The manufacturer of a CD player has found that the revenue R(in dollars) is R(p)= -4p^2+1460p, whenthe unit price is p dollars. If the manufacturer sets the price p to maximize revenue, what is the maximum revenue to the nearest whole dollar?
2007-10-11 10:40:03 · 3 answers · asked by Pinkigal 2 in Science & Mathematics ➔ Mathematics
This problem really amounts to finding the vertex of a parabola.
R(p) = -4p² + 1460p
R = -4(p² - 365p)
R - 4(365/2)² = -4[p² - 365p + (365/2)²]
R - 133,225 = -4(p - 365/2)²
The maximum revenue is $133,225 at price
p = 365/2 = $182.50.
2007-10-11 10:52:34 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
$133225
Setup Y=-4x^2+1460x in the graphing calculator.
Setup windows as 0, 400, 1, 0, 200000, 1, 1, respectively.
Go to calculate function and select maximum.
Go to the left side of the curve,hit enter.
Go to the right, hit enter.
Eyeball the center, hit enter.
Use Y as the maximum revenue in dollars.
2007-10-11 10:59:25 · answer #2 · answered by Jessie 3 · 0⤊ 0⤋
R = -4p^2 + 1460p
dR/dp = -8p +1460
for maximum dR/dp = 0
-8p + 1460 = 0
p = 1460 / 8 = 182.5
the maximum revenue would be:
R = -4*(182.5)^2 + 1460*(182.5) = 133,225
2007-10-11 10:49:56 · answer #3 · answered by landonastar 3 · 0⤊ 0⤋