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 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 length and width of the base of the open box be x and its depth be y. Also let volume of box be V. So, $V=x^{2}y$ $\Rightarrow y=\frac{V}{x^2}$ ....(i)

Let k (in Rs) be the cost per sq.units

So, the cost of material used , $C=k(x^2+4xy)=k\{x^2+\frac{4V}{x}\}$ (By (i)

$\Rightarrow \frac{dC}{dx}=k\{2x-\frac{4V}{x^2}\}=2k\{x-\frac{2V}{x^2}\},\frac{d^{2}C}{d{x}^2}=2k\{1+\frac{4V}{x^3}\}$

For $\Rightarrow \frac{dC}{dx}=0,2k\{x-\frac{2V}{x^2}=0\}{x}^3=2V$ i.e., $x=(2V{)}^{\frac{1}{3}}$

Clearly ${\begin{array}{c}\therefore \frac{d^{2}C}{d{x}^2}\end{array}]}_{atx=2V^{\frac{1}{3}}}$ = $2k\{1+\frac{4V}{2V}\}=6k>0$ so, C is least when $x=(2V{)}^{\frac{1}{3}}$

Now $V=x^{2}y$ $\Rightarrow \frac{x^3}{2}=x^{2}y$ $\Rightarrow y=\frac{x}{2}$ i.e., $\text{depth of box}=\frac{\text{width of box}}{2}$ Any appropraite value.