#### 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 ?

#### Solution

\[\text { Let l, b and V be the length, breadth and volume of the rectangle, respectively . Then, }\]

\[2\left( l + b \right) = 36\]

\[ \Rightarrow l = 18 - b . . . \left( 1 \right)\]

\[\text { Volume of the cylinder when revolved about the breadth, V } = \pi l^2 b\]

\[ \Rightarrow V = \pi \left( 18 - b \right)^2 b .............\left[\text{From eq. }\left( 1 \right) \right]\]

\[ \Rightarrow V = \pi\left( 324b + b^3 - 36 b^2 \right)\]

\[ \Rightarrow \frac{dV}{db} = \pi\left( 324 + 3 b^2 - 72b \right)\]

\[\text { For the maximum or minimum values of V, we must have }\]

\[\frac{dV}{db} = 0\]

\[ \Rightarrow \pi\left( 324 + 3 b^2 - 72b \right) = 0\]

\[ \Rightarrow 324 + 3 b^2 - 72b = 0\]

\[ \Rightarrow b^2 - 24b + 108 = 0\]

\[ \Rightarrow b^2 - 6b - 18b + 108 = 0\]

\[ \Rightarrow \left( b - 6 \right)\left( b - 18 \right) = 0\]

\[ \Rightarrow b = 6, 18\]

\[\text { Now,} \]

\[\frac{d^2 V}{d b^2} = \pi\left( 6b - 72 \right)\]

\[\text { At }b = 6: \]

\[\frac{d^2 V}{d b^2} = \pi\left( 6 \times 6 - 72 \right)\]

\[ \Rightarrow \frac{d^2 V}{d b^2} = - 36\pi < 0\]

\[\text{ At } b= 18: \]

\[\frac{d^2 V}{d b^2} = \pi\left( 6 \times 18 - 72 \right)\]

\[ \Rightarrow \frac{d^2 V}{d b^2} = 36\pi > 0\]

\[\text { Substitutingthe value of b in eq. } \left( 1 \right),\text { we get }\]

\[l = 18 - 6 = 12\]

\[\text { So, the volume is maximum when l = 12 cm and b = 6 cm }. \]