Answer to Question #281858 in Microeconomics for kalehiwot

Question

5. Suppose you have the following production function: Q = f (L, K) = 10L ^½ K ^½ . In addition, the price of labor is $1 and the price of capital is $4

a. What is the optimal amount of labor and capital if you want to produce 20 units?

b. What is the level of minimum cost ?( Ans L=4 and K=1,Min C=$8)

6. Suppose the short run production function can be represented by Q = 60,000L ² – 1000L ³ . Then, determine

a. The level labor employment that maximizes the level of output

b. The level of employment that maximizes AP_L and the maximum AP_L

Answer 1

Solution:

a.). Cost minimization:

MRTS = $\frac{MP_{L} }{MP_{K}} = \frac{w }{r}$

TC = 10L ^½ K^½

MP_L = $\frac{\partial Q} {\partial L}$ = 5L^-0.5 K^0.5

MP_K = $\frac{\partial Q} {\partial K}$ = 5L^0.5 K^-0.5

$\frac{MP_{L} }{MP_{K}} = \frac{w }{r}$

w = 20

r = 80

$\frac{5L^{-0.5} K^{0.5} }{5L^{0.5} K^{-0.5}} = \frac{20 }{80}$

$\frac{K }{L} = 0.25$

K = 0.25L

Q = 10L ^½ K^½

20 = 10(L^0.5) (0.25L^0.5)

L = 8

K = 0.25L = 0.25(8) = 2

The optimal amount of labor and capital to produce 2000 units (L,K) = (8, 2)

Labor = 8

Capital = 2

b.). Minimum cost:

C = wL + rK

C = (20 $\times$ 8) + (80 $\times$ 2) = 160 + 160 = 320

Minimum cost = 320

b.). The level of employment that maximizes AP_L and the maximum AP_L:

AP_L = $\frac{Q}{L}$ = 60,000L – 1,000L²

Maximize AP_L:

$\frac{\partial AP_{L} } {\partial L}$ = 60,000 – 2,000L

Set AP_L = 0

60,000 – 2,000L = 0

60,000 = 2,000L

L = 30

The level of employment that maximizes AP_L = 30 units

AP_L = 60,000L – 1,000L² = 60,000(30) – 1000(30²) = 1,800,000 – 900,000 = 900,000

AP_L = 900,000