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

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

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Let r be the radius of the base of cylinder and h be the height of the cylinder
we know that the surface area of the cylinder (A) = 2\pi r(r+h)
h = \frac{A-2\pi r^2}{2\pi r}
Volume of cylinder
 (V) = \pi r^2 h\\ = \pi r^2 \left ( \frac{A-2\pi r^2}{2\pi r} \right ) = r \left ( \frac{A-2\pi r^2}{2 } \right )
V^{'}(r)= \left ( \frac{A-2\pi r^2}{2} \right )+(r).(-2\pi r)\\ = \frac{A-2\pi r^2 -4\pi r^2}{2} = \frac{A-6\pi r^2}{2}
V^{'}(r)= 0 \\ \frac{A-6\pi r^2}{2} = 0\\ r = \sqrt{\frac{A}{6\pi}}
Hence, r = \sqrt{\frac{A}{6\pi}} is the critical point
Now,
V^{''}(r) = -6\pi r\\ V^{''}(\sqrt{\frac{A}{6\pi}}) = - 6\pi . \sqrt{\frac{A}{6\pi}} = - \sqrt{A6\pi} < 0
Hence,  r = \sqrt{\frac{A}{6\pi}} is the point of maxima 
h = \frac{A-2\pi r^2}{2\pi r} = \frac{2-2\pi \frac{A}{6\pi}}{2\pi \sqrt \frac{A} {6\pi}} = \frac{4\pi \frac{A}{6\pi}}{2\pi \sqrt \frac{A} {6\pi}} = 2\pi \sqrt \frac{A} {6\pi} = 2r
Hence,  the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter(D = 2r) of the base

Posted by

Gautam harsolia

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