SOLUTION: suppose that the manufacturer of a dvd player has found that, when the unit price is p dollars, the revenue r (in dollars) as a function of the price p is r(p)=-2.5p^2+800p.
(a)
Algebra ->
Customizable Word Problem Solvers
-> Numbers
-> SOLUTION: suppose that the manufacturer of a dvd player has found that, when the unit price is p dollars, the revenue r (in dollars) as a function of the price p is r(p)=-2.5p^2+800p.
(a)
Log On
Question 1115177: suppose that the manufacturer of a dvd player has found that, when the unit price is p dollars, the revenue r (in dollars) as a function of the price p is r(p)=-2.5p^2+800p.
(a) for what price will the revenue be maximized?
(b) what is the maximum revenue? Answer by ikleyn(52787) (Show Source):
The revenue will be maximized when the quadratic function
r(p) = =-2.5p^2 + 800p
will get the maximum.
It will happen when p = = = 160 dollars.
It is optimal price for dvd player.
And the maximum revenue then will be then -2.5*160^2 + 800*160 = 64000 dollars.
