Why is MC = ATC the same point for both the breakeven point and an investor maximizing return?

Question

Let $\pi(y) = R(y) - C(y)$ be profits where $R(y)$ is revenue and $C(y)$ is costs. Let $R(y) = p_y y$. Then

\begin{align*} \frac{\partial \pi }{\partial y} &= 0\\ \frac{\partial x}{\partial y}(p_y y -C(y)) &= 0\\ C'(y) &= p_y\\ \end{align*}

Thus, $MR = MC$ in this case.

Now define $r(y) = \frac{\pi(y)}{C(y)}$ to be the return. Intuitively, suppose you own the business. $C(y)$ is the amount you put in and $\pi(y)$ is what you get back. You want to maximize your return.

\begin{align*} \frac{\partial r}{\partial y} &=0\\ \frac{\partial}{\partial y} \frac{p_y y -C(y)}{C(y)} &=0\\ \frac{C(y)p_y - p_y y C'(y)}{[C(y)]^2} &=0\\ C(y)p_y - p_y y C'(y) &=0\\ C'(y) &=\frac{C(y)}{y}\\ \end{align*}

Thus, $MC = ATC$.

But when I googled this, I unexpectedly found this passage:

The point at which marginal cost equals average total cost (MC = ATC) is known as the break-even point. When the MR = P line crosses through this point, as is highlighted by the black circle on the graph, the product is said to be selling at its break-even price because the marginal revenue will exactly offset the marginal cost of production, and total revenue will exactly offset total cost. In this situation, the firm will break even: it will not be earning any profits, but it will not be losing money either. If the MR = P line lies above the break-even point, the firm will be operating at a profit, since the revenue earned on each unit of output sold will exceed the average cost of producing a unit of output, and thus total revenue will exceed total cost. If the MR = P line lies below the break-even point, the firm will be operating at a loss because the revenue earned on each unit of output will be less than the average cost of producing a unit of output, and so total revenue will be less that total cost.

I don't understand why, if you are maximizing the return for a founder investing in their business, why they would be getting the best return when they are breaking even. That doesn't make sense to me.

Can someone explain why $MC = ATC$ is both the point where an investor would be getting their max return and the breakeven point?

Answer 1

You write about two separate optimization problems, $$ 1. \max_y \pi(y) \\ 2. \max_y r(y) $$ The first problem's optimum yields $MC(y_1) = p_y$, while the second problem's optimum yields $MC(y_2) = ATC(y_2)$. Since these belong to solutions of two separate problems, the $y$ values may differ as well. Thus $$ p_y = MC(y_1) \ \ ?=? \ \ MC(y_2) = ATC(y_2), $$ there is no guarantee that $p_y = ATC(y_2)$ and you are at the break-even point.

On the quoted passage:
This passage states that if the price $p_y$ is exactly such that the above equation holds, then even if you try to maximize profits, the best you can do is break even, achieve zero profits. This is only true for a given price level, namely $$ p_y = \min_y ATC(y). $$

Answer 2

This is not a general result this is a result that holds only for certain market structures.

For example, suppose market is competitive so firms are price takers meaning $p(y)=p$. Then when you set up your profit equation:

$$\pi = R(y)-C(y) = p(y)y - C(y) = py - C(y)$$

when you optimize the profit you get:

$$\pi_y' = p - C'(y) = 0 \implies p = C'(y) $$

As you can see above, in the competitive case benefit/return/profit for the owner is maximized when $p=MC = ATC \implies \pi=0$. However, note $MC$ and $ATC$ includes opportunity cost so this is not the same as getting zero accounting profit. Here the investor's books would still be in green, but investor breaks even in terms of economic profit.

Next, suppose that this firm now has some market power so $p(y)$ will remain some function of $y$. Now we have:

$$\pi = p(y)y - C(y)$$

and hence optimum will be found where

$$\pi_y' = p'(y)y + p(y)- C'(y) = 0 \implies p'(y)y + p(y) = C'(y)$$

In this case it is trivial to see that (aside from corner solutions) at the profit maximizing point the profit will be $\pi> 0$.