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.

Answers (1)

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

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