Question

Find the number of CDs that will produce maximum revenue.

Answer 1

Given data:

Price of CD is,

$p(x)=90-\frac{x}{6}$

The total revenue is,

$R(x)=90x-\frac{x^2}{6}$

First find the derivative of revenue function and then equate it to zero we have,

$\begin{gathered} R^{\prime}(x)=0 \\ 90-\frac{2x}{6}=0 \end{gathered}$$\begin{gathered} \frac{x}{3}=90 \\ x=90\times3 \\ x=270 \end{gathered}$

Now, to prove the maximize find the double derivative of revenue function

$\begin{gathered} R^{\doubleprime}(x)<0 \\ \frac{-2}{6}=\frac{-1}{3}<0 \end{gathered}$

Thus, 270 CD's will produce maximum revenue.

Answer: Option (c) that is 270.