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?

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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 $R_max$ .

$color(magenta)(BONUS :$

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

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

The manager of a CD store has found that if the price of a CD is p(x)= 75-x/6p(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^2/6R(x) =75x-x^2/6 How do you find the number of CDs that will produce maximum revenue?