The manager of a CD store has found that if the price of a CD is #p(x)= 75-x/6# then x CDs will be sold. An expression for the total revenue from the sale of x CDs is #R(x) =75x-x^2/6# How do you find the number of CDs that will produce maximum revenue?

Question

5808 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-10T13:57:03

#225# CDs will produce the maximum revenue.

We know from Calculus that, for #R_(max)#, we must have,

#R'(x)=0, and, R''(x) lt 0#.

Now, #R(x)=75x-x^2/6 rArr R'(x)=75-1/6*2x=75-x/3#.

#:. R'(x)=0 rArr x/3=75, or, x=75*3=225#.

Further, #R'(x)=75-x/3 rArr R''(x)=-1/3 lt 0," already"#.

Hence, #x=225" gives "R_(max)#.

Thus, #225# CDs will produce the maximum revenue #R_max#.

#color(magenta)(BONUS :#

#R_max=R(225)=75*225-225^2/6=8437.5, and #

#"Price of a CD="p(225)=75-225/6=37.5#.