# Show that the marginal cost of the total output equals the marginal cost of individual plant's outputs

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?

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

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 $$$$q_1^m(q)+q_2^m(q)=q$$$$ $$$$c_1'(q_1^m(q))=c_2'(q_2^m(q))$$$$

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

$$$$c_1'(\hat{q}_1)=c_2'(\hat{q}_2)=MR(\hat{q}_1+\hat{q}_2)$$$$

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

$$c_1(q_1)+c_2(q_2)+\lambda(q-q_1-q_2)$$

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

$$\frac{d}{dq}c(\hat{q})=c_1'(\hat{q}_1)=c_2'(\hat{q}_2)$$

• 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! Nov 3, 2022 at 18:39