The production function for a company is given by f(x, y) = 100x^0.6y^0.4 where
ID: 2880786 • Letter: T
Question
The production function for a company is given by f(x, y) = 100x^0.6y^0.4 where x is the number of units of labor (at $60 per unit) and y is the number of units of capital (at $48 per unit). The total cost for labor and capital cannot exceed 5100,000. (a) Find the maximum production level for this manufacturer. (Round your answer to the nearest integer.) units (b) Find the marginal productivity of money. (Round your answer to three decimal places.) (c) Use the marginal productivity of money to find the maximum number of units that can be produced when $135,000 is available for labor and capital. units (d) Use the marginal productivity of money to find the maximum number of units that can be produced when $310,000 is available for labor and capital. unitsExplanation / Answer
Solution:
The production function, P(x,y) represents the number of units produced when the quantities of labour (x) and
capital (y) are available, where
P(x,y)=100x0.6y0.4
For the resources, we would like to minimize the total cost, which can be represented by the cost function,
C(x,y)=60x+48y
Maximum/minimum cost will be at the point where differentiation of the function will have zero value
f(x,y) =100x0.6y0.4
d(f(x,y) ) = d(100x0.6y0.4)/dx +d(100x0.6y0.4)/dy
=100*0.6x(-0.4)y0.4 +100*0.4x0.6y(-0.6)
= 0.6(y/x)(0.4) = - 0.4(x/y)(0.6)
let x/y =p
0.6(1/p)(0.4) = - 0.4(p)(0.6) (-0.6/0.4)
=p(0.6)/(1/p)(0.4) -1.5
=p(0.6)*p(0.4) p
=-1.5 x/y
=-1.5 x
= -1.5y f(x,y)
=100x(0.6)*y(0.4)
=100x(0.6)*(-1.5x)(0.4) 100,000
=100(-1.5)(0.4)x(0.4+0.6) 1000
=1.17607x
x= $850.283
y =1.5*x = $1275.42
Total cost
C(x,y)=60x+48y
=60*850.283 + 48*1275.42
= $112237.14
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