# An open tank is to be constructed

Question:

An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expenses of lining with lead with be least, if depth is made half of width.

Solution:

Let $l, h, V$ and $S$ be the length, height, volume and surface area of the tank to be constructed.

Since volume, $V$ is constant,

$l^{2} h=V$

$\Rightarrow h=\frac{V}{l^{2}}$                .....(1)

Surface area, $S=l^{2}+4 l h$

$\Rightarrow S=l^{2}+\frac{4 V}{l}$      $\left[\begin{array}{ll}\text { From eq. } & (1)\end{array}\right]$

$\Rightarrow \frac{d S}{d l}=2 l-\frac{4 V}{l^{2}}$

For $S$ to be maximum or minimum, we must have

$\frac{d S}{\eta}=0$

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

$\Rightarrow 2 l^{3}-4 V=0$

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

$\Rightarrow l^{3}=2 V$

Now,

$\frac{d^{2} S}{d l^{2}}=2+\frac{8 V}{l^{3}}$

$\Rightarrow \frac{d^{2} S}{d l^{2}}=2+\frac{8 V}{2 V}=6>0$

Here, surface area is minimum.

$h=\frac{V}{l^{2}}$

Substituting the value of $V=\frac{l^{3}}{2}$ in eq. $(1)$, we get

$h=\frac{l^{3}}{2 l^{2}}$

$\Rightarrow h=\frac{l}{2}$

Hence proved.