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 $x^2y$ 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 $V$.
$\therefore V = x^2\, y\quad ...(i)$
The cost of the material will be least if the total surface area is least. Let $S$ denote the total surface area. Then,
$S = x^2 + 4xy \quad ...(ii)$
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}{dx^{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, $\frac{d^{2}S}{dx^{2}} = 2 + \frac{8V}{x^{3}} > 0$ for all $x$.
Hence, $S$ is minimum when $x -2\, y$ i.e. the depth (height) of the tank is half of its width.