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

<p>It is not always the case that</p>

<p>$$TR(Q) &gt; TC(Q) \ \forall Q$$</p>

<p>If it were, we would have</p>

<p>$$\{Q | TR(Q) = TC(Q)\} = \emptyset$$</p>

<p>Even if we had</p>

<p>$$TR(Q) &gt; (or \ge) \ TC(Q) \ \forall Q$$</p>

<p>meaning our profit is positive for any quantity Q:</p>

<p>$$\pi(Q) := TR(Q) - TC(Q) &gt; 0$$</p>

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

<hr>

<p>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</p>

<hr>

<p>This is not done by solving</p>

<p>$$TR(Q) = TC(Q)$$</p>

<p>which merely gives the quantities that give us zero profit ie</p>

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

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

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

<p>Btw, solving</p>

<p>$$0 = MR(Q) - MC(Q)$$</p>

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

CC BY-SA 3.0
2015-12-13T22:08:13.040Z
2015-12-13T22:08:13.040Z
BCLC
<p>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. </p>

CC BY-SA 3.0
2015-12-13T17:25:21.113Z
2015-12-13T17:25:21.113Z
ml0105
<p>I know that I should optimize production by solving $MR = MC$ with respect to $Q$.</p>

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

CC BY-SA 3.0
2015-12-13T13:33:06.733Z
Because $TR = TC$ means zero profit. You want to maximize profit and positive profit might be feasible.
Giskard
CC BY-SA 3.0
2015-12-13T13:36:09.837Z
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 $TC$.
Jamgreen
CC BY-SA 3.0
2015-12-13T13:36:58.227Z
I think you are confusing the marginal and total concepts.
Giskard
CC BY-SA 3.0
2015-12-13T20:36:09.850Z
@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.
DornerA
<optimization><cost><profit-maximization><profits>
Optimize by MR = MC vs TR = TC
Jamgreen
# Optimize by MR = MC vs TR = TC

(Asked by: username_0 on 2015-12-13)

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 $Q$?

**Tags:** optimization, cost, profit-maximization, profits

> **Comments:**
> **username_1**: Because $TR = TC$ means zero profit. You want to maximize profit and positive profit might be feasible.  
> **username_0**: 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 $TC$.  
> **username_1**: I think you are confusing the marginal and total concepts.  
> **username_2**: @username_0 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.  

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## 2 answers

### Answer by username_3 on 2015-12-13. Score: 1

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

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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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### Answer by username_4 on 2015-12-13. Score: 0

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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