<p>You have found $\frac{dP}{dQ}=-1$ and $\frac{dQ}{dP}=-1$ </p>

<p>You have $R=PQ$ and $Q=120-P$ (equivalent to $R=120P-P^2$ or $R=120Q-Q^2$)</p>

<p>so the rate of marginal revenue for a small increase in quantity is $\frac{d(PQ)}{dQ}=Q\frac{dP}{dQ}+P\frac{dQ}{dQ} = -Q+P = 120-2Q$ leading to zero when $Q=60$  </p>

<p>while the rate of marginal revenue for a small increase in price is $\frac{d(PQ)}{dP}=Q\frac{dP}{dP}+P\frac{dQ}{dP} = Q-P = 120-2P$ leading to zero when $P=60$</p>

<p>both giving the point on the demand curve $Q=60, P=60$ as revenue maximising for the monopolist </p>

CC BY-SA 3.0
2016-09-26T13:08:11.683Z
2016-09-26T13:08:11.683Z
Henry
<p>The example you gave is correct.</p>

<p>What you are missing is that you took the derivative of the demand function and not of the revenue function. What you should have done is:</p>

<p>$Revenue = R = Price*Quantity = (-Q + 120) * Q = -Q^2 + 120Q$</p>

<p>$Marginal \; Revenue = MR=\frac{dR}{dQ} = \frac{d(P*Q)}{dQ} = 120 - 2Q$</p>

<p>What you did was:</p>

<p>$MR = \frac{dP}{dQ}=-1$.</p>

<p>Hence, you took the derivative of the Price and not of Revenue to get marginal revenue, which is wrong.</p>

CC BY-SA 3.0
2016-09-26T13:07:28.933Z
2016-09-26T13:07:28.933Z
BB King
<p>Good day</p>

<p>In my Microeconomics course, we are handling the 4 imperfect competition models.</p>

<p>Currently we are discussing the Cournot Model, however I am unsure about a certain example. This is given as a lecture example (all information provided).</p>

<blockquote>
  <p>Let us first assume that we start with a single firm (monopoly).</p>
  
  <p>This monopoly is the owner of a costless spring (and sells the water).</p>
  
  <p>I.e., MC = 0</p>
  
  <p>Start with our demand function:</p>
</blockquote>

<pre><code>Q = 120 – P
Now, determine the profit maximizing price/output combination.
MR = MC
MC = 0
Q = 120 – P (Demand function) 
P = -Q + 120
Thus, MR = -2Q + 120
Thus -2Q + 120 = 0
Thus Q = 60
P = 60
Profits = 3600
</code></pre>

<p>I am familiar with the concept of calculating each firm's demand function, what confuses me is, in this example, we have a a single firm (monopoly) and when deriving its mariginal revenue function, one should end up with the following:</p>

<pre><code>P = -Q + 120
</code></pre>

<p><code>d/dQ = P' = -1</code> (using chainrule for -Q => -1 and dQ of constant = 0)</p>

<p>thus <code>MR = P' = -1</code></p>

<p>but in the example given, we arrive at:</p>

<pre><code>MR = -2Q + 120
</code></pre>

<p>What am I missing or is this example incomplete (not all information given)?</p>

<monopoly><competition>
Cournot Model Monopoly example
CybeX
# Cournot Model Monopoly example

(Asked by: username_0 on 2016-09-26)

Good day
In my Microeconomics course, we are handling the 4 imperfect competition models.
Currently we are discussing the Cournot Model, however I am unsure about a certain example. This is given as a lecture example (all information provided).

Let us first assume that we start with a single firm (monopoly).
This monopoly is the owner of a costless spring (and sells the water).
I.e., MC = 0
Start with our demand function:

Q = 120 – P
Now, determine the profit maximizing price/output combination.
MR = MC
MC = 0
Q = 120 – P (Demand function) 
P = -Q + 120
Thus, MR = -2Q + 120
Thus -2Q + 120 = 0
Thus Q = 60
P = 60
Profits = 3600

I am familiar with the concept of calculating each firm's demand function, what confuses me is, in this example, we have a a single firm (monopoly) and when deriving its mariginal revenue function, one should end up with the following:
P = -Q + 120

d/dQ = P' = -1 (using chainrule for -Q => -1 and dQ of constant = 0)
thus MR = P' = -1
but in the example given, we arrive at:
MR = -2Q + 120

What am I missing or is this example incomplete (not all information given)?

**Tags:** monopoly, competition

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## 2 answers

### Answer by username_1 on 2016-09-26. Score: 1 | Accepted answer

You have found $\frac{dP}{dQ}=-1$ and $\frac{dQ}{dP}=-1$

You have $R=PQ$ and $Q=120-P$ (equivalent to $R=120P-P^2$ or $R=120Q-Q^2$)

so the rate of marginal revenue for a small increase in quantity is $\frac{d(PQ)}{dQ}=Q\frac{dP}{dQ}+P\frac{dQ}{dQ} = -Q+P = 120-2Q$ leading to zero when $Q=60$

while the rate of marginal revenue for a small increase in price is $\frac{d(PQ)}{dP}=Q\frac{dP}{dP}+P\frac{dQ}{dP} = Q-P = 120-2P$ leading to zero when $P=60$

both giving the point on the demand curve $Q=60, P=60$ as revenue maximising for the monopolist


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### Answer by username_2 on 2016-09-26. Score: 1

The example you gave is correct.

What you are missing is that you took the derivative of the demand function and not of the revenue function. What you should have done is:

$Revenue = R = Price\*Quantity = (-Q + 120) \* Q = -Q^2 + 120Q$

$Marginal \; Revenue = MR=\frac{dR}{dQ} = \frac{d(P\*Q)}{dQ} = 120 - 2Q$

What you did was:

$MR = \frac{dP}{dQ}=-1$.

Hence, you took the derivative of the Price and not of Revenue to get marginal revenue, which is wrong.


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