Answer
$4000$ $cm^{3}$
Work Step by Step
Let $b$ be the length of the base of the box and $h$ the height.
The surface area is
$1200$ = $b^{2}+4hb$
$h$ = $\frac{1200-b^{2}}{4b}$
The volume is
$V$ = $b^{2}h$ = $b^{2}\left(\frac{1200-b^{2}}{4b}\right)$ = $300b-\frac{b^{3}}{4}$
$V'(b)$ = $300-\frac{3}{4}b^{2}$ = $0$
$300$ = $\frac{3}{4}b^{2}$
$b$ = $20$
$V'(b)$ $\gt$ $0$ for $0$ $\lt$ $b$ $\lt$ $20$
$V'(b)$ $\lt$ $0$ for $b$ $\gt$ $20$
Therefore there is an absolute maximum when $b$ = $20$ by the First Derivative Test for Absolute Extreme Values.
If $b$ = $20$, then $h$ = $\frac{1200-20^{2}}{4(20)}$ = $10$
so the largest possible volume is
$V_{max}$ = $b^{2}h$ = $(20)^{2}(10)$ = $4000$ $cm^{3}$
