The surface area of an open box with a square base is 36 square units. It's maximum volume (in cubic units) is:

$12\sqrt{3}$

Let base square has edge = x

and height of box = y 

so surface area = $4×xy+x^2=36$ (given)

volume = $yx^2=V$

$S=4xy+x^2=36⇒xy=\frac{36-x^2}{4}$

so $V=xy×x=(\frac{36-x^2}{4})(x)$

$V(x)=9x-\frac{x^3}{4}$ so $V(x)=9-\frac{3x^2}{4}⇒V'(x)=0$ ar $x^2=12$

$x=2\sqrt{3}$ as $x≥0$

$V^n(x)=\frac{-6x}{4}⇒V''(2\sqrt{3})<0$

so $2\sqrt{3}$ → point of local maxima

$V(2\sqrt{3}=9(2\sqrt{3})-\frac{(2\sqrt{3})^3}{4}$

$=18\sqrt{3}-\frac{8×3\sqrt{3}}{4}$

$=12\sqrt{3}$ sq. units