3
$\begingroup$

I'm a bit unsure about how to derive a long-run cost function. Suppose my production function was $X(L, K)=L^a K^b$, where $a+b>1$.

I'm thinking about doing the following, but I'm not sure it's correct. If it's not, what am I not considering?


The beginning of my solution:

Our production function is $X=L^a K^b$ and our cost equation is $C=wL+rK$. So, we must solve $\max L^a K^b \text{ s.t. } C=wL+rK$. Therefore, our Lagrangian function is $\mathcal{L}=L^a K^b + \lambda(C-wL-rK)$.

The first order conditions are: (1) $aL^{a-1}K^b-\lambda w=0$, (2) $bL^aK^{b-1} -\lambda r=0$, and (3) $C-wL-rK=0$.

Then, we solve the maximization problem (dividing conditions 1 and 2, solving for $L$ and $K$, and then plugging $L$ and $K$ into the cost equation). Does this seem like the right way to do it?

$\endgroup$
3
$\begingroup$

You're right. Divide Eq (1) by Eq (2):

$$ \frac{a L^{a-1}K^b}{bL^aK^{b-1}} = \frac{aK}{bL} = \frac{w}{r} ~~~\Rightarrow~~~ L = \frac{ar}{bw}K \tag{4} $$

Now use this in Eq. (3)

$$ C = wL + rK = \left(\frac{a}{b} + 1\right)rK ~~~\Rightarrow~~~ K = \frac{C}{r(a/b + 1)} \tag{5} $$

Replace this in Eq. (4) to get $L$

$\endgroup$
  • $\begingroup$ Thank you for the help! So I plugged that into (4) to get $L=\displaystyle\frac{a C}{w(a + b)}$, but I'm not exactly sure where to go from there. Should I take $K$ from (5) and the $L$ I just found and plug them in somewhere? I'm not sure how to derive the cost function, since $C$ is a variable in both of the expressions I found for $L$ and $K$. $\endgroup$ – pril Oct 15 '17 at 18:43
  • $\begingroup$ @pril The cost function is a function of prices and the level of output. Under the optimal conditions you can express output as a function of $K$ only. $\endgroup$ – Alecos Papadopoulos Oct 15 '17 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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