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# 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. - CBSE (Science) Class 12 - Mathematics

#### 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

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

$2\left( l + b \right) = 36$

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

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

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

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

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

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

$\frac{dV}{db} = 0$

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

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

$\Rightarrow b^2 - 24b + 108 = 0$

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

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

$\Rightarrow b = 6, 18$

$\text { Now,}$

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

$\text { At }b = 6:$

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

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

$\text{ At } b= 18:$

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

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

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

$l = 18 - 6 = 12$

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

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Solution 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. Concept: Graph of Maxima and Minima.
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