<p>The long run is defined as the time period in which all the inputs can be changed as per the firm's desire i.e., for a production function of the form <span class="math-container">$f(K,L)$</span> both <span class="math-container">$K$</span> and <span class="math-container">$L$</span> are variable in determining output.</p>
<p>The long-run cost minimization problem is:
<span class="math-container">$$\begin{align}
\min_{K,L} \quad &amp;wL+rK \\
\textrm{s.t.} \quad&amp; \min(3K,2L)\geq \bar Y 
\end{align}$$</span></p>
<p>The constraint binds at optimum and solving the optimization problem yields conditional input demands <span class="math-container">$K^d(w,r,\bar Y)=\frac{\bar Y}{3}$</span> and <span class="math-container">$L^d(w,r,\bar Y)=\frac{\bar Y}{2}$</span>. Therefore,
LR Cost function is: <span class="math-container">$C(w,r,\bar Y)=\bar Y(\frac{w}{2}+\frac{r}{3})$</span></p>
<p>The short run is defined as that time period in which one or more inputs are fixed.</p>
<p>The short-run cost minimization problem is:
<span class="math-container">$$\begin{align}
\min_{K} \quad &amp; w \bar L+r K \\
\textrm{s.t.} \quad &amp; \min(3K,2 \bar L)\geq\bar Y \\ 
\end{align}$$</span></p>
<p>for the constraint <span class="math-container">$\min(3K,2\bar L)\geq \bar Y$</span> to hold we need <span class="math-container">$3K\geq \bar Y$</span> and <span class="math-container">$2\bar L\geq \bar Y$</span></p>
<p>therefore we can rewrite the problem as:
<span class="math-container">$$\begin{align}
\min_{K \geq \frac{\bar{Y}}{3}} \quad &amp; w \bar L+r K \\
\end{align}$$</span></p>
<p>notice that <span class="math-container">$w\bar L+rK$</span> is increasing in <span class="math-container">$K$</span> therefore in order to solve the above problem we set <span class="math-container">$K$</span> to the lowest value it can take.<br />
Thus, <span class="math-container">$K^d_{SR}(w,r,\bar L, \bar Y)=\frac{\bar Y}{3}$</span>
and  consequently SR cost function is: <span class="math-container">$C_{SR}(w,r,\bar L, \bar Y)=w\bar L+r\frac{\bar Y}{3}$</span></p>

CC BY-SA 4.0
2023-05-24T10:27:34.980Z
In case we were given that $\bar L<\frac{\bar Y}{2}$ in that case to minimize cost, we would set $3K=2\bar L <\bar Y$ i.e., $K_{SR}^d(w,r,\bar L, \bar Y)=\frac{2\bar L}{3}$. So, I think a more complete solution to the SR cost minimization problem is: $K_{SR}^d(w,r,\bar L, \bar Y)=\begin{cases} \frac{\bar Y}{3} & \text{if } \bar L>\frac{\bar Y}{2} \\ \frac{2 \bar L}{3} & \text{if } \bar L\leq \frac{\bar Y}{2} \end{cases}$ but I'm not sure how correct this is, so decided to include this in the comments
mynameparv
CC BY-SA 4.0
2023-05-29T19:21:55.750Z
Thank you! I'm also not sure about the second part but thanks anyway!
Debbie
CC BY-SA 4.0
2023-05-24T10:01:22.497Z
2023-05-24T10:01:22.497Z
mynameparv
<p>Since <span class="math-container">$L$</span> is fixed at some <span class="math-container">$\overline{L} &gt; \frac{\overline{Y}}{2}$</span>, then <span class="math-container">$\overline{Y} &lt; 2L$</span>.</p>
<p>This implies the <span class="math-container">$\min$</span> term equals <span class="math-container">$3K$</span>.</p>
<p>Therefore, <span class="math-container">$f(K,\overline{L}) = 3K$</span>.</p>
<p>With this, the problem we want to solve is</p>
<p><span class="math-container">$\min_K w\overline{L} + rK$</span></p>
<p>s.t. <span class="math-container">$3K \geq \overline{Y}$</span></p>
<p>Solving for <span class="math-container">$K$</span> we get <span class="math-container">$K \geq \frac{\overline{Y}}{3}$</span></p>
<p>Clearly the objective function is minimized at equality, so we get <span class="math-container">$K = \frac{\overline{Y}}{3}$</span></p>
<p>Plugging into the objective function, the short run minimum cost function is</p>
<p><span class="math-container">$C(r,w,\overline{Y};\overline{L}) = w \overline{L} + \frac{r \overline{Y}}{3}$</span></p>

CC BY-SA 4.0
2023-05-23T13:37:47.687Z
2023-05-23T13:44:34.220Z
2023-05-23T13:44:34.220Z
Nicolas Torres
<p>Given the production function <span class="math-container">$f(K, L)=\min\{3K,2L\}$</span>, the procedure to find the long-run cost function would be to use the condition: <span class="math-container">$3K=2L=Y$</span> where <span class="math-container">$K=\frac{\overline{Y}}{3}$</span> and <span class="math-container">$L=\frac{\overline{Y}}{2}$</span>.
<span class="math-container">$K$</span> is units of capital; <span class="math-container">$L$</span> is units of labor; <span class="math-container">$Y$</span> is units of output.</p>
<p>Now, I am being asked to find the short-run cost function where labor is fixed at <span class="math-container">$L&gt;\frac{\overline{Y}}{2}$</span>. I am not sure how to approach the problem in this case.</p>

CC BY-SA 4.0
2023-05-23T13:39:42.280Z
Since you say labor is fixed, I guess you need the short run cost function where $K$ is the factor that represents the variable cost, while $L$ representing the fixed cost.
Nicolas Torres
CC BY-SA 4.0
2023-05-29T20:20:17.747Z
Don't forget you can accept an answer by clicking on the checkmark if you found it helpful!
Nicolas Torres
<microeconomics><optimization><producer-theory><cost-functions>
How to derive the short run cost function
Debbie
# How to derive the short run cost function

(Asked by: username_0 on 2023-05-20)

Given the production function $f(K, L)=\min\{3K,2L\}$, the procedure to find the long-run cost function would be to use the condition: $3K=2L=Y$ where $K=\frac{\overline{Y}}{3}$ and $L=\frac{\overline{Y}}{2}$.
$K$ is units of capital; $L$ is units of labor; $Y$ is units of output.

Now, I am being asked to find the short-run cost function where labor is fixed at $L>\frac{\overline{Y}}{2}$. I am not sure how to approach the problem in this case.

**Tags:** microeconomics, optimization, producer-theory, cost-functions
> **Comments:**
> **username_1**: Since you say labor is fixed, I guess you need the short run cost function where $K$ is the factor that represents the variable cost, while $L$ representing the fixed cost.  
> **username_1**: Don't forget you can accept an answer by clicking on the checkmark if you found it helpful!  

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## 2 answers

### Answer by username_2 on 2023-05-24. Score: 0 | Accepted answer

The long run is defined as the time period in which all the inputs can be changed as per the firm's desire i.e., for a production function of the form $f(K,L)$ both $K$ and $L$ are variable in determining output.

The long-run cost minimization problem is:
$$\begin{align}
\min\_{K,L} \quad &wL+rK \\
\textrm{s.t.} \quad& \min(3K,2L)\geq \bar Y
\end{align}$$

The constraint binds at optimum and solving the optimization problem yields conditional input demands $K^d(w,r,\bar Y)=\frac{\bar Y}{3}$ and $L^d(w,r,\bar Y)=\frac{\bar Y}{2}$. Therefore,
LR Cost function is: $C(w,r,\bar Y)=\bar Y(\frac{w}{2}+\frac{r}{3})$

The short run is defined as that time period in which one or more inputs are fixed.

The short-run cost minimization problem is:
$$\begin{align}
\min\_{K} \quad & w \bar L+r K \\
\textrm{s.t.} \quad & \min(3K,2 \bar L)\geq\bar Y \\
\end{align}$$

for the constraint $\min(3K,2\bar L)\geq \bar Y$ to hold we need $3K\geq \bar Y$ and $2\bar L\geq \bar Y$

therefore we can rewrite the problem as:
$$\begin{align}
\min\_{K \geq \frac{\bar{Y}}{3}} \quad & w \bar L+r K \\
\end{align}$$

notice that $w\bar L+rK$ is increasing in $K$ therefore in order to solve the above problem we set $K$ to the lowest value it can take.  
Thus, $K^d\_{SR}(w,r,\bar L, \bar Y)=\frac{\bar Y}{3}$
and consequently SR cost function is: $C\_{SR}(w,r,\bar L, \bar Y)=w\bar L+r\frac{\bar Y}{3}$

> **Comments:**
> **username_2**: In case we were given that $\bar L<\frac{\bar Y}{2}$ in that case to minimize cost, we would set $3K=2\bar L <\bar Y$ i.e., $K_{SR}^d(w,r,\bar L, \bar Y)=\frac{2\bar L}{3}$. So, I think a more complete solution to the SR cost minimization problem is: $K_{SR}^d(w,r,\bar L, \bar Y)=\begin{cases} \frac{\bar Y}{3} & \text{if } \bar L>\frac{\bar Y}{2} \\ \frac{2 \bar L}{3} & \text{if } \bar L\leq \frac{\bar Y}{2} \end{cases}$ but I'm not sure how correct this is, so decided to include this in the comments  
> **username_0**: Thank you! I'm also not sure about the second part but thanks anyway!  

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### Answer by username_1 on 2023-05-23. Score: 1

Since $L$ is fixed at some $\overline{L} > \frac{\overline{Y}}{2}$, then $\overline{Y} < 2L$.

This implies the $\min$ term equals $3K$.

Therefore, $f(K,\overline{L}) = 3K$.

With this, the problem we want to solve is

$\min\_K w\overline{L} + rK$

s.t. $3K \geq \overline{Y}$

Solving for $K$ we get $K \geq \frac{\overline{Y}}{3}$

Clearly the objective function is minimized at equality, so we get $K = \frac{\overline{Y}}{3}$

Plugging into the objective function, the short run minimum cost function is

$C(r,w,\overline{Y};\overline{L}) = w \overline{L} + \frac{r \overline{Y}}{3}$


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