Let S represent the amount of steel produced (in tons). Steel production is rela
ID: 3122998 • Letter: L
Question
Let S represent the amount of steel produced (in tons). Steel production is related to the amount of labor used (L) and the amount of capital used (C) by the following function:
S = 10 L0.4 C 0.6
In this formula L represents the units of labor input and C the units of capital input. Each unit of labor costs $60, and each unit of capital costs $100.
(a) Formulate an optimization problem that will determine how much labor and capital are needed in order to produce 40,000 tons of steel at minimum cost. Min L + C L0.4 C 0.6 - Select your answer -<>=Item 4 L, C - Select your answer -<>=Item 6 (b) Solve the optimization problem you formulated in part a. Hint: When using Excel Solver, start with an initial L > 0 and C > 0. If required, round your answers to two decimal places. L = $ C = $ Cost = $Explanation / Answer
Let S represent the amount of steel produced (in tons). Steel production is related to the amount of labor used (L) and the amount of capital used (C) by the following function:
S = 10 L^0.4 C ^0.6
In this formula L represents the units of labor input and C the units of capital input. Each unit of labor costs $60, and each unit of capital costs $100.
P = 60*L + 100*C - (40,000 - 10L^0.30 C ^0.70)
dP/dL = 60 - 3 (C/L) ^0.70
dP/dC = 100 - 7(L/C)^0.30
dP/d = 40,000 - 10L^0.30 C ^0.70
Putting dP/dL = dP/dC = dP/d = 0
60/3(C/L) ^0.70 = 100/7(L/C)^0.30
L/C = (100/7*20
L = 0.714*C
4000 = (0.714)^0.4*C
C = 4000/(0.714)^0.4 = 4576.988
L = 0.714*4577 = 3267.97
(a) Formulate an optimization problem that will determine how much labor and capital are needed in order to produce 40,000 tons of steel at minimum cost. Min $60 L + $100 C Subject to L^0.30 C ^0.70 = 40,000/10 = 4000 L, C 0P = 60*L + 100*C - (40,000 - 10L^0.30 C ^0.70)
dP/dL = 60 - 3 (C/L) ^0.70
dP/dC = 100 - 7(L/C)^0.30
dP/d = 40,000 - 10L^0.30 C ^0.70
Putting dP/dL = dP/dC = dP/d = 0
60/3(C/L) ^0.70 = 100/7(L/C)^0.30
L/C = (100/7*20
L = 0.714*C
4000 = (0.714)^0.4*C
C = 4000/(0.714)^0.4 = 4576.988
L = 0.714*4577 = 3267.97
(b) Solve the optimization problem you formulated in part a. Hint: When using Excel Solver, start with an initial L > 0 and C > 0. If required, round your answers to two decimal places. L = $196078.2 C = $457698.87 Cost = $653777.07Related Questions
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