Question

Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm³, which has the minimum surface area?

Answer 1

$$\begin{gathered} \mathrm{Let}r\mathrm{and}h\mathrm{be}\mathrm{the}\mathrm{radius}\mathrm{and}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{cylinder},\mathrm{respectively}.\mathrm{Then}, \\ \mathrm{Volume}\left(V\right)\mathrm{of}\mathrm{the}\mathrm{cylinder}=\mathrm{\pi }{r}^{2}h \\ \Rightarrow 100=\mathrm{\pi }{r}^{2}h \\ \Rightarrow h=\frac{100}{\mathrm{\pi }{r}^2} \\ \mathrm{Surface}\mathrm{area}\left(S\right)\mathrm{of}\mathrm{the}\mathrm{cylinder}=2\mathrm{\pi }{r}^2+2\mathrm{\pi }rh=2\mathrm{\pi }{r}^2+2\mathrm{\pi }r\times \frac{100}{\mathrm{\pi }{r}^2} \\ \Rightarrow S=2\mathrm{\pi }{r}^2+\frac{200}{r} \\ \therefore \frac{dS}{dr}=4\mathrm{\pi }r-\frac{200}{r^2} \\ \mathrm{For}\mathrm{the}\mathrm{maximum}\mathrm{or}\mathrm{minimum},\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dS}{dr}=0 \\ \Rightarrow 4\mathrm{\pi }r-\frac{200}{r^2}=0 \\ \Rightarrow 4\mathrm{\pi }{r}^3=200 \\ \Rightarrow r={\left(\frac{50}{\mathrm{\pi }}\right)}^{\frac{1}{3}} \\ \mathrm{Now}, \\ \frac{{\mathrm{d}}^2\mathrm{S}}{\mathrm{d}{\mathrm{r}}^2}=4\mathrm{\pi }+\frac{400}{{\mathrm{r}}^3} \\ \Rightarrow \frac{d^{2}S}{d{r}^2}>0\mathrm{when}r={\left(\frac{50}{\mathrm{\pi }}\right)}^{\frac{1}{3}} \\ \mathrm{Thus},\mathrm{the}\mathrm{surface}\mathrm{area}\mathrm{is}\mathrm{minimum}\mathrm{when}r={\left(\frac{50}{\mathrm{\pi }}\right)}^{\frac{1}{3}}. \\ \mathrm{At}r={\left(\frac{50}{\mathrm{\pi }}\right)}^{\frac{1}{3}}: \\ h=\frac{100}{\mathrm{\pi }{\left({\displaystyle \frac{50}{\mathrm{\pi }}}\right)}^{\frac{2}{3}}}=2{\left(\frac{50}{\mathrm{\pi }}\right)}^{\frac{1}{3}} \end{gathered}$$