# Help trying to find Economic Profit for a Monopoly

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.

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

• Do you mean $p = 60 - 0.002q$ is your inverse demand function? Generally, $q$ is the output and $p$ is the price. Aug 11, 2015 at 4:06
• Price is on the y-axis. Aug 11, 2015 at 4:10
• That is an inverse demand function then, not a demand function. Aug 11, 2015 at 4:10
• First year econ. student, they've taught it as nothing other than a demand function lol. Aug 11, 2015 at 4:11
• Hopefully this is corrected in your upper division courses. For future reference: en.wikipedia.org/wiki/Inverse_demand_function :-) Aug 11, 2015 at 4:14

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 $q$.

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.