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  • Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R by square root of 3 . Also find the maximum volume.

17) Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is   \frac{2 R }{\sqrt 3 } . Also, find the maximum volume.

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The volume of the cylinder (V) = \pi r^2 h
By Pythagoras theorem in \Delta OAB
OA = \sqrt{R^2-r^2}
h = 2OA
h = 2\sqrt{R^2-r^2}
V = 2\pi r^2\sqrt{R^2-r^2}
V^{'}(r) = 4\pi r\sqrt{R^2-r^2}+2\pi r^2 . \frac{-2r}{2\sqrt{R^2-r^2}}\\ V^{'}(r) = 0\\ 4\pi r\sqrt{R^2-r^2}- \frac{2\pi r^3}{\sqrt{R^2-r^2}} = 0\\ 4\pi r (R^2-r^2 ) - 2\pi r^3 = 0\\ 6\pi r^3 = 4\pi rR^2\\ r =\frac{\sqrt6R}{3}
Now,
V^{''}(r) = 4\pi \sqrt{R^2-r^2}+4\pi r.\frac{-2r}{2\sqrt{R^2-r^2}}- \frac{6\pi r^2}{\sqrt{R^2-r^2}}.\frac{(-1)-2r}{2(R^2-r^2)\frac{3}{2}}\\ V^{''}(\frac{\sqrt6R}{3}) < 0
Hence, the point r = \frac{\sqrt6R}{3} is the point of maxima
h = 2\sqrt{R^2-r^2} = = 2\sqrt{R^2 - \frac{2R^2}{3}} =\frac{2R}{\sqrt3}
Hence,  the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is   \frac{2 R }{\sqrt 3 }
and maximum volume is V = \pi r^2 h = \pi \frac{2R^2}{3}.\frac{2R}{\sqrt3} = \frac{4\pi R^3}{3\sqrt3}

Posted by

Gautam harsolia

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