## University Calculus: Early Transcendentals (3rd Edition)

Hence, the maximum area is $32$ for a rectangle with dimensions of 4 and 8.
Let $A(x)$ be the area. Then, we have $A(x)=2x(12-x^2)$ and $f'(x)=24x-6x^2$ for $x \gt 0$ Maximum value of $A(x)$ will be $A'(x)=0$ when $x=2$ Thus, we have a local maximum at $x=2$ (when $x \gt 0$) and the maximal area is when $x=2$; with dimensions $2x \implies (2)(2)=4$ and $12-(2)^2=12-4=8$. Thus, $A(2)= 4 \cdot 8=32$ Hence, the maximum area is $32$ for a rectangle with dimensions of 4 and 8.