I know that I should optimize production by solving $MR = MC$ with respect to $Q$.

But if $TR > TC$, I am making a profit. Why is not enough to just solve $TR = TC$ with respect to $Q$?

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    $\begingroup$ Because $TR = TC$ means zero profit. You want to maximize profit and positive profit might be feasible. $\endgroup$ – Giskard Dec 13 '15 at 13:33
  • $\begingroup$ I just though that $TR=TC$ must have multiple intersections, so I would choose that Q from where I have made a profit from all the previous units but no longer makes a profit because $TR$ now equals $TC$. $\endgroup$ – Jamgreen Dec 13 '15 at 13:36
  • $\begingroup$ I think you are confusing the marginal and total concepts. $\endgroup$ – Giskard Dec 13 '15 at 13:36
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    $\begingroup$ @Jamgreen TR = TC could have multiple intersections, but the story is the same at each one. You are making zero profit. At an intersection where MC < MR, you are making zero profit and you could make positive profit by producing more units. At an intersection where MC > MR, you are losing money on each unit, so if you produce further, your profit will be less than it would have been if you stopped at MR = MC, and again, you will be making zero profit. This is not optimal obviously. $\endgroup$ – DornerA Dec 13 '15 at 20:36

Let $TR(Q), TC(Q): [0, \infty) \to [0, \infty)$ be continuous and twice differentiable functions w/ resp derivatives $MR(Q), MC(Q)$.

It is not always the case that

$$TR(Q) > TC(Q) \ \forall Q$$

If it were, we would have

$$\{Q | TR(Q) = TC(Q)\} = \emptyset$$

Even if we had

$$TR(Q) > (or \ge) \ TC(Q) \ \forall Q$$

meaning our profit is positive for any quantity Q:

$$\pi(Q) := TR(Q) - TC(Q) > 0$$

we still want to find $Q^{*}$ s.t. $\pi(Q^{*}) = TR(Q^{*}) - TC(Q^{*})$ is maximised.

Analogy: $e^x > x \ \forall x \in \mathbb R$, but $f(x) := |e^x - x|$ is not constant. Some values of $x$ give greater distances between $e^x$ and $x$ than others

This is not done by solving

$$TR(Q) = TC(Q)$$

which merely gives the quantities that give us zero profit ie

$$\pi(Q) := TR(Q) - TC(Q) = 0$$

We want to maximise $\pi(Q)$ so we get the first derivative and set it to zero:

$$0 = \frac{d}{dQ} \pi(Q) = MR(Q) - MC(Q)$$

Btw, solving

$$0 = MR(Q) - MC(Q)$$

gives us some value $Q_0$, but this is not necessarily give $Q^{*}$. We must check the value of $\pi''(Q_0)$

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MR and MC are first derivatives of the Revenue and Cost functions respectively. If for every point greater than 0, we have Revenue greater than Cost, then no global maximum exists. This usually isn't the case though. And if we have multiple points where the tangency condition holds, we have to check each one to find the best.

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