Browse Questions

# An open box with square base is to be made of a given quantity of card board of area $c^2$.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 $f(x)$
• (i) Find $f'(x)$ and put $f'(x)=0$
• (ii) Obtain the points from $f'(x)=0$
• (iii) 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 $l, x\: and \: y$ 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