Question

A closed cylinder has volume 2156 cm³. What will be the radius of its base so that its total surface area is minimum.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{height},\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{base}\mathrm{and}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cylinder}\mathrm{be}h,r\mathrm{and}S,\mathrm{respectively}.\mathrm{Then}, \\ \mathrm{Volume}=\pi r^{2}h \\ \Rightarrow 2156=\pi r^{2}h \\ \Rightarrow 2156=\frac{22}{7}{r}^{2}h \\ \Rightarrow h=\frac{2156\times 7}{22r^2} \\ \Rightarrow h=\frac{686}{r^2}...\left(1\right) \\ \mathrm{Surface}\mathrm{area}=2\pi rh+2\pi r^2 \\ \Rightarrow S=\frac{4312}{r}+\frac{44r^2}{7}\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow \frac{dS}{dr}=\frac{4312}{-r^2}+\frac{88r}{7} \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}S,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dS}{dr}=0 \\ \Rightarrow \frac{4312}{-r^2}+\frac{88r}{7}=0 \\ \Rightarrow \frac{4312}{r^2}=\frac{88r}{7} \\ \Rightarrow r^3=\frac{4312\times 7}{88} \\ \Rightarrow r^3=343 \\ \Rightarrow r=7\mathrm{cm} \\ \mathrm{Now}, \\ \frac{d^{2}s}{d{r}^2}=\frac{8624}{r^3}+\frac{88}{7} \\ \Rightarrow \frac{d^{2}s}{d{r}^2}=\frac{8624}{343}+\frac{88}{7} \\ \Rightarrow \frac{d^{2}s}{d{r}^2}=\frac{176}{7}>0 \\ \mathrm{So},\mathrm{the}\mathrm{surface}\mathrm{area}\mathrm{is}\mathrm{minimum}\mathrm{when}r=7\mathrm{cm}. \end{gathered}$$