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?

  • $\begingroup$ Do you mean $p = 60 - 0.002q$ is your inverse demand function? Generally, $q$ is the output and $p$ is the price. $\endgroup$ – ml0105 Aug 11 '15 at 4:06
  • $\begingroup$ Price is on the y-axis. $\endgroup$ – Romaion Aug 11 '15 at 4:10
  • 1
    $\begingroup$ That is an inverse demand function then, not a demand function. $\endgroup$ – ml0105 Aug 11 '15 at 4:10
  • $\begingroup$ First year econ. student, they've taught it as nothing other than a demand function lol. $\endgroup$ – Romaion Aug 11 '15 at 4:11
  • $\begingroup$ Hopefully this is corrected in your upper division courses. For future reference: en.wikipedia.org/wiki/Inverse_demand_function :-) $\endgroup$ – ml0105 Aug 11 '15 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 $q$.


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