The manufacturer of a gas clothes dryer has found that when

The manufacturer of a gas clothes dryer has found that when the unit price is p dollars, the revenue, in dollars, is R(p) = -4p^2 + 4000 P. What price will maximize revenue? What is the maximum revenue?

Solution

R(p) = -4p² + 4000p

Maximum Revenue ==> R \'(p) = 0 and R \'\'(p) < 0

R \'(p) = -4(2p) + 4000(1)           since d/dx xⁿ = nx^n-1

==> R \'(p) = -8p + 4000

R \'(p) = 0 ==> -8p + 4000 = 0

==> p = 4000/8 = 500

Hence at price $ 500 , revenue is maximum

Maximum revenue = R(500) = -4(500)² + 4000(500)

= $ 1000000