Step 1

Since the box is of square base,

Let the length, breadth and height of the box be $x, x\: and \: y$ units respectively.

Given: Area of the box = $36$ sq. metres.

We know that area of the open box=$x^2+4xy$

$ \therefore x^2+4xy=36 \Rightarrow y=\large\frac{36-x^2}{4x}$

Let $V$ be the volume of the box, then

$V=x^2y$

$ \Rightarrow V=x^2 \bigg(\large\frac{36-x^2}{4x}\bigg)$

$V=\large\frac{36}{4}$$x-\large\frac{x^3}{4}$

Differentiate w.r.t $x$ we get,

$\Rightarrow\: \large\frac{dV}{dx}$$=9-\large\frac{3x^2}{4}$

Differentiate again $\large\frac{dV}{dx }$ w.r.t $x$ we get,

$\Rightarrow\: \large\frac{d^2V}{dx^2}=-\large\frac{3x}{2}$

For maximum or minimum , we must have,

$ \large\frac{dV}{dx}$$=0 \Rightarrow 9-\large\frac{3x^2}{4}$$=0$

$ \Rightarrow \large\frac{3x^2}{4}$$=9$

$ \Rightarrow x=2\sqrt 3$

$ \bigg( \large\frac{d^2V}{dx^2} \bigg)_{x=2\sqrt 3}$$=-3\sqrt 3 <0$

Thus, $V$ is maximum when $x=2\sqrt 3$

Put $x=2\sqrt 3$, we get

$ y=\sqrt 3$

$ \therefore$ The maximum volume of the box is given by

$ V= x^2y=12\sqrt 3$ cubic metres.