<p>To complement @AdamBailey to-the-point answer, the purpose of this post was to alert interested readers to the consequences of changing decision-variables in our thinking.</p>
<p>We are accustomed to think of Demand as either &quot;price depending on quantity&quot; or &quot;quantity depending on price&quot;. But on the production-cost side, we automatically tend to think of cost depending on quantity, not on selling price.</p>
<p>Therefore, being even a bit tediously explicit with notation pays off (ask the guys over dynamic optimization, e.g. <a href="https://rads.stackoverflow.com/amzn/click/com/0521603684" rel="nofollow noreferrer" rel="nofollow noreferrer">Caputo's book</a>). In the specific example, the symbols <span class="math-container">$TR$</span>, <span class="math-container">$MR$</span>, <span class="math-container">$MC$</span>, do not reveal the decision variable, and this is where the ruse was based. But if, we wrote</p>
<p><span class="math-container">$$\max \pi = TR[Q(P)] - C[Q(P)]$$</span></p>
<p>we would clearly signaled that our ultimate decision variable is price, and so</p>
<p><span class="math-container">$$f.o.c: \;\;\;MR(Q)\cdot \frac {\partial Q}{\partial P} - MC(Q)\frac {\partial Q}{\partial P} =0 $$</span></p>
<p><span class="math-container">$$\implies (MR(Q)  - MC(Q))\cdot \frac {\partial Q}{\partial P} =0 \implies MR(Q)  = MC(Q)$$</span></p>
<p>while also we would clearly see that</p>
<p><span class="math-container">$$\frac {\partial TR}{\partial P} = MR(Q)\cdot \frac {\partial Q}{\partial P} = \frac {\partial Q}{\partial P}Q + Q$$</span></p>
<p>and so that the requirement on the price elasticity of demand leads to</p>
<p><span class="math-container">$$\frac {\partial TR}{\partial P} = MR(P) = \frac {\partial Q}{\partial P}Q + Q &lt; 0 \implies MR(Q)\cdot \frac {\partial Q}{\partial P} &lt; 0 \implies MR(Q) &gt;0$$</span></p>
<p>(since <span class="math-container">$\frac {\partial Q}{\partial P} &lt;0$</span>). So at the optimal point, marginal revenue <em>with respect to quantity</em> should be positive, but marginal revenue <em>with respect to price</em> should be negative.</p>

CC BY-SA 3.0
2017-10-19T19:26:31.477Z
I like this type of tricky questions and/or small riddles. Maybe we should think of something like this every once in a while.  With a lower bound to how fast one can be, so that every one can think while there's still no answer in post.
An old man in the sea.
CC BY-SA 3.0
2017-10-19T22:57:14.443Z
@Anoldmaninthesea. If you like riddles, check my answer to this post, https://math.stackexchange.com/q/490851/87400 I have to say I am really proud of it.
Alecos Papadopoulos
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2017-10-21T21:30:14.087Z
What do you think of caputo's book? do you recommend it?
An old man in the sea.
CC BY-SA 3.0
2017-10-21T21:45:26.440Z
@Anoldmaninthesea. Absolutely. It may drive you nuts at the beginning, with all its maddening notation and its insistent on writing in detail all the arguments of every function present in the various relations, but if you get familiar with that you will realize how it helps to clearly understand everything. I really understood Hamilton-Jacobi-Bellman equation for the first time due to this book.
Alecos Papadopoulos
CC BY-SA 3.0
2017-10-21T21:48:51.850Z
Now, I must really read it. =)
An old man in the sea.
CC BY-SA 4.0
2017-10-19T16:53:54.317Z
2020-11-11T12:48:07.630Z
2020-11-11T12:48:07.630Z
Alecos Papadopoulos
<p>$PQ(P)=TR$, Total Revenue. </p>

<p>$\frac{∂Q}{∂P}P+Q$ is the derivative of $PQ(P)$ <em>with respect to $P$</em>.</p>

<p>$MR$, Marginal Revenue, is the derivative of $TR$ <em>with respect to $Q$</em>.</p>

<p>So in general $\frac{∂Q}{∂P}P+Q \neq MR$</p>

CC BY-SA 3.0
2017-10-18T21:49:04.260Z
That is the perfect "few-dozen-seconds-as-requested" answer!
Alecos Papadopoulos
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2017-10-18T21:55:05.843Z
@AlecosPapadopoulos  Thank you (mainly my luck to have logged in at the right time).
Adam Bailey
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2017-10-18T21:41:46.680Z
2017-10-18T21:41:46.680Z
Adam Bailey
<p>A little head-scratcher (and a good example why we should be careful with notation).</p>

<p>Consider a profit maximizing monopoly, that solves over price </p>

<p>$$\max \pi = PQ(P) - C(Q(P)) \tag{1}$$</p>

<p>Following the routine steps (<strong><a href="https://economics.stackexchange.com/a/1728/61">see this post</a></strong>)</p>

<p>we arrive at the important result that, at the profit maximizing price, the price elasticity of demand should be higher than $1$ in absolute terms, or lower than $-1$ in algebraic terms. Namely at the profit-maximizing price we have</p>

<p>$$\eta^* = \frac {\partial Q }{ \partial P}\cdot \frac {P}{Q} &lt;-1 \Rightarrow \frac {\partial Q }{ \partial P}P &lt;-Q$$</p>

<p>$$\Rightarrow \frac {\partial Q }{ \partial P}P +Q &lt;0 \tag{2}$$</p>

<p>But $\frac {\partial Q }{ \partial P}P +Q$ is the derivative of $PQ(P)$ and $PQ(P) = TR$, Total Revenue. So $\frac {\partial Q }{ \partial P}P +Q = MR$, Marginal Revenue and we just obtained that at the profit maximizing price and in order to have elasticity greater than $1$ in absolute terms, we must have $MR^* &lt;0$. </p>

<p>But we also now that at the profit maximizing point we have $MR^*=MC^*&gt;0$.</p>

<p><strong>So a solution does not exist, and therefore we conclude that monopolies are just a mathematical misunderstanding.</strong></p>

<p>Now, I went into the trouble(?) to write this smirking post, I hope somebody will go into the few dozens of seconds required to write a clear answer to point out where the trick lies.</p>

CC BY-SA 3.0
2017-10-18T21:11:30.847Z
I don't know much about anything, specially micro. Here's my try. Isn't MR <0 only when demand is inelastic. In the  post you link, you state that it's for elastic demand...
An old man in the sea.
CC BY-SA 3.0
2017-10-18T21:44:42.197Z
@Anoldmaninthesea. The issue is that my post "shows" that in order to have elastic demand, you have to have $MR<0$.
Alecos Papadopoulos
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2017-10-18T23:20:17.377Z
@AlecosPapadopoulos, excuse my unrelated comment, but how could this question get 220+ views in a few hours?
london
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2017-10-18T23:28:14.150Z
@london Due to its title.
Alecos Papadopoulos
CC BY-SA 3.0
2017-10-18T23:30:27.590Z
@london And then, there is the "hot questions" accelerating effect. it is currently in the hot questions sidebar over at the mathematics se site.
Alecos Papadopoulos
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2017-10-19T04:54:28.440Z
@london Yeah, it's the HNQ list that's bringing in a lot of views. It does that.
reirab
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2017-10-19T05:43:19.750Z
Do I understand correctly that you're deliberately posting trick questions?
410 gone
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2017-10-19T08:08:19.907Z
@EnergyNumbers Yes, this was a trick question, as is written in the last sentence of the post.
Alecos Papadopoulos
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2017-10-19T22:07:30.770Z
@AlecosPapadopoulos  Could I suggest adding a question mark to the title of the question (in case the title should be quoted out of context)?
Adam Bailey
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2017-10-19T22:54:37.730Z
@AdamBailey I would rather disagree. This post is all about mischief, so there is no dignity to be preserved here.
Alecos Papadopoulos
<monopoly><profit-maximization>
Monopolies are just a mathematical misunderstanding
Alecos Papadopoulos
# Monopolies are just a mathematical misunderstanding

(Asked by: username_0 on 2017-10-18)

A little head-scratcher (and a good example why we should be careful with notation).

Consider a profit maximizing monopoly, that solves over price

$$\max \pi = PQ(P) - C(Q(P)) \tag{1}$$

Following the routine steps (**[see this post](https://economics.stackexchange.com/a/1728/61)**)

we arrive at the important result that, at the profit maximizing price, the price elasticity of demand should be higher than $1$ in absolute terms, or lower than $-1$ in algebraic terms. Namely at the profit-maximizing price we have

$$\eta^\* = \frac {\partial Q }{ \partial P}\cdot \frac {P}{Q} <-1 \Rightarrow \frac {\partial Q }{ \partial P}P <-Q$$

$$\Rightarrow \frac {\partial Q }{ \partial P}P +Q <0 \tag{2}$$

But $\frac {\partial Q }{ \partial P}P +Q$ is the derivative of $PQ(P)$ and $PQ(P) = TR$, Total Revenue. So $\frac {\partial Q }{ \partial P}P +Q = MR$, Marginal Revenue and we just obtained that at the profit maximizing price and in order to have elasticity greater than $1$ in absolute terms, we must have $MR^\* <0$.

But we also now that at the profit maximizing point we have $MR^\*=MC^\*>0$.

**So a solution does not exist, and therefore we conclude that monopolies are just a mathematical misunderstanding.**

Now, I went into the trouble(?) to write this smirking post, I hope somebody will go into the few dozens of seconds required to write a clear answer to point out where the trick lies.

**Tags:** monopoly, profit-maximization
> **Comments:**
> **username_1**: I don't know much about anything, specially micro. Here's my try. Isn't MR <0 only when demand is inelastic. In the  post you link, you state that it's for elastic demand...  
> **username_0**: @Anoldmaninthesea. The issue is that my post "shows" that in order to have elastic demand, you have to have $MR<0$.  
> **username_2**: @AlecosPapadopoulos, excuse my unrelated comment, but how could this question get 220+ views in a few hours?  
> **username_0**: @london Due to its title.  
> **username_0**: @london And then, there is the "hot questions" accelerating effect. it is currently in the hot questions sidebar over at the mathematics se site.  
> **username_3**: @london Yeah, it's the HNQ list that's bringing in a lot of views. It does that.  
> **username_4**: Do I understand correctly that you're deliberately posting trick questions?  
> **username_0**: @EnergyNumbers Yes, this was a trick question, as is written in the last sentence of the post.  
> **username_5**: @AlecosPapadopoulos  Could I suggest adding a question mark to the title of the question (in case the title should be quoted out of context)?  
> **username_0**: @AdamBailey I would rather disagree. This post is all about mischief, so there is no dignity to be preserved here.  

---

## 2 answers

### Answer by username_5 on 2017-10-18. Score: 18 | Accepted answer

$PQ(P)=TR$, Total Revenue.

$\frac{∂Q}{∂P}P+Q$ is the derivative of $PQ(P)$ *with respect to $P$*.

$MR$, Marginal Revenue, is the derivative of $TR$ *with respect to $Q$*.

So in general $\frac{∂Q}{∂P}P+Q \neq MR$

> **Comments:**
> **username_0**: That is the perfect "few-dozen-seconds-as-requested" answer!  
> **username_5**: @AlecosPapadopoulos  Thank you (mainly my luck to have logged in at the right time).  

---

### Answer by username_0 on 2017-10-19. Score: 5

To complement @AdamBailey to-the-point answer, the purpose of this post was to alert interested readers to the consequences of changing decision-variables in our thinking.

We are accustomed to think of Demand as either "price depending on quantity" or "quantity depending on price". But on the production-cost side, we automatically tend to think of cost depending on quantity, not on selling price.

Therefore, being even a bit tediously explicit with notation pays off (ask the guys over dynamic optimization, e.g. [Caputo's book](https://rads.stackoverflow.com/amzn/click/com/0521603684)). In the specific example, the symbols $TR$, $MR$, $MC$, do not reveal the decision variable, and this is where the ruse was based. But if, we wrote

$$\max \pi = TR[Q(P)] - C[Q(P)]$$

we would clearly signaled that our ultimate decision variable is price, and so

$$f.o.c: \;\;\;MR(Q)\cdot \frac {\partial Q}{\partial P} - MC(Q)\frac {\partial Q}{\partial P} =0 $$

$$\implies (MR(Q) - MC(Q))\cdot \frac {\partial Q}{\partial P} =0 \implies MR(Q) = MC(Q)$$

while also we would clearly see that

$$\frac {\partial TR}{\partial P} = MR(Q)\cdot \frac {\partial Q}{\partial P} = \frac {\partial Q}{\partial P}Q + Q$$

and so that the requirement on the price elasticity of demand leads to

$$\frac {\partial TR}{\partial P} = MR(P) = \frac {\partial Q}{\partial P}Q + Q < 0 \implies MR(Q)\cdot \frac {\partial Q}{\partial P} < 0 \implies MR(Q) >0$$

(since $\frac {\partial Q}{\partial P} <0$). So at the optimal point, marginal revenue *with respect to quantity* should be positive, but marginal revenue *with respect to price* should be negative.

> **Comments:**
> **username_1**: I like this type of tricky questions and/or small riddles. Maybe we should think of something like this every once in a while.  With a lower bound to how fast one can be, so that every one can think while there's still no answer in post.  
> **username_0**: @Anoldmaninthesea. If you like riddles, check my answer to this post, https://math.stackexchange.com/q/490851/87400 I have to say I am really proud of it.  
> **username_1**: What do you think of caputo's book? do you recommend it?  
> **username_0**: @Anoldmaninthesea. Absolutely. It may drive you nuts at the beginning, with all its maddening notation and its insistent on writing in detail all the arguments of every function present in the various relations, but if you get familiar with that you will realize how it helps to clearly understand everything. I really understood Hamilton-Jacobi-Bellman equation for the first time due to this book.  
> **username_1**: Now, I must really read it. =)  

---

