Question

An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. The cost of the material will be least when depth of the tank is

• A. twice of its width
• B. half of its width
• C. equal to its width
• D. None of these

Answer 1

Answer: (B)

Let the length, width and height of the open tank be $x$, $x$ and $y$ units respectively. Then, its volume is $x^{2}y$ and the total surface area is $x^2+4xy$.
It is given that the tank can hold a given quantity of water. This means that its volume is constant. Let it be $V$.
$\therefore V=x^{2}y...(i)$
The cost of the material will be least if the total surface area is least. Let $S$ denote the total surface area. Then,
$S=x^2+4xy...(ii)$
We have to minimize S subject to the condition that the volume is constant.
Now, $S=x^2+4xy$
$\Rightarrow S=x^2+\frac{4V}{x}$ [Using $(i)$]
$\Rightarrow \frac{dS}{dx}=2x-\frac{4V}{x^2}$
and $\frac{d^{2}S}{d{x}^2}=2+\frac{8V}{x^3}$
For maximum or minimum values of $S$, we must have
$\frac{dS}{dx}=0$
$\Rightarrow 2x-\frac{4V}{x^2}=0$
$\Rightarrow 2x^3=4V$
$\Rightarrow 2x^3=4x^{2}y$
$\Rightarrow x=2y$
Clearly, $\frac{d^{2}S}{d{x}^2}=2+\frac{8V}{x^3}>0$ for all $x$.
Hence, $S$ is minimum when $x-2y$ i.e. the depth (height) of the tank is half of its width.