Question

The Return of Detroit City: Enpar manufactures engine parts for Ford using steel as an input. Enpar operates two plants, one in Detroit, MI and one in Greenville, SC. The production functions for the two Enpar plants are: Detroit: Qp = 20 5p -0.25 50? Greenville: QG = 40 SG -0.556? where Qp and Qo are the outputs of engine parts (in units) from the Detroit and Greenville plants, respectively. Sp and Sg are the amounts of steel for the two plants. The firm has 110 units of steel available. How much steel should be sent to Greenville? Enter as a value.

Answer 1

Total units of steel available is 110 units.

S_D is the amount of steel at Detroit plant.

S_G is the amount of steel at Greenville plant.

So,

S_D + S_G = 100

or,

S_D = 100 - S_G

Production function of Detroit Plant is as follows -

Q_D = 20S_D - 0.25S_D²

Calculate marginal product at Detroit plant -

MP_D = dQ_D/dS_D = d(20S_D - 0.25S_D²)/dS_D = 20 - 0.50S_D

Production function of Greenville plant is as follows -

Q_G = 40S_G - 0.5S_G²

Calculate the marginal product at Greenville plant -

MP_G = dQ_G/dS_G = d(40S_G - 0.5S_G²)/dS_G = 40 - S_G

At optimum production by Enpar, marginal product of two plants would be equal.

Assuming optimum production,

MP_D = MP_G

20 - 0.5S_D = 40 - S_G

20 - 0.5(100 - S_G) = 40 - S_G

20 - 50 + 0.5S_G = 40 - S_G

-30 + 0.5S_G = 40 - S_G

1.5S_G = 70

S_G = 70/1.5 = 46.67 or 47

Thus,

The steel that should be sent to Greenville is 47 units.