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I have derived the following first order condition

$$g(x,C'(x),..)=1,$$

where $x$ is the production level and $C(x)$ the cost of $x$, and $C'(x)$ is the first derivative of $C(x)$ (i.e. the marginal cost). By the implicit function theorem, we can derive how production changes with marginal cost:

$$\frac{\partial x}{\partial C'(x)}=-\frac{\partial g/\partial C'(x)}{\partial g/\partial x}$$

Where, the numerator on the RHS equals

$$\partial g/\partial C'(x)=\frac{\partial g/\partial x}{d C'(x)}$$

This then allows us to simplify the second equation

$$-\frac{\frac{\partial g/\partial x}{d C'(x)}}{\partial g/\partial x}=-\frac{1}{C''(x)}$$

And we then conclude that $\frac{\partial x}{\partial C'(x)}=-\frac{1}{C''(x)}$. This is clearly absurd, since with a concave cost function, the derivative of $x$ with respect to marginal cost will always be positive (i.e. production is increasing in marginal cost). What mistake did I make to derive something this absurd?

The step I think may be problematic is the third equation $$\partial g/\partial C'(x)=\frac{\partial g/\partial x}{d C'(x)}.$$ I asked about this elsewhere, and the step seemed fine to some.

pafnuti
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    I don't think $\partial x / \partial C'(x)$ makes sense by any convention I'm aware of for interpreting the symbols. Recall that weird notation like $\partial g / \partial C'(x)$ is meant to be read as "the derivative of $g$ in its second place", and that reading is meant to be inferred from the fact that $g$ is being used as shorthand for $g(x, C'(x), \ldots)$, and $C'(x)$ is the formula appearing in the second plane. –  Jan 22 '18 at 00:24
  • What are the arguments of $g(.)$? Normally the arguments of a production function are the inputs of factors of production.. For example $f(K,L)=1$ is one way of defining the unit isoquant. If you're defining a cost function, then $C(q,w,r)$ would be the minimum cost of producing $q$ units of output when factor prices are $w$ and $r$. Please clarify what you're writing down. – Eric Fisher Jan 22 '18 at 00:24
  • @EricFisher $g()$ is the function that specifies the first order condition for profit maximisation. Here $C(x)$ is some function that defines the cost of producing $x$ units. – pafnuti Jan 22 '18 at 00:27
  • Are factor prices considered fixed? – Eric Fisher Jan 22 '18 at 00:29
  • @EricFisher not necessarily. Cost is simply given by some arbitrary function $C: \mathbb{R_+} \rightarrow \mathbb{R_+}$. – pafnuti Jan 22 '18 at 00:30
  • @Hurkyl Thanks for your comment. Not sure if I agree that the derivative makes no sense. I believe it makes mathematical sense. See for example https://math.stackexchange.com/questions/954073/derivative-of-a-function-with-respect-to-another-function – pafnuti Jan 22 '18 at 00:32
  • Your question doesn't make sense. Here's a simple example. Let the production function be $Q=2L$. Let wages be given as $w$. Then the cost function is $C(q;w)=2wq$. So marginal cost is $2w$. In this case, what's your $g(.)$? – Eric Fisher Jan 22 '18 at 00:32
  • I think the foc would be $2/2w=1$. But an example for which a FOC might not work does not show that my FOC is nonsensical. – pafnuti Jan 22 '18 at 00:35
  • @pafnuti: The post you link is about differentials, which are much better behaved algebraically than the usual notation for partial derivatives. –  Jan 22 '18 at 11:48

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Your problem comes from assuming that the cost function is concave.

With perfectly competitive markets, the firm’s profit maximisation problem cannot be solved using first-order conditions. With concave costs, the second derivative of the profit function will be convex, so you can’t solve it via first-order conditions. Attempting to do so gives you a solution that minimises profits. (This might also explain your nonsensical result.)

Perhaps you intended to assume that production costs are convex. This would give you the intuitive result that output is decreasing in marginal cost.

Theoretical Economist
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