It seems that an equilibrium where MC<MR is preferred to one where MC=MR since in the case of the latter the firm still makes a profit.

Put differently, why would a firm produce a unit at all if it will receive no profit from it?

  • $\begingroup$ I think you meant to say the former instead of latter? $\endgroup$ Dec 18, 2023 at 5:37

3 Answers 3


Because profit is maximized at MC=MR. If MC<MR, then firm could still earn more profit by producing little bit more. Also firms do earn positive profit at point where MC=MR.

Practical example:

Firm profit is given by:

$$ \Pi = p(q) q - c(q)$$

Where $\Pi$ is profit $p(q)$ price which depends on quantity sold, $q$ is quantity, and $c(q)$ are costs.

Now lets assume $p(q)$ is given by $p=100-q$ and $c(q)=q^2$.

So we have:

$$\Pi = (100-q) q - q^2$$

To find profit maximizing quantity we take the derivative wrt quantity (and equate it to zero):

$$100-2q -2q= 0 \implies \underbrace{100 -2q }_{MR}= \underbrace{2q}_{MC} $$

So this first order condition for profit maximization literally says that $MR=MC$. Now lets continue, the profit maximizing quantity will be given by:

$$25 =q^*$$

This firm will produce at point where MC=MR when quantity is 25. Now lets compare profit where firm produces 25 and 1 unit less as you suggested (i.e. 24).

Profit at MC=MR

$$\Pi= (100-25) 25 - 25^2 = 1250 $$

Profit at 1 unit less (MC<MR)

$$\Pi= (100-24) 24 - 24^2 = 1248 $$

Hence as you can clearly see the profit is maximized at point where MC=MR, despite the fact that there MC=MR. Literally by going one less from point where MC=MR firm would lose money. The reward for operating at MC=MR, as opposed to one unit less than MC=MR, is not zero. In the case above it is 2 extra dollars!

  • $\begingroup$ How did you identify whether the LHS or RHS of the FOC for profit maximization corresponds to either MR or MC? $\endgroup$ May 18, 2022 at 21:33
  • 1
    $\begingroup$ @StatsScared definitionally it does. $p(q)q$ is the revenue so $\frac{d}{dq}[p(q)q]$ is by definition marginal revenue and $c(q)$ is total cost so $\frac{d}{dw}[c(q)]$ by definition is marginal cost. You can verify that the derivative of revenue is 100-2q and derivative of cost is 2q $\endgroup$
    – 1muflon1
    May 18, 2022 at 22:15
  • $\begingroup$ Great thanks. One last follow up. So was it my mistake to assume that the marginal profit from one extra unit of production tells us anything useful about the firm's overall profit? In the sense that MR vs MC is about the marginal profit/loss and not the firm's overall profit. Is this a take away? $\endgroup$ May 18, 2022 at 23:34
  • $\begingroup$ @StatsScared yes, here is a good analogy. Imagine profit as a mountain and marginal profit MR-MC as mountain elevation. On the top of the mountain slope is zero yet you are at the very top. If you would take one small step back slope is positive but you are not at the top of the mountain. You could still increase your profit (area of the mountain below your feet) by taking that extra step where slope is zero. Hence even though we are at point MR=MC it’s still the top of profit “mountain” $\endgroup$
    – 1muflon1
    May 19, 2022 at 3:30
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    $\begingroup$ This answer, even though already accepted, mixes a continuous production set example with a discrete units explanation, which is somewhat problematic. I think the answer should refer to the discrete units model that is implied by the question, so I added one. $\endgroup$
    – VARulle
    May 19, 2022 at 8:59

We note that until MC actually meets MB, all production is profitable (by logic you've already seen).

Since we agree you want to produce everything up until MC=MB, let's point out that might occur in the middle of a unit. So a firm would like to produce the first half of the unit - but not the latter half (by the logic of MC=MB). Since we are allowing for breaks in production like this, let's say this break happens nearly at the end of the unit. 99.999999999% of the way done with production, to be exact. That's indistinguishable from actually producing the unit.

Cases where your professor says "Oh, they just make this last unit" can get there by saying you are making the overwhelming majority of some fractional unit. And if you allow calculus along continuous production function, the problem disappears entirely, since there is only an instantaneous moment where MC=MB.


Since your question is posed in terms of a firm producing discrete units of a quantity, the answer has to take this discreteness into account. With discrete units, "marginal" refers to the next unit produced. For simplicity let's assume that the profit function has a unique maximizer. Generically, a quantity $Q$ with $MC(Q)=MR(Q)$ will not exist. However, profit is clearly maximized if the revenue from selling the last unit produced exceeded the costs of producing it, but the revenue from selling the next unit would already fall short of the costs of producing it. Therefore the profit maximizing quantity $Q$ is the one which solves

$$MC(Q-1)\le MR(Q-1)\quad\text{and}\quad MC(Q)\ge MR(Q).$$


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