Monopolies are just a mathematical misunderstanding

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)

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.

• @AlecosPapadopoulos, excuse my unrelated comment, but how could this question get 220+ views in a few hours? Oct 18 '17 at 23:20
• @london Due to its title. Oct 18 '17 at 23:28
• @london And then, there is the "hot questions" accelerating effect. it is currently in the hot questions sidebar over at the mathematics se site. Oct 18 '17 at 23:30
• Do I understand correctly that you're deliberately posting trick questions? Oct 19 '17 at 5:43
• @EnergyNumbers Yes, this was a trick question, as is written in the last sentence of the post. Oct 19 '17 at 8:08

$PQ(P)=TR$, 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 $\frac{∂Q}{∂P}P+Q \neq MR$

• That is the perfect "few-dozen-seconds-as-requested" answer! Oct 18 '17 at 21:49
• @AlecosPapadopoulos Thank you (mainly my luck to have logged in at the right time). Oct 18 '17 at 21:55

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). 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.

• 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. Oct 19 '17 at 19:26
• @Anoldmaninthesea. If you like riddles, check my answer to this post, math.stackexchange.com/q/490851/87400 I have to say I am really proud of it. Oct 19 '17 at 22:57
• What do you think of caputo's book? do you recommend it? Oct 21 '17 at 21:30
• @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. Oct 21 '17 at 21:45
• Now, I must really read it. =) Oct 21 '17 at 21:48