# An open box with a square base is to be made out of a given quantity of cardboard of area $$c^2$$ square units. Show that the maximum volume of the box is $$\large\frac{c^3}{6\sqrt 3}$$ cubic units.

Toolbox:
• To obtain the absolute maxima or minima for the function $f(x)$
• Find $f'(x)$ and put $f'(x)=0$
• Obtain the points from $f'(x)=0$
• By Phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1:
Let the length, breadth and height of the box be $l, x\: and \: y$ respectively.
Area = $c^2$ sq. units.
$\therefore x^2+4xy=c^2$
$y=\large\frac{c^2-x^2}{4x}$
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}$$x-\large\frac{x^3}{4} Step 2: Differentiate w.r.t x we get, \large\frac{dv}{dx}=\large\frac{c^2}{4}-\large\frac{3x^2}{4} Again differentiate w.r.t x we get, \large\frac{d^2v}{dx^2}=-\large\frac{3x}{2} Step 3: For maximum or minimum , we must have, \large\frac{dv}{dx}=$$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 $x=\large\frac{c}{\sqrt 3}$
Step 4:
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