#### 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

\[\text { Let l, h, V and S be the length, height, volume and surface area of the tank to be constructed }. \]

\[\text { Since volume, V is constant,} \]

\[ l^2 h = V\]

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

\[\text { Surface area, S = } l^2 + 4lh\]

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

\[ \Rightarrow \frac{dS}{dl} = 2l - \frac{4V}{l^2}\]

\[\text { For S to be maximum or minimum, we must have }\]

\[\frac{dS}{dl} = 0\]

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

\[ \Rightarrow 2 l^3 - 4V = 0\]

\[ \Rightarrow 2 l^3 = 4V\]

\[ \Rightarrow l^3 = 2V\]

\[\text { Now, }\]

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

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

\[\text { Here, surface area is minimum }. \]

\[h = \frac{V}{l^2}\]

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

}\]

\[h = \frac{l^3}{2 l^2}\]

\[ \Rightarrow h = \frac{l}{2}\]

\[\text { Hence proved }.\]