[hmovva]

The equation of weekly quantity demanded, q, and price p is estimated to be
p = 60 - q/100,000, where p is measured in dollars and q is measured in thousands of kilowatt hours. The company has fixed costs of $7,000,000 per week and vairable costs of $30 per thousand kilowatt hours, so cost
C(x) = 7,000,000 + 30q. At what price is profit a maximum ?

[ArcSine]

• First form a Total Revenue function r(q) by noting that revenue = price x units, so your revenue function is thus the result of multiplying your given price function by q.

• Next form a Profit function π(q) which is the result of

Revenue less Total Cost = r(q) - C(q).

r(q) is what you derived in the previous step; the cost function is given.

Finally, note that the profit function π(q) is a concave quadratic, in this case. Max profit is found either by finding the vertex of the quadratic (if you like the algebraic approach) or by finding where the 1st derivative of π(q) = 0 (if the calculus approach is more to your liking).

(Note, though, that the algebraic approach really comes up short on other (non-quadratic) functions, while the calculus / derivatives approach delivers the goods on a wide variety of functions.)

Start plugging away with those steps, and check back in for more guidance if need be.