A firm produces an output with the production function Q = KL, where Q is the number of units of output per hour when the firm uses K machines and hires L workers each hour. The marginal products for this production function are MPK = L and MPL = K. The factor price of K is 4 and the factor price of L is 2. The firm is currently using K = 16 and just enough L to produce Q = 32. How much could the firm save if it were to adjust K and L to produce 32 units in the least costly way possible?
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Currently the firm must be using L = Q/K = 32/16 = 2 units of labor. Let the factor prices of capital and labor be, respectively, r and w.
Its total expenditure is C = wL + rK = 2(2) + 4(16) = 68.
If it were to minimize cost, it would hire L and K so that (1) MPK/r = MPL/w, or L/4 = K/2, or L = 2K and (2) Q = LK.
(1) and (2) imply that Q = 2K2, or 32 = 2K2, and thus K = 4 and L = 8.
So Q = 32 can be produced efficiently with a cost of C = wL + rK = 2(8) + 4(4) = 32.
The firm could save 68 – 32 = 36 by producing efficiently.
- A firm operates with the production function Q = K2L. Q is the number of units of output per day when the firm rents K units of capital and employs L workers each day. The manager has been given a production target: Produce 8,000 units per day. She knows that the daily rental price of capital is $400 per unit. The wage rate paid to each worker is $200 day.
- a) Currently the firm employs at 80 workers per day. What is the firm’s daily total cost if it rents just enough capital to produce at its target?
- b) Compare the marginal product per dollar sent on K and on L when the firm operates at the input choice in part (a). What does this suggest about the way the firm might change its choice of K and L if it wants to reduce the total cost in meeting its target?
- c) In the long run, how much K and L should the firm choose if it wants to minimize the cost of producing 8,000 units of output day? What will the total daily cost of production be?
- a) Suppose that the firm is operating in the short run, with L = 80. To produce Q = 8000, how much K will it require? From the production function we observe that 8,000 = K2 (80) => K = 10.
The total cost would be C = wL + rK = $200(80) + $400(10) = $2,000 per day.
- b) Let’s examine the “bang for the buck” for K and L when K = 10 and L = 80.
For capital: MPK / r = 2KL / 400 = 2(10)(80) / 400 = 4
For labor: MPL / w = K2 / 200 = 102 / 200 = 0.5
So the marginal product per dollar spent on capital exceeds that of labor. The firm would like to rent more capital and hire fewer workers.
- c) Because the production function is Cobb-Douglas, we know that it has diminishing MRTSL,K and that the isoquants do not intersect either the K or L axis. Thus the cost reducing basket (K,L) will be interior (with K > 0 and L > 0). To find the optimum, we use the two conditions:
(1) Tangency condition: MPK / MPL = r / w => 2KL/K2 = 400 / 200 => K = L
(2) Production Requirement: K2L = 8,000
Together equations (1) and (2) tell us that K = 20 and L = 20.
The total cost would be C = wL + rK = $200(20) + $400(20) = $12,000 per day.
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