<p>Profit is revenue minus costs. So we have $q = 12500$ and $p = 35$. (Note that I'm assuming you've correctly figured this out.) Then we have revenue:</p>

<p>$$R = \int_{0}^{12500} MR dq = \int_{0}^{12500} (60 - 0.004q)dq$$</p>

<p>And costs:</p>

<p>$$C = 300000 + \int_{0}^{12500} 10dq = 300000 + 125000 = 425000$$</p>

<p>The profit is $\Pi = R - C$. </p>

<p>The average cost is $C/12500$. </p>

<p>I leave you to crunch the integrals.</p>

<p><strong>Edit</strong>: In looking at your problem, $p(q) = 60 - 0.002q$ is definitely an inverse demand function. The revenue $R = qp(q) = (60q - 0.002q^{2})$ (quantity times prices). This yields the profit maximization problem (maximizing with respect to quantity produced):</p>

<p>$$\max_{q} qp(q) - c(q) = (60q - 0.004q^{2}) - 10q - 300000$$</p>

<p>The First Order Conditions are:</p>

<p>$$60 - 0.004q - 10 = 0 \implies 60 -0.004q = 10$$</p>

<p>You did the work from there to solve for $q$.</p>

CC BY-SA 3.0
2015-08-11T04:11:21.017Z
2015-08-11T04:21:46.820Z
2015-08-11T04:21:46.820Z
ml0105
<p>The demand curve is: </p>

<p><span class="math-container">$$p = 60 - 0.002q$$</span></p>

<p>Marginal cost (<span class="math-container">$C'(q)$</span>) is a constant 10 dollars and fixed costs (<span class="math-container">$k$</span>) are 300,000 dollars. Hence,</p>

<p><span class="math-container">$$C(q) = \int_{0}^{q} 10 \ \mathrm{d}x + k = 10q + k$$</span></p>

<p>where <span class="math-container">$k = 300000$</span>. Thus, </p>

<p><span class="math-container">$$C(q) = 10q + 300000$$</span></p>

<p>The monopolist’s problem is</p>

<p><span class="math-container">$$\max_{q} π = pq - C(q)$$</span></p>

<p>In this case, that is:</p>

<p><span class="math-container">$$\max_{q} q(60 - 0.002q) - (10q + 300000)$$</span> </p>

<p>Equivalently:</p>

<p><span class="math-container">$$\max_{q} -0.002q^2 + 50q - 300000$$</span></p>

<p>The first order condition is:</p>

<p><span class="math-container">$$-0.004q + 50 = 0 \implies -0.004q = -50 \implies q = \frac{-50}{-0.004} = 12500$$</span></p>

<p>Hence, the profit maximizing quantity is 12,500. (You can verify the second order condition as well if you’d like to.)</p>

<p><strong>Economic Profit</strong></p>

<p>Substituting this into the equation for profit, we have:</p>

<p><span class="math-container">$$π = -0.002(12500)^2 + 50(12500) - 300000 = 12500$$</span></p>

<p>So funnily enough, by producing 12,500 units, the monopolist earns an economic profit of $12,500, which is the maximum profit the monopolist can earn given this demand curve.</p>

<p><strong>Average Cost</strong></p>

<p>Average cost at <span class="math-container">$q = q_0$</span> is defined as</p>

<p><span class="math-container">$$AC(q_0) := \frac{C(q_0)}{q_0}$$</span></p>

<p>In this case, <span class="math-container">$q = 12500$</span>. Hence, <span class="math-container">$C(q) = C(12500) = 425000$</span>. This means, from the above definition of average cost:</p>

<p><span class="math-container">$$AC(12500) = \frac{425000}{12500} = 34$$</span></p>

<p>Hence, the average cost is $34. Hope that helps.</p>

CC BY-SA 4.0
2019-12-25T19:39:43.577Z
2019-12-25T19:46:15.023Z
2019-12-25T19:46:15.023Z
Tejas Subramaniam
<p>I'm given a demand function ( $p = 60-.002q$), Marginal Cost (\$10) and Fixed Costs (\$300,000) for a monopoly. </p>

<p>I've figured the profit maximizing output and price:</p>

<p>Marginal revenue function $MR = 60 -.004q$.</p>

<p>$10 = 60-.004q$  # To find monopoly output.</p>

<p>$q = 12500 \implies$ Price of monopolist is $35$  #  $p = 60-.002(12500)$. </p>

<p>Now, I usually figure out the economic profit by taking the area of the rectangle formed by the bounds of the demand function (at monopolist output) down and stopping where it meets the AC function, and then across to the y-intercept of the graph; but they didn't give me an average cost function. </p>

<p>How can I figure out the economic profit? Please help and thanks for your time. </p>

<p>P.S. Is there a way of finding what Average Cost is at output(12500) with the given information?</p>

CC BY-SA 3.0
2015-08-11T04:06:25.500Z
Do you mean $p = 60 - 0.002q$ is your inverse demand function? Generally, $q$ is the output and $p$ is the price.
ml0105
CC BY-SA 3.0
2015-08-11T04:10:00.410Z
Price is on the y-axis.
Romaion
CC BY-SA 3.0
2015-08-11T04:10:30.223Z
That is an inverse demand function then, not a demand function.
ml0105
CC BY-SA 3.0
2015-08-11T04:11:26.480Z
First year econ. student, they've taught it as nothing other than a demand function lol.
Romaion
CC BY-SA 3.0
2015-08-11T04:14:41.947Z
Hopefully this is corrected in your upper division courses. For future reference: https://en.wikipedia.org/wiki/Inverse_demand_function :-)
ml0105
CC BY-SA 3.0
2015-08-11T04:23:23.940Z
I updated my answer a little at the end to clarify regarding the profit maximization problem and the role of the inverse demand function. I would encourage you to clarify this point with your instructor. In either case, use the terminology you need to in order to get the grade. Hopefully my edit is helpful in clarifying. Best of luck!
ml0105
<microeconomics><monopoly>
Help trying to find Economic Profit for a Monopoly
Romaion
# Help trying to find Economic Profit for a Monopoly

(Asked by: username_0 on 2015-08-11)

I'm given a demand function ( $p = 60-.002q$), Marginal Cost (\$10) and Fixed Costs (\$300,000) for a monopoly.

I've figured the profit maximizing output and price:

Marginal revenue function $MR = 60 -.004q$.

$10 = 60-.004q$ # To find monopoly output.

$q = 12500 \implies$ Price of monopolist is $35$ # $p = 60-.002(12500)$.

Now, I usually figure out the economic profit by taking the area of the rectangle formed by the bounds of the demand function (at monopolist output) down and stopping where it meets the AC function, and then across to the y-intercept of the graph; but they didn't give me an average cost function.

How can I figure out the economic profit? Please help and thanks for your time.

P.S. Is there a way of finding what Average Cost is at output(12500) with the given information?

**Tags:** microeconomics, monopoly
> **Comments:**
> **username_1**: Do you mean $p = 60 - 0.002q$ is your inverse demand function? Generally, $q$ is the output and $p$ is the price.  
> **username_0**: Price is on the y-axis.  
> **username_1**: That is an inverse demand function then, not a demand function.  
> **username_0**: First year econ. student, they've taught it as nothing other than a demand function lol.  
> **username_1**: Hopefully this is corrected in your upper division courses. For future reference: https://en.wikipedia.org/wiki/Inverse_demand_function :-)  
> **username_1**: I updated my answer a little at the end to clarify regarding the profit maximization problem and the role of the inverse demand function. I would encourage you to clarify this point with your instructor. In either case, use the terminology you need to in order to get the grade. Hopefully my edit is helpful in clarifying. Best of luck!  

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## 2 answers

### Answer by username_1 on 2015-08-11. Score: 2

Profit is revenue minus costs. So we have $q = 12500$ and $p = 35$. (Note that I'm assuming you've correctly figured this out.) Then we have revenue:

$$R = \int\_{0}^{12500} MR dq = \int\_{0}^{12500} (60 - 0.004q)dq$$

And costs:

$$C = 300000 + \int\_{0}^{12500} 10dq = 300000 + 125000 = 425000$$

The profit is $\Pi = R - C$.

The average cost is $C/12500$.

I leave you to crunch the integrals.

**Edit**: In looking at your problem, $p(q) = 60 - 0.002q$ is definitely an inverse demand function. The revenue $R = qp(q) = (60q - 0.002q^{2})$ (quantity times prices). This yields the profit maximization problem (maximizing with respect to quantity produced):

$$\max\_{q} qp(q) - c(q) = (60q - 0.004q^{2}) - 10q - 300000$$

The First Order Conditions are:

$$60 - 0.004q - 10 = 0 \implies 60 -0.004q = 10$$

You did the work from there to solve for $q$.


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### Answer by username_2 on 2019-12-25. Score: 0

The demand curve is:

$$p = 60 - 0.002q$$

Marginal cost ($C'(q)$) is a constant 10 dollars and fixed costs ($k$) are 300,000 dollars. Hence,

$$C(q) = \int\_{0}^{q} 10 \ \mathrm{d}x + k = 10q + k$$

where $k = 300000$. Thus,

$$C(q) = 10q + 300000$$

The monopolist’s problem is

$$\max\_{q} π = pq - C(q)$$

In this case, that is:

$$\max\_{q} q(60 - 0.002q) - (10q + 300000)$$

Equivalently:

$$\max\_{q} -0.002q^2 + 50q - 300000$$

The first order condition is:

$$-0.004q + 50 = 0 \implies -0.004q = -50 \implies q = \frac{-50}{-0.004} = 12500$$

Hence, the profit maximizing quantity is 12,500. (You can verify the second order condition as well if you’d like to.)

**Economic Profit**

Substituting this into the equation for profit, we have:

$$π = -0.002(12500)^2 + 50(12500) - 300000 = 12500$$

So funnily enough, by producing 12,500 units, the monopolist earns an economic profit of $12,500, which is the maximum profit the monopolist can earn given this demand curve.

**Average Cost**

Average cost at $q = q\_0$ is defined as

$$AC(q\_0) := \frac{C(q\_0)}{q\_0}$$

In this case, $q = 12500$. Hence, $C(q) = C(12500) = 425000$. This means, from the above definition of average cost:

$$AC(12500) = \frac{425000}{12500} = 34$$

Hence, the average cost is $34. Hope that helps.


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