I'm trying to maximize a firm's profit given the production function $F(L,K)=L^\alpha K^\beta$ (where $L$ is labor and $K$ is capital) and that $\alpha + \beta \neq 1$.


So, I know that this maximization problem can be written as $\text{max }pF(L,K)-w_1 L-w_2 K$.

Since $pMP_L (L^*,K^*)=w_1$, $p\alpha(L^*)^{\alpha-1}(K^*)^\beta=w_1$. And since $pMP_K (L^*,K^*)=w_2$, $p\beta(L^*)^{\alpha}(K^*)^{\beta-1}=w_2$.

By dividing these functions and simplifying, we get $\displaystyle\frac{\alpha K^*}{\beta L^*}=\displaystyle\frac{w_1}{w_2}$.

I'm unsure how to proceed from here, though. Should I solve for $L^*$ by separating $K^*$ from the equation and plugging into $pMP_L$? Wouldn't this yield a very complicated solution?

  • How about actually maximizing with respect to the decision variables? (I.e. take the derivatives, set to equal zero, etc.) – denesp Oct 8 '17 at 20:13
  • the system you are trying to solve has two equations namely the derivatives of the profit function wrt to each of the inputs; what you have obtained above is simply the result of dividing your first equation with your second equation; you still have one more equation you are not taking advantage of... – user14471 Oct 8 '17 at 20:27
  • @user43282 Ah, I see that I'm not taking advantage of the $pF(L,K)-w_1 L - w_2 K$. Is that what you meant to point out? If so, would it make sense to set up a Lagrange function to solve this? – pril Oct 8 '17 at 21:49
  • @pril: you are not making use of the second equation you have listed above; setting up a Lagrangian would imply that your optimization problem is constrained, which doesn't seem to be the case here unless you are solving something like max $p F(K,L)-w_1L-w_2K$ s.t. $F(K,L)=c$ – user14471 Oct 9 '17 at 6:19
  • 2
    I'm voting to close this question as off-topic because based on the comments the OPs problems are with math, not economics. – denesp Oct 9 '17 at 6:20

Should I solve for $L^∗$ by separating $K^∗$ from the equation and plugging into $pMP_{L}$

Yep, that's about it.

Wouldn't this yield a very complicated solution?

Somewhat. The math is available at many places, like section 4 here. But you can surely do it yourself!

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.