what is the maximum revenue to the nearest whole​ dollar

Question

The manufacturer of a CD player has found that the revenue R​ (in dollars) is

R(p)=−5p2+1,520p,

when the 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

Answer 1

R(p) = -5p²+1520p = ap²+bp

This is a parabola opening down. The vertex of the parabola is: p=-b/(2a)

p = -1520/-10 = 152

R(152) = -5(152)²+1520(152) = -115520 + 231040 = $115520

Another way to look at it:

R = -5p(p-304), roots are p = 0 and 304 found when R=0

The vertex is half way between the x-axis crossings. (304+0)/2 = 152.

Answer 2

Hello, Gabriela,

The maximum of R(p)=−5p2+1,520p can be found either by graphing the equation, or by taking the first derivative. Try the Desmos graphing site and enter y=-5x²+1520x. You'll find a maximum at (152, 115529). At a price of $152, the revenue is $115,529.

Or we can differentiate:

dy/dy = -10x-1520

Set this equal to 0, the slope at the the maximum point when the curve reverses.

0 = -10x - 1520

x = $152

Enter $152 for p in the original equation to find the revenue for this price.

R(p)=−5p2+1,520p

R(152) = -5(152)^2 + (1520)*(152)

R(152) = $115,520

Bob