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. Show that the cost of the material will be least when depth of the tank is half of its width.

Answer 1

Let the length, width and height of the open tank be $x$, $x$ and $y$ with respectively. Then its volume is $x^{2}y$ and total surface area is $x^2+4xy$.

It is given that the tank can be hold a given quantity of water. This mean that its volume is constant. Let it be $V$.

$V=x^{2}y$

The cost of the material will be least if the total surface area is least. Let's denote total surface area

$S=x^2+4xy$

We have to minimize $S$ subjects to the condition that the volume is constant.

$S=x^2+4xy$

$S=x^2+\frac{4V}{x}$

$\Rightarrow \frac{dS}{dx}=2x-\frac{4V}{x^2}$

$\Rightarrow \frac{d^{2}S}{{dx}^2}=2+\frac{8V}{x^3}$

For maximum or minimum value of $S$

$\frac{dS}{dx}=0$

$\Rightarrow 2x-\frac{4V}{x^2}=0$

$\Rightarrow {2x}^3=4V$

$\Rightarrow {2x}^3={4x}^{2}y$

$\Rightarrow x=2y$

Clearly $\frac{d^{2}x}{{dx}^2}=2+\frac{8V}{x^3}>0$ for all $x$.

Here $S$ is minimum when $x=2y$ depth of $tanh$ is half of width.