Question

An open tank with 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 material will be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question ?

Answer 1

Let $x$ be the side of the square base and $y$ be the height, i.e., vertical side of the water tank.

$\therefore$ Volume of the tank $\left(V\right)=x^{2}y$

$\Rightarrow y=\frac{V}{x^2}$

Also, surface area $=x^2+4xy$

$=x^2+4x\cdot \frac{V}{x^2}=x^2+4V{x}^{-1}$

If $k$ be the rate of metal sheet used per square units.


$\therefore$ Total cost of sheet $\left(C\right)=k(x^2+4V{x}^{-1})$


$\frac{dC}{dx}=k(2x-4V{x}^{-2})=k(2x-\frac{4V}{x^2})$

$\frac{d^{2}C}{d{x}^2}=k(2+8V{x}^{-3})=k(2+\frac{8V}{x^3})$

For minimum cost, put $\frac{dC}{dx}=0$

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

$\Rightarrow 2k(x^3-2V)=0$

$\Rightarrow x^3=2V=2\times x^{2}y$ .....$\left[i\right]$

$\Rightarrow x=2y$

$\Rightarrow y=\frac{1}{2}x$

Now, $\frac{d^{2}C}{d{x}^2}=k$ $(2+\frac{8V}{x^3})$

=k $(2+\frac{8V}{2V})$ = 6$k$ = positive from $\left[i\right]$

Cost is least when $y=\frac{1}{2}x$.

Hence, the cost of material will be least when depth of the tank is half of its width.