Given data:
Price of CD is,
[tex]p(x)=90-\frac{x}{6}[/tex]The total revenue is,
[tex]R(x)=90x-\frac{x^2}{6}[/tex]First find the derivative of revenue function and then equate it to zero we have,
[tex]\begin{gathered} R^{\prime}(x)=0 \\ 90-\frac{2x}{6}=0 \end{gathered}[/tex][tex]\begin{gathered} \frac{x}{3}=90 \\ x=90\times3 \\ x=270 \end{gathered}[/tex]Now, to prove the maximize find the double derivative of revenue function
[tex]\begin{gathered} R^{\doubleprime}(x)<0 \\ \frac{-2}{6}=\frac{-1}{3}<0 \end{gathered}[/tex]Thus, 270 CD's will produce maximum revenue.
Answer: Option (c) that is 270.