stuck on this multivariable calculus problem

Question

Suppose a Cobb-Douglas Production function is given by the following:

P(L,K)= 50L^(0.75) *K^(0.25)

where L is units of labor, K is units of capital, and P(L,K) is total units that can be produced with this labor/capital combination. Suppose each unit of labor costs $400 and each unit of capital costs $2,400. Further suppose a total of $576,000 is available to be invested in labor and capital (combined).

A) How many units of labor and capital should be "purchased" to maximize production subject to your budgetary constraint?

Units of labor, L = _____

Units of capital, K = ____

B) What is the maximum number of units of production under the given budgetary conditions? (Round your answer to the nearest whole unit.)

Max production = ___  units

Answer 1

A) The equation from budgetary constraint is

400L+ 2400 K=576000

=> L=1440 - 6K ................(1)

Substituting the expression of L in terms of K in Cobb-Douglas Production function,

P(K) = 50 (1440-6K)^0.75 K^0.25

Now, for maximum production, we need to find the derivative of production function w.r.t to K.

Using chain rule,

dp/dK = 50[0.75(-6)(1440-6K)^-0.25 K^0.25+0.25(1440-6K)^0.75 K^-0.75]

To find maxima,

dp/dK =0

=> 50[0.75(1440-6K)^-0.25 K^0.25)(-6)+0.25(1440-6K)^0.75 K^-0.75]=0

=> 0.75(1440-6K)^-0.25 K^0.25(6)=0.25(1440-6K)^0.75 K^-0.75

Multiplying the equation with (1440-6K)^0.25 K^0.75, we get

18K = 1440-6K

=> 24K=1440 => K= 60

From equation (1)

L= 1440-6(60) = 1080

So for maximum production 60 labor units and 1080 capital units should be purchased.

B) Putting value of L and K in Cobb-Douglas Production function, we get

P= 50 (1080)^0.75 60^0.25 = 26216.6 ≈26217

So max production is 26217 units.

N.B. One can also write K in terms of L and find the derivative w.r.t. to L in A. The final answer will be same.