Question

7. Suppose the hourly wage is S10, the price of each unit of capital is S20, and the price of output is S60 per unit. Assume that the firm cannot affect any of these prices. The production function of the firm is a. Find the MPE (4 pts) b. If the firm is maximizing profits, what must the ratio of MPE to MPK be? What must the ratio of E to K be? Hint MPK-dQ/dK 1/3K-23E13 (2 points) c. Find the FOCE. Use the FOCE and the ratio of E to K to find the optimal level of E. Then find the optimal level of K. (3 points)

Answer 1

Given, $\LARGE Q = E^{\frac{1}{3}}K^{\frac{1}{3}}$

Ce = $ 10 per hour, Ck = $ 20, price of output = $ 60

1. Determine MPE

Differentiating Q wrt E we get

$\LARGE \frac{dQ}{dE} = \frac{d(E^{\frac{1}{3}}K^{\frac{1}{3}})}{dE}$

$\LARGE MP_E = \frac{1}{3}E^{\frac{1}{3}-1}K^{\frac{1}{3}}$

$\LARGE MP_E = \frac{1}{3}*\frac{E^{\frac{1}{3}}K^{\frac{1}{3}}}{E}$

$\LARGE MP_E = \frac{Q}{3E}$

Similarly we can deter mine the MPk

$\LARGE MP_K = \frac{Q}{3K}$

The firm can maximize profit by satisfying the equimarginal criterion

$\LARGE \frac{MP_E}{MP_K }= \frac{C_E}{C_K}$

Plug in the values we get

$\LARGE \frac{\frac{Q}{3E}}{\frac{Q}{3K}}= \frac{10}{20}$

$\large \frac{K}{E}= \frac{1}{2}$

K = 0.5E

Or, E = 2K

C. The first order condition is to equate the MC to P

In a competitive market MC = P

TC = 10E + 20K

TC = 20E

OR, TC = 40K

Differentiating above equation wrt Q we get

$\LARGE MC = 40\frac{dK}{dQ}$

$\LARGE MC = \frac{40}{MP_K}$

$\LARGE MC = 40*\frac{3K}{Q}$

$\LARGE MC = \frac{120K}{E^{\frac{1}{3}}K^{\frac{1}{3}}}$

Since, E = 2K

$\LARGE MC = \frac{120K^{\frac{2}{3}}}{E^{\frac{1}{3}}}$

$\LARGE MC = \frac{120K^{\frac{2}{3}}}{(2K)^{\frac{1}{3}}}$

$\LARGE MC = \frac{120K^{\frac{1}{3}}}{2^{\frac{1}{3}}}$

Now equate MC to price

$\LARGE \frac{120K^{\frac{1}{3}}}{2^{\frac{1}{3}}} = 60$

$\LARGE \frac{2K^{\frac{1}{3}}}{2^{\frac{1}{3}}} = 1$

$\LARGE K^{\frac{1}{3}}= \frac{2^{\frac{1}{3}}} {2}$

$\LARGE K^{\frac{1}{3}}= 2^{\frac{1}{3}-1}$

$\LARGE K^{\frac{1}{3}}= 2^{\frac{-2}{3}}$

$\LARGE K= 2^{-2} = 0.25$

E = 0.5

Please contact if having any query thank you.