Cobb-Douglas Calculus Problem!

Question

A company has 6000 dollars to invest, which must be divided between capital expenditures and labor. Each unit of labor costs the company 20 dollars, and each unit of capital costs 20 dollars. Therefore

20L+40K = 6000 ,

where L and K are the units of labor and capital, respectively. Production output is modeled by the Cobb-Douglas formula P=L^(0.1)K^(0.9)

Find the values of L and K which maximize production

L =

K =

Answer 1

This is an optimization problem!

The first thing is to calculate either L in terms of K (you can also do it K in terms of L).

20L = 6000 - 40K

L(K) = 300 - 2K

Now we substitute L(K) into P function.

P (K) = (300 - 2K)^0.1 * K^0.9

Next we take the derivate of P with respect to K

P' = (0.9 (300 - 2 K)^0.1)/K^0.1 - (0.2 K^0.9)/(300 - 2 K)^0.9

Now we set the P' function equal to zero (this is the maximum value of P(K) function, slope is zero)

(0.9 (300 - 2 K)^0.1)/K^0.1 - (0.2 K^0.9)/(300 - 2 K)^0.9 = 0

Solve for K.

K = 135

Then we take our original L(K) function to solve L

L = 300 - 2*135

L = 30