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

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

$$TR(Q) &gt; TC(Q) \ \forall Q$$

If it were, we would have

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

Even if we had

$$TR(Q) &gt; (or \ge) \ TC(Q) \ \forall Q$$

meaning our profit is positive for any quantity Q:

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

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



Analogy: $e^x &gt; x \ \forall x \in \mathbb R$, 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 $\pi''(Q_0)$