Question

A sheet of area $40m^2$ is used to make an open tank with square base. The side of the base such that the volume of this tank is maximum, is

• A. $\sqrt{\frac{80}{3}}m$
• B. $\sqrt{\frac{40}{3}}m$
• C. $\sqrt{\frac{20}{3}}m$
• D. $\sqrt{\frac{10}{3}}m$

Answer 1

The correct option is B $\sqrt{\frac{40}{3}}m$

Let the length of base be $xm$ and height be $ym$.

Volume, $V=x^{2}y$
Again $x$ and $y$ are related to the surface area of this tank which is equal to $40m^2$
$\Rightarrow x^2+4xy=40$
$\Rightarrow y=\frac{40-x^2}{4x}$
$\Rightarrow V\left(x\right)=x^2\left(\frac{40-x^2}{4x}\right)=\frac{40x-x^3}{4}$
$V^{\prime }\left(x\right)=\frac{40-3x^2}{4}$
$V^{\prime \prime }\left(x\right)=-\frac{6x}{4}$
Maximizing volume,
$V^{\prime }\left(x\right)=\frac{40-3x^2}{4}=0$
$\Rightarrow x=\sqrt{\frac{40}{3}}m$
$V^{\prime \prime }\left(x\right)=\frac{-6x}{4}\Rightarrow V^{\prime \prime }\left(\sqrt{\frac{40}{3}}\right)<0$
$\Rightarrow$ Volume is maximum at $x=\sqrt{\frac{40}{3}}m$