Step 1

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

Area = $c^2$ sq. units.

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

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

$v=x^2y$

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

$v=\large\frac{c^2}{4}x-\large\frac{x^3}{4}$

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

$ \large\frac{dv}{dx}=\large\frac{c^2}{4}-\large\frac{3x^2}{4}$

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

$ \large\frac{d^2v}{dx^2}=-\large\frac{3x}{2}$

For maximum or minimum , we must have,

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

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

$ \Rightarrow x=\large\frac{c}{\sqrt 3}$

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

Thus, $v$ is maximum when $x=\large\frac{c}{\sqrt 3}$

Put $x=\large\frac{c}{\sqrt 3}$, we get

$ y=\large\frac{c}{2\sqrt 3}$

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

$ v= x^2y$

$=\large\frac{c^2}{3} \times \large\frac{c}{2\sqrt 3}$

$ = \large\frac{c^3}{6\sqrt 3}$ cubic units