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 = 15 L0.2 C 0.8 In this formula L represents the units of labor input and C the units of capital input. Each unit of labor costs $50, 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 50,000 tons of steel at minimum cost. Min L + C L0.2 C 0.8 L, C (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 = $

Answer 1

Solution

S = 15 x $L^{0.2}$ x $C^{0.8}$

Total cost, T = wL + rC = 50L + 100C

Total revenue, R = Output price (P) x Quantity = P x 15 x $L^{0.2}$x $C^{0.8}$

(a)

Optimization problem will be:

Max R = P x 15 x $L^{0.2}$ x $C^{0.8}$

Subject to T = 50L + 100C

(b) When S = 50,000

Cost is minimized when (MPL / MPC) = w / r

MPL = $\partial$R / $\partial$L = P x 15 x 0.2 x $(C / L)^{0.8}$ = P x 3 x $(C / L)^{0.8}$

MPC = $\partial$R / $\partial$C = P x 15 x 0.8 x $(L / C)^{0.2}$ = P x 12 x $(L / C)^{0.2}$

MPL / MPC = (3/12) x (C / L) = 50/100

C / 4L = 1/2

4L = 2C

2L = C

Substituting in production function,

15 x $L^{0.2}$ x $C^{0.8}$ = S

15 x$L^{0.2}$ x $(2L)^{0.8}$ = 50,000

15 x $2^{0.8}$ x $L^{0.2}$ x $L^{0.8}$ = 50,000

L = 50,000 / (15 x 20.8)

L = 1,914.50

C = 2L = 3,829.00

Total cost ($) = 50 x 1,914.50 + 100 x 3,829.00 = 95,725.00 + 382,900 = 478,625.00

Note: This optimization problem can be solved without using Solver too, as shown here.