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 born by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question?

Answer 1

According to the question, the tank must hold a given quantity of water, so the volume of water can be assumed to be fixed. Let its value be denoted as $C$ cubic units, a real constant.

The base of the tank is square, so, let the length of its side be $a$ units and the depth of the tank be $x$ units (a variable quantity whose value in terms of $a$ is to be determined).

The volume of the tank is clearly area of base times depth $=a^{2}x=C$(fixed).

Since the tank is open at the top, its surface area is: (bottom area + 4 walls)

$a^2+4ax$ square units.

If the cost of sheet metal material per square meter is $p$ (fixed), then the cost of building the tank is:

$p(a^2+4ax)$

So, the problem is to minimize $p(a^2+4ax)$ subject to the constraint $a^{2}x=C$.

$\Rightarrow a=\sqrt{\frac{C}{x}}$

$\Rightarrow p(a^2+4ax)=p(\frac{C}{x}+4\sqrt{Cx})$

Differentiating the above expression with respect to x and equating to zero to find the minimum, we get:

$p(-\frac{C}{x^2}+\frac{2\sqrt{C}}{\sqrt{x}})=0\Rightarrow x=\sqrt[3]{3\frac{C}{4}}\Rightarrow a=\sqrt[3]{32C}$

Clearly, we can see from the above that $x=\frac{a}{2}$

So, the cost of material is minimized when the depth is half of the width.