Answer
$a.\quad V$ has a maximum of $144$ when $x=2.$
$ b.\quad$
The largest volume of the box is achieved when $x=2$ units. This volume is $144$ cubic units.
Work Step by Step
$V(x)=x(160-20x-32x+4x^{2})=4x^{3}-52x^{2}+160x,\quad x\in(0,5)$
$V'(x)=12x^{2}-104x+160$
$V'(x)$ is defined for all $x\in(0,5)$: no critical points caused.
$V'(x)=0$ when
$12x^{2}-104x+160=0$
$ 3x^{2}-26x+40=0\qquad$
$x=\displaystyle \frac{26\pm\sqrt{26^{2}-4(3)(40)}}{2(3)}=\frac{26\pm\sqrt{196}}{6}=\frac{26\pm 14}{6}$
$x=\displaystyle \frac{26+14}{6}=\frac{20}{3}\not\in(0,5)$, so it is not a critical value.
$x=\displaystyle \frac{26+14}{6}=2\in(0,2) \qquad$, which is a critical value.
At $x=2,$ $V$ has a maximum of $f(2)=4(8)-52(4)+160(2)=144$.
$(b)$
In terms of the volume of the box:
The largest volume of the box is achieved when $x=2$ units. This volume is $144$ cubic units.
