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,
$x^2+4xy=c^2\dots \left(i\right)$
Let $V$ be the volume of the box. Then,
$V=x^{2}y\dots \left(ii\right)$
$\Rightarrow V=x^2\left(\frac{c^2-x^2}{4x}\right)$ [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 $\frac{d^{2}V}{d{x}^2}=-\frac{3x}{2}$
For maximum or minimum, we must have
$\frac{dV}{dx}=0$
$\Rightarrow \frac{c^2}{4}-\frac{3x^2}{4}=0$
$\Rightarrow x=\frac{c}{\sqrt{3}}$
$\left[\text{neglecting}x=-\frac{c}{\sqrt{3}}\because \frac{d^{2}V}{d{x}^2}{|}_{x=-\frac{c}{\sqrt{3}}}>0\right]$
Now, ${\left(\frac{d^{2}V}{d{x}^2}\right)}_{x=-\frac{c}{\sqrt{3}}}=\frac{-3c}{2\sqrt{3}}<0$
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