Answer
a. At $x=2$, $V(x)=144$ maximum.
b. See explanations.
Work Step by Step
a. Given the function $V(x)=x(10-2x){16-2x}$, we can find its derivative as $V'(x)=(10-2x){16-2x}+x(-2){16-2x}+x(10-2x){-2}=4((x-5)(x-8)+x(x-8)+x(x-5))=4(x^2-13x+40+x^2-8x+x^2-5x)=4(3x^2-26x+40)=4(3x-20)(x-2)$
The critical points can be found when $V'=0$ or undefined which happens when $x=2, 20/3$. The domain of the function $0\lt x\lt 5$ does not contain any endpoints. Discard $x=20/3$ as it is not in the domain.
At $x=2$, $V(x)=2(10-4)(16-4)=144$ and we can identify this as a maximum (use local test points as necessary) as shown in the figure.
b. Consider the volume as the product of three sides and $x$ is the variable which changes the volume. The maximum value $144$ obtained in part (b) represents the maximum volume that can be attained at $x=2$ for all the possible $x$ values in the domain.
