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Marginal costs MC is defined as $MC=\frac{dC}{dq}$. Taking into account that $C=wL+rK$,

$$MC=\frac{dC}{dq}=w\frac{dL}{dq}+r\frac{dK}{dq}$$

Recall that marginal product of labor $MP_{L}=\frac{\partial q}{\partial L}$ and marginal product of capital $MP_{K}=\frac{\partial q}{\partial K}$.

Question: is the following correct

$$\frac{dL}{dq}=1/\frac{\partial q}{\partial L},\;\frac{dK}{dq}=1/\frac{\partial q}{\partial K}$$

which implies

$$MC=w\frac{1}{MP_{L}}+r\frac{1}{MP_{K}}$$

If no, then no need to read further.

If yes, then, consider profit maximization of a firm.

$$\max_{L,K}pq\left(L,K\right)-wL-rK$$

FOC:

$$\begin{cases}pMP_{L}=w\\pMP_{K}=r\end{cases}\Rightarrow\begin{cases}MP_{L}=\frac{w}{p}\\MP_{K}=\frac{r}{p}\end{cases}$$

Therefore,

$$MC=w\frac{1}{MP_{L}}+r\frac{1}{MP_{K}}=w\frac{1}{w/p}+r\frac{1}{r/p}=p+p=2p$$

The result is wrong for sure. I wonder, at what step of derivation I made a mistake?

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  • $\begingroup$ What's your basis for thinking that dL/dq might equal 1/(dq/DL)? dL/dq is a derivative, not a fraction $\endgroup$ – EnergyNumbers Jun 15 '15 at 9:14
  • $\begingroup$ Perloff in his textbook on micro, uses what you just wrote when discussing short run. For unknown reason I voluntarily wrote in my notes more general version for long run, i.e. using partial derivatives. But after reading the chapter on demand on factors of production, I updated my notes and realized that there must be a mistake somewhere. $\endgroup$ – ji borrob Jun 15 '15 at 9:25
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    $\begingroup$ @EnergyNumbers, btw, the use of what you wrote is valid thanks to inverse function theorem. $\endgroup$ – ji borrob Jun 15 '15 at 9:27
  • $\begingroup$ Can you show the conditions for that, hold? (in particular, can you show that there are no zeros where there shouldn't be) $\endgroup$ – EnergyNumbers Jun 15 '15 at 11:11
  • $\begingroup$ (and be careful with partial derivatives - I think you want to take the inverse of the matrix, rather than of just an individual element in the matrix, don't you - sorry if this is a red-herring, it's been a while since I worked through this) $\endgroup$ – EnergyNumbers Jun 15 '15 at 11:19
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Question: is the following correct ?

$$\frac{dL}{dq}=1/\frac{\partial q}{\partial L},\;\frac{dK}{dq}=1/\frac{\partial q}{\partial K}$$

In general, no. Since $q= f(L,K)$ is a multivariable, single-valued function, then by the implicit function theorem applied on the implied equation $H = f(L,K)-q=0$, what we can say is only that

$$\frac {\partial L}{\partial q} = -\frac {\partial H/\partial q}{\partial H /\partial L} = -\frac {-1}{MP_L} = \frac {1}{\partial q /\partial L}$$

Essentially, here we treat $q$ as a univariate function (by taking partial derivatives, we keep all other variables fixed).

But total derivatives are another matter.
For $q = f(L,K)$, we have the implicit relation $L = L(q,K)$ and so by taking the total differential and divide throughout by $dq$ we get

$$\frac {d L(q,K)}{d q} = \frac {\partial L(q,K)}{\partial q} + \frac {\partial L(q,K)}{\partial K}\frac {dK}{dq}$$

But $\frac {\partial L}{\partial K} \neq 0$, because we implicitly keep $q$ constant in order to calculate this partial derivative, and so by changing $K$ we have also to change $L$. The other possibility would be that $dK/dq$ is zero. If we do not have such a production function (or we are not in a range where such a thing holds), we obtain

$$\frac {dL}{dq} \neq \frac {\partial L}{\partial q} = \frac {1}{\partial q/\partial L}$$

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It is valid but only in the short run with the assumption that the capital is fixed.

By assuming that output depends on labor and capital you can write

$$q=q(L,K)$$

Now taking the total derivative $$dq=\frac{\partial q}{\partial L}dL+\frac{\partial q}{\partial K}dK$$

In the short run capital is fixed such as $dK=0$ $$\longrightarrow\quad dq=\frac{\partial q}{\partial L}dL\longrightarrow \frac{dL}{dq}=1/\frac{\partial q}{\partial L}$$

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