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?


1 Answer 1


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 $L$

  • $\begingroup$ 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
    Commented Oct 15, 2017 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. $\endgroup$ Commented Oct 15, 2017 at 21:46

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