Question

ONLY NEED ANSWERS FOR i AND j

To produce traps for capturing humans, the Coyote Cooperative requires both capital, K, and labour, L. Suppose that the production technology is given by by the production function q=20L^0.5K^0.5, where q is the number of traps, MPL =10L^-0.5K^0.5, and MPK=10L^0.5K^-0.5. a) What are the returns to scale for this production function? b) What is the equation of the Cooperative’s isoquant? c) What is the equation for a slope of its isoquant? d) What is this called? e) What does it indicate? f) Set up the cost minimization problem and solve for the conditional capital and labour demands as functions of w (the labour costs), r (the capital costs), and q (number of traps). g) What is the equation of the cooperative’s expansion path? Discuss your findings. h) Discuss the demand functions you derived in d).Are the production inputs normal inputs? What happens to the optimal amount of labour as w increases? What happens to the optimal amount of capital as r decreases

i) Now assume that the cost of the labour is $10 while cost of capital is $40. What is your "optimal production plan"?

j) Derive the cost function and mathematically simplify it as much as you can

Answer 1

q = 20L^0.5K^0.5

MPL = 10 x (K/L)^0.5

MPK = 10 x (L/K)^0.5

(i)

Production is optimal when MPL/MPK = w/r = 10/40 = 1/4

MPL/MPK = K/L = 1/4

L = 4K

Substituting in production function,

20 x (4K)^0.5K^0.5 = q

20 x 4^0.5 x K^0.5 x K^0.5 = q

20 x 2 x K = q

40K = q

K = q/40

L = 4 x (q/40) = q/10

(j)

MPL/MPK = K/L = w/r, therefore

K = wL/r = (w/r) x L

Substituting in production function,

20 x L^0.5(wL/r)^0.5 = q

20 x (w/r)^0.5 x L^0.5 x L^0.5 = q

20 x (w/r)^0.5 x L = q

L = (q/20) x (r/w)^0.5

K = (w/r) x (q/20) x (r/w)^0.5 = (q/20) x (w/r)^0.5

Total cost (C) = wL + rK = w x (q/20) x (r/w)^0.5 + r x (q/20) x (w/r)^0.5

C = [(q/20) x (wr)^0.5 + (q/20) x (wr)^0.5]

C = (q/20) x 2 x (wr)^0.5

C = (q/10) x (wr)^0.5

When w = 10 and r = 40 [as in part (i)],

C = (q/10) x (10 x 40)^0.5

C = (q/10) x (400)^0.5

C = (q/10) x 20

C = 2q