<p>I'm just gonna try doing it. The problem is:</p>

<p><span class="math-container">$$\text{min } C = rK + wL \\
\text{s.t. } q = K + \ln(L) $$</span></p>

<p>The implied constraint, of course, is <span class="math-container">$L \in [1, \infty)$</span> (non-negativity constraint, as <span class="math-container">$q &gt; 0$</span>, and <span class="math-container">$K = 0$</span> is, of course, possible).</p>

<p>We can try solving this using Lagrange multipliers. Define the Lagrangian:</p>

<p><span class="math-container">$$\mathcal{L} = (rK + wL) + \lambda(q - K - \ln(L))$$</span></p>

<p>The first order conditions are:</p>

<p><span class="math-container">$$\mathcal{L}_{K} = r - \lambda = 0$$</span>
<span class="math-container">$$\mathcal{L}_{L} = w - \lambda(\frac{1}{L}) = 0$$</span>
<span class="math-container">$$\mathcal{L}_{\lambda} = q - K - \ln(L) = 0$$</span></p>

<p>Using some algebra, we get <span class="math-container">$r = wL$</span> and <span class="math-container">$L = e^{q-K}$</span>. Substituting into the original cost equation:</p>

<p><span class="math-container">$$C = K(we^{q-K}) + we^{q-K} = we^{q-K}(K + 1) $$</span></p>

<p>I don’t think this can be simplified much further.</p>

<p>That seems to be the same answer that you got (I just expressed it in terms of wages and capital). So not sure – maybe I’m making a mistake. </p>

CC BY-SA 4.0
2019-12-24T16:33:22.070Z
2019-12-24T16:33:22.070Z
Tejas Subramaniam
<p>I am confused about the Short Run Total Cost function of this problem. </p>

<p>If the firm's production function is <span class="math-container">$F(K,L) = K+\ln(L)$</span> derive the short run total cost function. </p>

<p>I was able to solve the Short run total cost function, but my TA had 2 cases for her function whereas I only had 1 part of her function.</p>

<p>I got <span class="math-container">$TC= W*e^{q-K} + r\bar{K}$</span> for all <span class="math-container">$q$</span> but my TA had 2 cases for <span class="math-container">$q$</span>. Specifically, when <span class="math-container">$q&lt;\bar{K}$</span> and <span class="math-container">$q&gt;\bar{K}$</span></p>

<p>Thank you.</p>

CC BY-SA 4.0
2019-11-22T19:52:10.690Z
Well, how much labor will you hire when $q < \bar{K}$?
Giskard
CC BY-SA 4.0
2019-11-22T22:47:22.223Z
Hello! Labor would be 0 right? If so, if L=0 won't it be undefined because ln(0)=undefined?
susu
CC BY-SA 4.0
2019-11-23T04:40:48.780Z
You're right, this production function does not make much sense because $F(K = 0,L) < 0$ for $L < 1$. You can  have an additional constraint that at least one unit of labor must be hired i.e. $L\geq 1$ to get away with this problem.
Amit
<microeconomics><cost-functions>
Short Run Total Cost function Question
susu
# Short Run Total Cost function Question

(Asked by: username_0 on 2019-11-22)

I am confused about the Short Run Total Cost function of this problem.

If the firm's production function is $F(K,L) = K+\ln(L)$ derive the short run total cost function.

I was able to solve the Short run total cost function, but my TA had 2 cases for her function whereas I only had 1 part of her function.

I got $TC= W\*e^{q-K} + r\bar{K}$ for all $q$ but my TA had 2 cases for $q$. Specifically, when $q<\bar{K}$ and $q>\bar{K}$

Thank you.

**Tags:** microeconomics, cost-functions

> **Comments:**
> **username_1**: Well, how much labor will you hire when $q < \bar{K}$?  
> **username_0**: Hello! Labor would be 0 right? If so, if L=0 won't it be undefined because ln(0)=undefined?  
> **username_2**: You're right, this production function does not make much sense because $F(K = 0,L) < 0$ for $L < 1$. You can  have an additional constraint that at least one unit of labor must be hired i.e. $L\geq 1$ to get away with this problem.  

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## 1 answers

### Answer by username_3 on 2019-12-24. Score: 1

I'm just gonna try doing it. The problem is:

$$\text{min } C = rK + wL \\
\text{s.t. } q = K + \ln(L) $$

The implied constraint, of course, is $L \in [1, \infty)$ (non-negativity constraint, as $q > 0$, and $K = 0$ is, of course, possible).

We can try solving this using Lagrange multipliers. Define the Lagrangian:

$$\mathcal{L} = (rK + wL) + \lambda(q - K - \ln(L))$$

The first order conditions are:

$$\mathcal{L}\_{K} = r - \lambda = 0$$
$$\mathcal{L}\_{L} = w - \lambda(\frac{1}{L}) = 0$$
$$\mathcal{L}\_{\lambda} = q - K - \ln(L) = 0$$

Using some algebra, we get $r = wL$ and $L = e^{q-K}$. Substituting into the original cost equation:

$$C = K(we^{q-K}) + we^{q-K} = we^{q-K}(K + 1) $$

I don’t think this can be simplified much further.

That seems to be the same answer that you got (I just expressed it in terms of wages and capital). So not sure – maybe I’m making a mistake.


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