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.

Answer 1

$$\begin{gathered} \mathrm{Let}l,h,V\mathrm{and}S\mathrm{be}\mathrm{the}\mathrm{length},\mathrm{height},\mathrm{volume}\mathrm{and}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{tank}\mathrm{to}\mathrm{be}\mathrm{constructed}. \\ \mathrm{Since}\mathrm{volume},V\mathrm{is}\mathrm{constant}, \\ {l}^{2}h=V \\ \Rightarrow h=\frac{V}{l^2}...\left(1\right) \\ \mathrm{Surface}\mathrm{area},S=l^2+4lh \\ \Rightarrow S=l^2+\frac{4V}{l}\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow \frac{dS}{dl}=2l-\frac{4V}{l^2} \\ \mathrm{For}S\mathrm{to}\mathrm{be}\mathrm{maximum}\mathrm{or}\mathrm{minimum},\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dS}{dl}=0 \\ \Rightarrow 2l-\frac{4V}{l^2}=0 \\ \Rightarrow 2l^3-4V=0 \\ \Rightarrow 2l^3=4V \\ \Rightarrow l^3=2V \\ \mathrm{Now}, \\ \frac{d^{2}S}{d{l}^2}=2+\frac{8V}{l^3} \\ \Rightarrow \frac{d^{2}S}{d{l}^2}=2+\frac{8V}{2V}=6>0 \\ \mathrm{Here},\mathrm{surface}\mathrm{area}\mathrm{is}\mathrm{minimum}. \\ h=\frac{V}{l^2} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}V=\frac{l^3}{2}\mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ h=\frac{l^3}{2l^2} \\ \Rightarrow h=\frac{l}{2} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$