SOLUTION: Maximizing Revenue Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p dollars, the revenue R (in $) is R (p) = -4p^2 + 4000p What unit

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Maximizing Revenue Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p dollars, the revenue R (in $) is R (p) = -4p^2 + 4000p What unit      Log On


   

Question 54927: Maximizing Revenue Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p dollars, the revenue R (in $) is
R (p) = -4p^2 + 4000p
What unit price should be established for the dryer to maximize revenue? What is the maximum revenue?
Textbook College Algebra 7th ed. Sullivan. Pearson Prentice Hall
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
The equation is a parabola in the form R%28p%29+=+ap%5E2+%2Bbp+%2Bc that opens downward, and it's maximum will occur at the value where p=-b%2F%282a%29.

In this case, a=-4 and b=4000, so p=+-b%2F%282a%29+=+-4000%2F%28-8%29=500 dollars

Maximum revenue will be
R%28500%29+=+-4%28500%29%5E2+%2B+4000%28500%29=1000000 dollars

Did I do the math right? I left my calculator in the other room!!

By the way, can you use a graphing calculator on this???

R^2 at SCC