A firm has two plants that produce outputs of three different goods. Its total labour force is fixed. When a fraction ? of its labour force is allocated to its first plant and a fraction 1 - ? to its...

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A firm has two plants that produce outputs of three different goods. Its total labour force is fixed. When a fraction ? of its labour force is allocated to its first plant and a fraction 1 - ? to its second plant (with 0 = ? = 1), the total output of the three different goods are given by the vector ?(8, 4, 4) + (1 - ?)(2, 6, 10) = (6? + 2, -2? + 6, -6? + 10).


(a) Is it possible for the firm to produce either of the two output vectors a = (5, 5, 7) and b = (7, 5, 5) if output cannot be thrown away?


(b) How do your answers to part (a) change if output can be thrown away?


(c) How will the revenue-maximizing choice of the fraction ? depend upon the selling prices (p1, p2, p3) of the three goods? What condition must be satisfied by these prices if both plants are to remain in use?



Answered Same DayDec 27, 2021

Answer To: A firm has two plants that produce outputs of three different goods. Its total labour force is...

Robert answered on Dec 27 2021
126 Votes
Let the fraction that went to first plant be x and the other plant be
(1 – x)
Output vector = (
6 * x + 2 , -2 * x + 6, -6 * x + 10)
For a = (5,5,7) to be present we must have
6 * x + 2 = 5
 x = ½
Putting this value of x we get
-2 * x + 6 = 5
-6 *x + 10 = 7
Also for b = (7, 5, 5)
It must satisfy
6 * x + 2 = 7
 x = 5/6
Putting this value of x we get
-2 * x + 6 = 13/3
-6 * x + 10 = 6
which is not equal to the required value of 5 so the vector b is not
possible
This occurs if output cannot be thrown away.
(b) When output can be thrown away the vector a...
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