Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal to the diameter of its base.

Solution

let r and h be the radius and height of the cylinder. Then,

$A = 2 π r h + 2 π r^{2} (Given) \Rightarrow h = \frac{A - 2 π r^{2}}{2 π r} Now, Volume (V) = π r^{2} h \Rightarrow V = π r^{2} (\frac{A - 2 π r^{2}}{2 π r}) = \frac{1}{2} A r - 2 π r^{3} \Rightarrow \frac{dv}{dr} = \frac{1}{2} A - 6 π r^{2} . . . . . . . . . . . . . . (i) \Rightarrow \frac{d^{2} v}{{dr}^{2}} = \frac{1}{2} - 12 π r . . . . . . . . . . . . . . (i)$

$Now, \frac{dV}{dr} = 0 \Rightarrow \frac{1}{2} A - 6 π r^{2} = 0 \Rightarrow r^{2} = \frac{A}{6 π} \Rightarrow r = \sqrt{\frac{A}{6 π}} Now, {|\frac{d^{2} V}{{dr}^{2}}|}_{r = \sqrt{\frac{A}{6 π}}} = \frac{1}{2} (- 12 π \sqrt{\frac{A}{6 π}}) < 0 Therefore, Volume is maximum at r = \sqrt{\frac{A}{6 π}}$

$\Rightarrow r^{2} = \frac{A}{6 π} \Rightarrow 6 π r^{2} = A \Rightarrow 6 π r^{2} = 2 π r h + 2 π r^{2} \Rightarrow 4 π r^{2} = 2 π r h \Rightarrow 2 r = h$

Hence, the volume is maximum if its height is equal to its diameter.

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Relations And Functions

Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal to the diameter of its base.

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