Answer
The largest possible volume of the box is $~~4000~cm^3$
Work Step by Step
Let $x$ be the length of one side of the base and let $h$ be the height of the box.
Note that $x \geq 0$ and $h \geq 0$
We can express $h$ in terms of the area $A$ and $x$:
$A = x^2+4xh$
$h = \frac{A-x^2}{4x}$
We can write an expression for the volume:
$V = x^2h = x^2(\frac{A-x^2}{4x}) = \frac{Ax}{4}- \frac{x^3}{4}$
We can find the point where $V'(x) = 0$:
$V(x) = \frac{Ax}{4}- \frac{x^3}{4}$
$V'(x) = \frac{A}{4}- \frac{3x^2}{4} = 0$
$A-3x^2 = 0$
$x^2 = \frac{A}{3}$
$x = \sqrt{\frac{A}{3}}$
$x = \sqrt{\frac{1200~cm^2}{3}}$
$x = \sqrt{400~cm^2}$
$x = 20~cm$
When $0 \lt x \lt 20$, then $L'(x) \gt 0$
When $x \gt 20$, then $L'(x) \lt 0$
Thus, $x=20~cm$ is the point where $V$ is a maximum.
We can find the largest possible volume:
$V(20) = \frac{Ax}{4}- \frac{x^3}{4}$
$V(20) = \frac{(1200~cm^2)(20~cm)}{4}- \frac{(20~cm)^3}{4}$
$V(20) = 6000~cm^3-2000~cm^3$
$V(20) = 4000~cm^3$
The largest possible volume of the box is $~~4000~cm^3$
