<p>The reason your solution is wrong is because this assumption is wrong Qd =/= AR.</p>

<p>The comments from your question has already stated that point. But lets put it into perspective.</p>

<p>The way to arrive at Average Revenue (AR) is by taking Total Revenue dividing by the quantity you sold.</p>

<p>Total revenue is derived from the price per unit multiplying by the number of units you sold (quantity, Q)</p>

<p>Essentially, this becomes AR = (P*Q)/Q = P</p>

<p>Therefore, AR = P</p>

<p>Now that we have established that, you can go on to use TR = Q*AR which will eventually bring you through the steps of the second solution which you went through in the classroom.</p>

<p>For your second question, I do not really get where the confusion is coming from where Qd = Q.</p>

<p>I assume you get confused because you wrote TR = P * Q</p>

<p>But read out the whole expression to yourself, the expression is saying Total revenue is equals to Price multiplied by the Quantity sold (AKA quantity demanded)</p>

<p>AND, the quantity demanded is a function of the price which is provided to you in the question, Qd = 600 - 2P</p>

<p>So at a particular price (P), you have a particular quantity demanded (Qd) which leads you to a particular Total Revenue. The Qd is equal to Q then, in your terms. </p>

<p>Hope this helps!</p>

CC BY-SA 3.0
2017-09-19T11:47:16.247Z
2017-09-19T11:47:16.247Z
MH.Q
<p>The reason why the answers are different is because the way you solved the for marginal revenue on your own includes <span class="math-container">$P$</span> in your solution.</p>
<p>The reason why this is problematic is because <span class="math-container">$MR=P$</span> for a monopolist. You'll end up with some mathematically impossible statements when attempting to find the optimal quantity produced.</p>
<p>I.e.</p>
<blockquote>
<p><span class="math-container">$$MR = \frac{\partial TR}{\partial Q} = 600 - 2P$$</span>
<span class="math-container">$$MR = 600 - 2MR$$</span>
<span class="math-container">$$ MR = 200$$</span></p>
<p>the monopolist prodouces where <span class="math-container">$MC=MR$</span>  therefore
<span class="math-container">$$\color{red}{20=200} $$</span></p>
</blockquote>
<p>As for the second point <span class="math-container">$Q_d$</span> is <span class="math-container">$Q$</span> because we can only have a single quantity supplied and demanded in equilibrium.</p>

CC BY-SA 3.0
2017-09-18T21:12:05.930Z
John I have no clue what you are getting at but for sure it is not true that $MR = P$.
Giskard
CC BY-SA 3.0
2017-09-19T20:34:46.053Z
@denesp the derivative if total revenue  is marginal revenue. Thus MR=P http://evlm.stuba.sk/~partner7/DBfiles/Modules/Differentiation/DifferentiationApp.pdf
EconJohn
CC BY-SA 3.0
2017-09-18T18:11:55.247Z
2017-09-18T20:26:18.930Z
2020-06-17T09:40:54.643Z
EconJohn
<blockquote>
  <p>Suppose that monopoly produces product $X$. Let $MC = 20$, Quantity demanded
  function is given by equation $Q_d = 600 - 2P$ (where $Q_d$ - demand curve, 
  $P$ - price). Calculate the amount of production and the price in which the 
  profit will be maximized.</p>
</blockquote>

<p><strong>I'd like to know why is the following solution wrong:</strong></p>

<p>$MC = 20 ,\ Q_d = 600 - 2P$</p>

<p>I know that</p>

<p>$(Q_d = AR $ &amp; $ TR = Q*AR) \implies TR = Q*Q_d$</p>

<p>$TR = Q*(600-2P)$</p>

<p>$TR = 600Q-2PQ + C$</p>

<p>$MR = \frac{\partial TR}{\partial Q} = 600 - 2P$</p>

<p>since the profit is maximsed when $MR = MC$:</p>

<p>$MR = MC$</p>

<p>$20 = 600 -2P \implies P = 290$</p>

<p>since $Qd = 600-2P,$ &amp; $ P^* = 290 \implies Q^* = 20$</p>

<p>so the solution should be:</p>

<p>$
\left\{ 
\begin{array}{c}
P^* =  290 \\ 
Q^* = 20 \\
\end{array}
\right. 
$</p>

<p><strong>But when we solved this task in classroom, the correct solution was:</strong></p>

<p>$MC = 20 ,\ Q_d = 600 - 2P$</p>

<p>$Q_d = 600 - 2P \implies P=-\frac{Q_d}{2} + 300$</p>

<p>since $TR = P*Q \implies TR = (-\frac{Q_d}{2} + 300)*Q = -\frac{Q_d^2}{2} + 300Q_d$ (I don't understand why the $Q = Q_d$ here)</p>

<p>we know that the profit is maximsed when $MR = MC$, so:</p>

<p>$MR = MC$</p>

<p>$-Q+300 = 20 \implies Q = 280$</p>

<p>$P=\frac{Q_d}{2} + 300 \implies P = 160$</p>

<p>so the solution is:</p>

<p>$
\left\{ 
\begin{array}{c}
P^* =  160 \\ 
Q^* = 280 \\
\end{array}
\right. 
$</p>

<p>Why the answer is different when we solve this exercise in two different ways? What am I doing wrong in the first solution? Why is $Q_d = Q$?</p>

CC BY-SA 3.0
2017-09-16T16:36:04.703Z
average revenue is price, not quantity. your solution makes no sense.
Fato
CC BY-SA 3.0
2017-09-16T17:03:42.227Z
Unfortunately it's not correct. Please check the following sources  https://www.quora.com/Why-is-the-average-revenue-curve-the-same-as-the-demand-curve and https://www.acted.co.uk/forums/index.php?threads/demand-curve-ar-or-mr.9296/
user3130324
CC BY-SA 3.0
2017-09-19T03:17:27.050Z
Unfortunately you did not understand your own sources correctly. Try to grasp the concept instead of taking the literal meaning of AR is the same as demand curve. What they meant was in graphical representation, since they both plot price against quantity, although the more correct term they should have used is inverse demand curve. Reading your own sources would also have led you to conclude that they actually meant AR = P, not AR = Q
MH.Q
<microeconomics><monopoly>
Microeconomy, monopoly, maximum profit - two different answers when solving exercise in two different ways
user3130324
# Microeconomy, monopoly, maximum profit - two different answers when solving exercise in two different ways

(Asked by: username_0 on 2017-09-16)

> Suppose that monopoly produces product $X$. Let $MC = 20$, Quantity demanded
> function is given by equation $Q\_d = 600 - 2P$ (where $Q\_d$ - demand curve,
> $P$ - price). Calculate the amount of production and the price in which the
> profit will be maximized.

**I'd like to know why is the following solution wrong:**

$MC = 20 ,\ Q\_d = 600 - 2P$

I know that

$(Q\_d = AR $ & $ TR = Q\*AR) \implies TR = Q\*Q\_d$

$TR = Q\*(600-2P)$

$TR = 600Q-2PQ + C$

$MR = \frac{\partial TR}{\partial Q} = 600 - 2P$

since the profit is maximsed when $MR = MC$:

$MR = MC$

$20 = 600 -2P \implies P = 290$

since $Qd = 600-2P,$ & $ P^\* = 290 \implies Q^\* = 20$

so the solution should be:

$
\left\{
\begin{array}{c}
P^\* = 290 \\
Q^\* = 20 \\
\end{array}
\right.
$

**But when we solved this task in classroom, the correct solution was:**

$MC = 20 ,\ Q\_d = 600 - 2P$

$Q\_d = 600 - 2P \implies P=-\frac{Q\_d}{2} + 300$

since $TR = P\*Q \implies TR = (-\frac{Q\_d}{2} + 300)\*Q = -\frac{Q\_d^2}{2} + 300Q\_d$ (I don't understand why the $Q = Q\_d$ here)

we know that the profit is maximsed when $MR = MC$, so:

$MR = MC$

$-Q+300 = 20 \implies Q = 280$

$P=\frac{Q\_d}{2} + 300 \implies P = 160$

so the solution is:

$
\left\{
\begin{array}{c}
P^\* = 160 \\
Q^\* = 280 \\
\end{array}
\right.
$

Why the answer is different when we solve this exercise in two different ways? What am I doing wrong in the first solution? Why is $Q\_d = Q$?

**Tags:** microeconomics, monopoly
> **Comments:**
> **username_1**: average revenue is price, not quantity. your solution makes no sense.  
> **username_0**: Unfortunately it's not correct. Please check the following sources  https://www.quora.com/Why-is-the-average-revenue-curve-the-same-as-the-demand-curve and https://www.acted.co.uk/forums/index.php?threads/demand-curve-ar-or-mr.9296/  
> **username_2**: Unfortunately you did not understand your own sources correctly. Try to grasp the concept instead of taking the literal meaning of AR is the same as demand curve. What they meant was in graphical representation, since they both plot price against quantity, although the more correct term they should have used is inverse demand curve. Reading your own sources would also have led you to conclude that they actually meant AR = P, not AR = Q  

---

## 2 answers

### Answer by username_2 on 2017-09-19. Score: 1

The reason your solution is wrong is because this assumption is wrong Qd =/= AR.

The comments from your question has already stated that point. But lets put it into perspective.

The way to arrive at Average Revenue (AR) is by taking Total Revenue dividing by the quantity you sold.

Total revenue is derived from the price per unit multiplying by the number of units you sold (quantity, Q)

Essentially, this becomes AR = (P\*Q)/Q = P

Therefore, AR = P

Now that we have established that, you can go on to use TR = Q\*AR which will eventually bring you through the steps of the second solution which you went through in the classroom.

For your second question, I do not really get where the confusion is coming from where Qd = Q.

I assume you get confused because you wrote TR = P \* Q

But read out the whole expression to yourself, the expression is saying Total revenue is equals to Price multiplied by the Quantity sold (AKA quantity demanded)

AND, the quantity demanded is a function of the price which is provided to you in the question, Qd = 600 - 2P

So at a particular price (P), you have a particular quantity demanded (Qd) which leads you to a particular Total Revenue. The Qd is equal to Q then, in your terms.

Hope this helps!


---

### Answer by username_3 on 2017-09-18. Score: 0

The reason why the answers are different is because the way you solved the for marginal revenue on your own includes $P$ in your solution.

The reason why this is problematic is because $MR=P$ for a monopolist. You'll end up with some mathematically impossible statements when attempting to find the optimal quantity produced.

I.e.

> $$MR = \frac{\partial TR}{\partial Q} = 600 - 2P$$
> $$MR = 600 - 2MR$$
> $$ MR = 200$$
>
> the monopolist prodouces where $MC=MR$ therefore
> $$\color{red}{20=200} $$

As for the second point $Q\_d$ is $Q$ because we can only have a single quantity supplied and demanded in equilibrium.

> **Comments:**
> **username_4**: John I have no clue what you are getting at but for sure it is not true that $MR = P$.  
> **username_3**: @denesp the derivative if total revenue  is marginal revenue. Thus MR=P http://evlm.stuba.sk/~partner7/DBfiles/Modules/Differentiation/DifferentiationApp.pdf  

---

