Question

An open tank with a square base and vertical side is to be constructed 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

Solution:-

Let l be the length and width of square base and h be the depth of given tank.

$\therefore$ Volume of the tank, $V=l^{2}h$

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

Total surface area, $S=l^2+4lh$

$\Rightarrow S=l^2+4l\times \frac{V}{l^2}\left(\text{From}\left(1\right)\right)$

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

Differentiating above equation w.r.t. l, we have

$\frac{dS}{dl}=2l+\left(\frac{-4V}{l^2}\right)$

$\Rightarrow \frac{dS}{dl}=2l-\frac{4V}{l^2}..........\left(2\right)$

$\frac{dS}{dl}=0$, we have

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

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

$\Rightarrow l^3=2V..........\left(3\right)$

Now, from ${eq}^n\left(1\right)\&\left(3\right)$, we have

$h=\frac{l}{2}$, i.e., depth of the tank is half of its width

Now, differentiating ${eq}^n\left(2\right)$ w.r.t. l, we have

$\frac{d^{2}S}{d{l}^2}=2-\left(\frac{-2\times 4V}{l^3}\right)$

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

$\Rightarrow \frac{d^{2}S}{d{l}^2}=2+\frac{8V}{2V}\left(\text{From}\left(3\right)\right)$

$\Rightarrow \frac{d^{2}S}{d{l}^2}=2+4=6>0$

$\therefore$ S is minimum.

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