Question

Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible when revolved about one of its sides.

Answer 1

$$\begin{gathered} \mathrm{Let}l,b\mathrm{and}V\mathrm{be}\mathrm{the}\mathrm{length},\mathrm{breadth}\mathrm{and}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{rectangle},\mathrm{respectively}.\mathrm{Then}, \\ 2\left(l+b\right)=36 \\ \Rightarrow l=18-b...\left(1\right) \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cylinder}\mathrm{when}\mathrm{revolved}\mathrm{about}\mathrm{the}\mathrm{breadth},V=\pi l^{2}b \\ \Rightarrow V=\pi {\left(18-b\right)}^{2}b\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow V=\pi \left(324b+b^3-36b^2\right) \\ \Rightarrow \frac{dV}{db}=\pi \left(324+3b^2-72b\right) \\ \mathrm{For}\mathrm{the}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}V,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dV}{db}=0 \\ \Rightarrow \pi \left(324+3b^2-72b\right)=0 \\ \Rightarrow 324+3b^2-72b=0 \\ \Rightarrow b^2-24b+108=0 \\ \Rightarrow b^2-6b-18b+108=0 \\ \Rightarrow \left(b-6\right)\left(b-18\right)=0 \\ \Rightarrow b=6,18 \\ \mathrm{Now}, \\ \frac{d^{2}V}{d{b}^2}=\pi \left(6b-72\right) \\ \mathrm{At}b=6: \\ \frac{d^{2}V}{d{b}^2}=\pi \left(6\times 6-72\right) \\ \Rightarrow \frac{d^{2}V}{d{b}^2}=-36\pi <0 \\ \mathrm{At}b=18: \\ \frac{d^{2}V}{d{b}^2}=\pi \left(6\times 18-72\right) \\ \Rightarrow \frac{d^{2}V}{d{b}^2}=36\pi >0 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value\; of}b\mathit{}\mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ l=18-6=12 \\ \mathrm{So},\mathrm{the}\mathrm{volume}\mathrm{is}\mathrm{maximum}\mathrm{when}l=12\mathrm{cm}\mathrm{and}b=6\mathrm{cm}. \end{gathered}$$