<p>You're right. Divide Eq (1) by Eq (2):</p>

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

<p>Now use this in Eq. (3)</p>

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

<p>Replace this in Eq. (4) to get $L$</p>

CC BY-SA 3.0
2017-10-15T18:43:15.863Z
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 $K$.
pril
CC BY-SA 3.0
2017-10-15T21:46:55.403Z
@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.
Alecos Papadopoulos
CC BY-SA 3.0
2017-10-15T03:42:16.017Z
2017-10-15T03:42:16.017Z
caverac
<p>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 $a+b&gt;1$.</p>

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

<hr>

<p><strong>The beginning of my solution</strong>:</p>

<p>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 $\mathcal{L}=L^a K^b + \lambda(C-wL-rK)$.</p>

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

<p>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?</p>

<production-function><self-study><cost-functions>
Deriving long-run cost function
pril
# Deriving long-run cost function

(Asked by: username_0 on 2017-10-15)

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?

**Tags:** production-function, self-study, cost-functions

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## 1 answers

### Answer by username_1 on 2017-10-15. Score: 4

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 $L$

> **Comments:**
> **username_0**: 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 $K$.  
> **username_2**: @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.  

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