified that output can’t be achieved without throwing away any

goods. Hence if this output can be achieved, then some goods must be thrown

away.

First, output of G1, has to be greater that 7.

2 +6x >= 7

x >= 5/6

But note that 0 <= x <= 1, hence 5/6 <= x <= 1.

We need to check whether other goods, G2 and G3 can be produced in excess.

output of G2 : 6 -2x has to be more than 5, for atleast one value in the x.

if 5/6 <= x <= 1

then 5/3 <= 2x <= 2

then -2 <= -2x <= -5/3

then 4 <= 6-2x <= 13/3 ( 6- 5/3 = 13/3 )

Hence, Output of G2 cannot exceed 13/3, which is less than 5.

Hence even if goods can be thrown out, the output vector cannot be achieved.

c) If prices of G1, G2 and G3 are p1,p2 and p3 respectively.

Revenue from each good is output of good * price of the good.

Revenue of G1 : R1 = (2+6x)·p1

Revenue of G2 : R2 = (6 -2x)·p2

Revenue of G3 : R3 = (10 -6x).p3

Total Revenue : R = R1 + R2 +R3

So, R = (2·p1 +3·p2 +10·p3) + x (6·p1- 2·p2 - 6·p3)

Let A =(2·p1 +3·p2 +10·p3) and B = (6·p1- 2·p2 - 6·p3)

Note A and B are constants.

  

Then R = A +xB

  

if B = 0, then R is constant, and R = A.

if B > 0, then R is maximum when x = 1.

if B < 0, then R is maximum when x = 0.

unless B = 0, one of the plants has to be closed. Hence the condition for both

the plants to be in use is - B =0 or (6·p1- 2·p2 - 6·p3) = 0.

  

  

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