Question

An open box with a square base is to be made out of a given quantity of a cardboard of area $c^2$ square units. The maximum volume of the box is (in cubic units)

• A. $\frac{c^{2}}{2\sqrt{3}}$
• B. $\frac{c^{3}}{6\sqrt{3}}$
• C. $4c/5$
• D. $6c^2$

Answer 1

Answer: (B)

Let the length, breadth and height of the box $x$, $x$ and $y$ units respectively. Then,
$x^{2 }+ 4xy = c^{2} \quad\ldots\left(i\right)$
Let $V$ be the volume of the box. Then,
$V = x^{2}y \quad\ldots\left(ii\right)$
$\Rightarrow V = x^{2}\left(\frac{c^{2}-x^{2}}{4x}\right)$ [Using $\left(i\right)$]
$\Rightarrow V = \frac{c^{2}}{4} x - \frac{x^{3}}{4}$
$\Rightarrow \frac{dV}{dx} = \frac{c^{2}}{4} - \frac{3x^{2}}{4} $
and $\frac{d^{2}V}{dx^{2}} = -\frac{3x}{2}$
For maximum or minimum, we must have
$\frac{dV}{dx} = 0$
$\Rightarrow \frac{c^{2}}{4} - \frac{3x^{2}}{4} = 0$
$\Rightarrow x = \frac{c}{\sqrt{3}}$
$\left[\text{neglecting}\, x = - \frac{c}{\sqrt{3}} \because \frac{d^{2}V}{dx^{2}}\bigg|_{x = - \frac{c}{\sqrt{3}}} > 0\right]$
Now, $\left(\frac{d^{2}V}{dx^{2}}\right)_{x = - \frac{c}{\sqrt{3}}} = \frac{-3c}{2\sqrt{3}} < 0$
Thus, $V$ is maximum when $x = \frac{c}{\sqrt{3}}$
Putting $x = \frac{c}{\sqrt{3}}$ in $\left(i\right)$,
we obtain $y =\frac{c}{2\sqrt{3}}$
The maximum volume of the box is given by
$V = x^{2}y = \frac{c^{2}}{3} \times \frac{c}{2\sqrt{3}}$
$ = \frac{c^{3}}{6\sqrt{3}}$ cubic units