Assume that the maxima/minima exists wherever referred (i.e., the necessary secondary conditions are satisfied). $p$ is the inverse-demand, $c_i(q_i)$ are the cost functions of the plants that a certain firm has, and $c(q)$ is the total cost function of the firm.

Suppose $q = q_1 + q_2$ and we have to maximize $\pi(q) = p(q)q - c_1(q_1) - c_2(q_2)$. The maximization holds at $(q_1, q_2, q)$ which satisfies MR$(q)$ = MC$_1(q_1)$ = MC$_2(q_2)$ such that $q = q_1 + q_2$.

I conjectured that at the point of optima, $c'(q) = c_1'(q_1) = c_2(q_2)$ holds where $c(q) = \min[c_1(q_1) + c_2(q_2)] \text{ subject to } q = q_1+q_2$. In other words, MC$(q)$ = MC$_1(q_1)$ = MC$_2(q_2)$ at the optimal point.

How do I prove this?

  • $\begingroup$ Are you good with derivatives and maximization? $\endgroup$
    – Giskard
    Nov 3, 2022 at 9:15
  • $\begingroup$ @Giskard Yes, I think I'll be able to follow your explanation if you provide one. $\endgroup$
    – Harris
    Nov 3, 2022 at 11:59

1 Answer 1


Consider $\min_{q_1,q_2} c_1(q_1)+c_2(q_2)$ st $q_1+q_2=q$. The solution to this problem is $(q_1^m(q),q_2^m(q))$ and they satisfy \begin{equation} q_1^m(q)+q_2^m(q)=q \end{equation} \begin{equation} c_1'(q_1^m(q))=c_2'(q_2^m(q)) \end{equation}

Consider $\max_{q_1,q_2}p(q)q-c_1(q_1)-q_2(q_2)$

Let $\hat{q}_1$ and $\hat{1}_2$ be the solution. We have

\begin{equation} c_1'(\hat{q}_1)=c_2'(\hat{q}_2)=MR(\hat{q}_1+\hat{q}_2) \end{equation}

Let $\hat{q}:=\hat{q}_1+\hat{q}_2$. What is $(q_1^m(\hat{q}),q_2^m(\hat{q}))$? I claim that it is $(\hat{q}_1,\hat{q}_2)$ to see this, plug $q=\hat{q}$ and $q_1^m=\hat{q}_1$ and $q_2^m=\hat{q}_2$ into the first two equations. They are satisfied. Therefore, $(q_1^m(\hat{q}),q_2^m(\hat{q}))=(\hat{q}_1,\hat{q}_2)$.

The last part of the question, is what is $\frac{d}{dq}c(q)$? This is just a straightforward application of the Envelope Theorem. The Lagrangian of the first problem is


The envelope theorem says that $\frac{d}{dq}c(q)=\lambda$. In addition, $c_1'(q_1^m(q))=c_2'(q_2^m(q))=\lambda$. Therefore, $\frac{d}{dq}c(q)=c_1'(q_1^m(q))=c_2'(q_2^m(q))$. Finally, with $q=\hat{q}$ we have


  • $\begingroup$ I have encountered the envelope theorem earlier but I haven't read it. I will do so in order to understand the proof. Thank you for the effort! $\endgroup$
    – Harris
    Nov 3, 2022 at 18:39

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