<p>Since</p>
<p><span class="math-container">$$TC(Q)=VC(Q)+FC$$</span></p>
<p>we have</p>
<p><span class="math-container">$$MC(Q)=TC'(Q)=VC'(Q)$$</span></p>
<p>To get <span class="math-container">$VC$</span>, just find the antiderivative of <span class="math-container">$MC$</span> which is zero when <span class="math-container">$Q=0$</span>. (Alternatively, to get <span class="math-container">$TC$</span> find the antiderivative of <span class="math-container">$MC$</span> which is equal to the fixed cost of <span class="math-container">$4$</span> when <span class="math-container">$Q=0$</span>.)</p>
<p>In this case <span class="math-container">$MC(Q)=2Q$</span>, and the antiderivatives are given by <span class="math-container">$Q^2+a$</span> for some constant <span class="math-container">$a$</span>. To ensure <span class="math-container">$0^2+a=0$</span>, we need to set <span class="math-container">$a=0$</span>. So <span class="math-container">$VC(Q)=Q^2$</span>, and we are given that <span class="math-container">$FC=4$</span>. Thus profit is</p>
<p><span class="math-container">$$pQ-TC(Q)=pQ-(Q^2+4)$$</span></p>
<p>For profit maximization <span class="math-container">$p=MC(Q)=2Q$</span>, so maximized profit is</p>
<p><span class="math-container">$$2Q^2-(Q^2+4)=Q^2-4$$</span></p>
<p>This is zero when <span class="math-container">$Q=2$</span>.</p>

CC BY-SA 4.0
2024-01-18T22:27:01.453Z
Well, actually $a = 4$.
Matteo Bulgarelli
CC BY-SA 4.0
2024-01-19T14:22:42.447Z
@MatteoBulgarelli No, $a=0$ if we want the antiderivative of the $MC$ function that represents the $VC$ function. On the other hand, $a=4$ if we want the antiderivative of the $MC$ function that represents the $TC$ function.
smcc
CC BY-SA 4.0
2024-01-20T08:40:56.980Z
I do agree, that's just a matter of notation, and of course your answer is perfectly fine.
 MY point is: you are basically telling OP that integrating the marginal costs, you a function that is composed of two parts: a variable part, that depends on the quantity you produce, and a fixed part, that is a constant. It would be more intuitive, in my view, to tell students "look, the part of this function that depends on  Q are variable costs, the other, constant part are fixed costs", rather then tellimg them "if you want to find the variable cost function, set the constant to 0.
Matteo Bulgarelli
CC BY-SA 4.0
2023-08-16T13:40:36.710Z
2023-08-16T13:40:36.710Z
smcc
<p><a href="https://i.sstatic.net/tpfoN.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/tpfoN.png" alt="enter image description here" /></a>
I'm struggling to find the correct answer which is (B). So far what I've done is say that P=MC as we're in a perfectly competitive market. So profit = PQ - TC which gives 0 = (2Q)Q -(VC+4) ? But we don't know VC so I doubt this method works...</p>

<marginal><profits><perfect-competition>
Finding Quantity when Profit = 0
Inès
# Finding Quantity when Profit = 0

(Asked by: username_0 on 2023-08-16)

[![enter image description here](https://i.sstatic.net/tpfoN.png)](https://i.sstatic.net/tpfoN.png)
I'm struggling to find the correct answer which is (B). So far what I've done is say that P=MC as we're in a perfectly competitive market. So profit = PQ - TC which gives 0 = (2Q)Q -(VC+4) ? But we don't know VC so I doubt this method works...

**Tags:** marginal, profits, perfect-competition


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## 1 answers

### Answer by username_1 on 2023-08-16. Score: 2

Since

$$TC(Q)=VC(Q)+FC$$

we have

$$MC(Q)=TC'(Q)=VC'(Q)$$

To get $VC$, just find the antiderivative of $MC$ which is zero when $Q=0$. (Alternatively, to get $TC$ find the antiderivative of $MC$ which is equal to the fixed cost of $4$ when $Q=0$.)

In this case $MC(Q)=2Q$, and the antiderivatives are given by $Q^2+a$ for some constant $a$. To ensure $0^2+a=0$, we need to set $a=0$. So $VC(Q)=Q^2$, and we are given that $FC=4$. Thus profit is

$$pQ-TC(Q)=pQ-(Q^2+4)$$

For profit maximization $p=MC(Q)=2Q$, so maximized profit is

$$2Q^2-(Q^2+4)=Q^2-4$$

This is zero when $Q=2$.

> **Comments:**
> **username_2**: Well, actually $a = 4$.  
> **username_1**: @username_2 No, $a=0$ if we want the antiderivative of the $MC$ function that represents the $VC$ function. On the other hand, $a=4$ if we want the antiderivative of the $MC$ function that represents the $TC$ function.  
> **username_2**: I do agree, that's just a matter of notation, and of course your answer is perfectly fine.  MY point is: you are basically telling OP that integrating the marginal costs, you a function that is composed of two parts: a variable part, that depends on the quantity you produce, and a fixed part, that is a constant. It would be more intuitive, in my view, to tell students "look, the part of this function that depends on  Q are variable costs, the other, constant part are fixed costs", rather then tellimg them "if you want to find the variable cost function, set the constant to 0.  

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