Answer
Maximal area $= 32$ for a rectangle with sides 4 and 8
Work Step by Step
Let $A(x)$ be the area. Then, we have $A(x)=2x(12-x^2)$ and $f'(x)=24-6x^2$ for $x \gt 0$
Need to find the maximal value of $A(x)$ .
when $A'(x)=0$ then $x=2$
This implies that the local maximum is at $x=2$ when $x \gt 0$.
Thus, the maximal area is when $x=2$ ; with dimensions $2x=(2)(2)=4$ and $12-2^2=8$. So, $A(2)=(4)(8)=32$
Hence, Maximal area $= 32$ for a rectangle with sides 4 and 8
