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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.

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    $\begingroup$ Well, how much labor will you hire when $q < \bar{K}$? $\endgroup$ – Giskard Nov 22 '19 at 19:52
  • $\begingroup$ Hello! Labor would be 0 right? If so, if L=0 won't it be undefined because ln(0)=undefined? $\endgroup$ – susu Nov 22 '19 at 22:47
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    $\begingroup$ 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. $\endgroup$ – Amit Nov 23 '19 at 4:40
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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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