Maximation of Cobb Douglas Function: Maximize and find total profit of 3aK^1/2 *L^3/4......?

Question

K= $3/unit of capital

L = $2/unit of capital

Sales price = $5/unit

I know to start by taking the 1st and 2nd order partial derivatives with respect to K and L but after that, I'm lost.

Thanks for any help! :o)

Answer 1

w - wages = $2

r - interest = $3

Q - output

Q=3aK^(1/2)*L^(3/4)

K^(1/2)=Q/(3aL^3/4)

K=(Q/(3aL^3/4))^2 = (Q/3a) ^2 / L^3/2

K= (Q/3a) ^2 / L^3/2

L^(3/4) =Q/(3aK^1/2)

L=(Q/(3aK^1/2))^4/3 = (Q/3a) ^4/3 / K^2/3

L=(Q/3a) ^4/3 / K^2/3

MPK=δQ/δK=(3a/2) * K^(-1/2) * L^(3/4) =

=3a * K^(1/2) * L^(3/4) / 2K = Q/2K

MPL=δQ/δL = (9a/4) * K^(1/2) * L^(-1/4) =

=3a * K^(1/2) * L^(3/4) *3/4L = 3Q/4L

Equilibrium condition:

MPL/w = MPK/r

MPL/2=MPK/3

MPL=3Q/4L

MPK=Q/2K

3Q/8L=Q/6K

8L/3=6K

4L/3=3K

K=4L/9

L=9K/4

K=(Q/3a) ^2 / L^3/2

K=4L/9

K=K

4L/9=(Q/3a) ^2 / L^3/2

L^(1+3/2)= (Q/3a) ^2 *9/4

L^(5/2)= (Q/2a) ^2

L= ((Q/2a) ^2)^2/5

L=(Q/2a)^4/5

L=(Q/3a) ^4/3 / K^2/3

L=9K/4

L=L

9K/4=(Q/3a) ^4/3 / K^2/3

K^(1+2/3)= (Q/3a) ^4/3 * 4/9

K^(5/3)= (Q/3a) ^4/3 * 4/9

K= ((Q/3a) ^4/3 * 4/9) ^ (3/5)

Or since K=4L/9 and L=(Q/2a)^4/5 then:

K=4/9 * (Q/2a)^4/5

TC= wL+rK

Profit=TR-TC

Maximum profit MR=MC

TR=Q*P=Q*5 = 5Q

MR=(TR)'=(5Q)'

MR=5

TC=wL+rK = 2L+3K =2* (Q/2a)^4/5 + 3* 4/9 * (Q/2a)^4/5 =

= 2* (Q/2a)^4/5 + 4/3 (Q/2a)^4/5 =

= 10/3 (Q/2a)^4/5

MC=(TC)'= 40/(15*(2a)^(4/5)*Q^1/5) =

=8/(3*(2a)^(4/5)*Q^1/5)

MC=MR

8/(3*(2a)^(4/5)*Q^1/5)=5

8/(15*(2a)^(4/5))=Q^1/5

Q= (8/(15*(2a)^(4/5)))^5

Q=(8/15)^5 / (2a)^4 =

=2048 / 759375*a^4

Profit=TR-TC=5Q- (10/3)*(Q/2a)^4/5

Q=2048 / 759375*a^4

Now substitute Q in profit formula and get maximal profit (depending on "a" parameter).

Answer 2

dF(K,L)/dK=3a 1/2 K^-1/2 L^3/4=Marginal Product of K

dF(K,L)/dL=3a K^1/2 L^-1/4=Marginal Product of L

Marginal Product of K=Price of K=3

Marginal Product of L=Price of L=2

You get two equations and two variables. You find K and L and plug their values in F(K,L) * 5=Total Income

Here we have got a constant return to scales as µ+ß>0

I think that´s the way to solve it.