<p>We note that until MC actually meets MB, all production is profitable (by logic you've already seen).</p>
<p>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.</p>
<p>Cases where your professor says &quot;Oh, they just make this last unit&quot; 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.</p>

CC BY-SA 4.0
2022-05-18T21:00:54.210Z
2022-05-18T21:07:13.183Z
2022-05-18T21:07:13.183Z
RegressForward
<p>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, &quot;marginal&quot; refers to the <em>next</em> unit produced. For simplicity let's assume that the profit function has a unique maximizer. Generically, a quantity <span class="math-container">$Q$</span> with <span class="math-container">$MC(Q)=MR(Q)$</span> will not exist. However, profit is clearly maximized if the revenue from selling the <em>last</em> unit produced exceeded the costs of producing it, but the revenue from selling the <em>next</em> unit would already fall short of the costs of producing it. Therefore the profit maximizing quantity <span class="math-container">$Q$</span> is the one which solves</p>
<p><span class="math-container">$$MC(Q-1)\le MR(Q-1)\quad\text{and}\quad MC(Q)\ge MR(Q).$$</span></p>

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2022-05-19T08:44:29.670Z
2022-05-19T08:51:21.407Z
2022-05-19T08:51:21.407Z
VARulle
<p>Because profit is maximized at MC=MR. If MC&lt;MR, then firm could still earn <strong>more</strong> profit by producing little bit more. Also firms do earn positive profit at point where MC=MR.</p>
<p>Practical example:</p>
<p>Firm profit is given by:</p>
<p><span class="math-container">$$ \Pi = p(q) q - c(q)$$</span></p>
<p>Where <span class="math-container">$\Pi$</span> is profit <span class="math-container">$p(q)$</span> price which depends on quantity sold, <span class="math-container">$q$</span> is quantity, and <span class="math-container">$c(q)$</span> are costs.</p>
<p>Now lets assume <span class="math-container">$p(q)$</span> is given by <span class="math-container">$p=100-q$</span> and <span class="math-container">$c(q)=q^2$</span>.</p>
<p>So we have:</p>
<p><span class="math-container">$$\Pi = (100-q) q - q^2$$</span></p>
<p>To find profit maximizing quantity we take the derivative wrt quantity (and equate it to zero):</p>
<p><span class="math-container">$$100-2q -2q= 0 \implies \underbrace{100 -2q }_{MR}= \underbrace{2q}_{MC} $$</span></p>
<p>So this first order condition for profit maximization literally says that <span class="math-container">$MR=MC$</span>. Now lets continue, the profit maximizing quantity will be given by:</p>
<p><span class="math-container">$$25 =q^*$$</span></p>
<p>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).</p>
<p><strong>Profit at MC=MR</strong></p>
<p><span class="math-container">$$\Pi= (100-25) 25 - 25^2 = 1250 $$</span></p>
<p><strong>Profit at 1 unit less (MC&lt;MR)</strong></p>
<p><span class="math-container">$$\Pi= (100-24) 24 - 24^2 = 1248 $$</span></p>
<hr />
<p>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 <strong>lose</strong> money. The reward for operating at MC=MR, as opposed to one unit less than MC=MR, is <strong>not</strong> zero. In the case above it is 2 extra dollars!</p>

CC BY-SA 4.0
2022-05-18T21:33:05.210Z
How did you identify whether the LHS or RHS of the FOC for profit maximization corresponds to either MR or MC?
StatsScared
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2022-05-18T22:15:56.910Z
@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
1muflon1
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2022-05-18T23:34:12.610Z
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?
StatsScared
CC BY-SA 4.0
2022-05-19T03:30:54.880Z
@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”
1muflon1
CC BY-SA 4.0
2022-05-19T04:26:41.907Z
@1muflon1, I think you meant the gradient/ slope is zero at the top? Elevation is usually something like 8000 FT above sea level. Great answer by the way, and amazing how quick you usually are.
Alex
CC BY-SA 4.0
2022-05-19T08:59:10.917Z
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.
VARulle
CC BY-SA 4.0
2022-05-19T09:01:39.903Z
@Alex yes sorry I will fix my comment
1muflon1
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2022-05-18T21:08:04.610Z
2022-05-18T21:08:04.610Z
1muflon1
<p>It seems that an equilibrium where MC&lt;MR is preferred to one where MC=MR since in the case of the latter the firm still makes a profit.</p>
<p>Put differently, why would a firm produce a unit at all if it will receive no profit from it?</p>

CC BY-SA 4.0
2023-12-18T05:37:59.610Z
I think you meant to say the former instead of latter?
Franck Zhang
<microeconomics><profit-maximization>
Why is a firm's profit maximized when MC=MR? Why not stop one unit before where it will still make a profit?
StatsScared
# Why is a firm's profit maximized when MC=MR? Why not stop one unit before where it will still make a profit?

(Asked by: username_0 on 2022-05-18)

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?

**Tags:** microeconomics, profit-maximization

> **Comments:**
> **username_1**: I think you meant to say the former instead of latter?  

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## 3 answers

### Answer by username_4 on 2022-05-18. Score: 4 | Accepted answer

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!

> **Comments:**
> **username_0**: How did you identify whether the LHS or RHS of the FOC for profit maximization corresponds to either MR or MC?  
> **username_4**: @username_0 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  
> **username_0**: 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?  
> **username_4**: @username_0 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”  
> **username_5**: @username_4, I think you meant the gradient/ slope is zero at the top? Elevation is usually something like 8000 FT above sea level. Great answer by the way, and amazing how quick you usually are.  
> **username_3**: 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.  
> **username_4**: @username_5 yes sorry I will fix my comment  

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### Answer by username_2 on 2022-05-18. Score: 1

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.


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### Answer by username_3 on 2022-05-19. Score: 0

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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