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

 

 

 

 
 
 
 
 

Answers (1)

Let 'x' be the diameter of the bare of the cylinder and Let 'h' be height of the cylinder. 

In \bigtriangleup ABC, we \: \: \: have

(BC)^2+(AB)^2=(AC)^2

h^2+x^2=(2R)^2

x^2=4R^2-h^2\: \: \: \: -(1)

The volume of cylinder, V=\pi r^2 h

\Rightarrow V=\pi \times \left ( \frac{x}{2} \right )^2h \\ = \pi \times \frac{x^2}{4}h \\ V=\pi\frac{(4R^2-h^2)}{4}\times h\: \: \: \: \: \: \: [using (i)]\\ = \frac{4\pi R^2h}{4}\times \frac{\pi h^3}{4}\\=\pi h R^2-\frac{\pi h^3}{4}\: \: \: -(ii)

On differentiating equation (ii) w.r.t h we get, 

\frac{\mathrm{d} v}{\mathrm{d} h}=d\frac{(\pi hR-\pi h^3/4)}{dh}\\ \Rightarrow \frac{\mathrm{d} v}{\mathrm{d} h}= \pi R^2 \frac{\mathrm{d} (h)}{\mathrm{d} (h)}-\frac{\pi d (h^3)}{4\: \: \: \: \: dh}\\ \Rightarrow \frac{\mathrm{d} v}{\mathrm{d} h}=\pi R^2- \frac{\pi}{4}(3h^2)\\ \Rightarrow \frac{\mathrm{d} v}{\mathrm{d} h}=\pi R^2 - \frac{3\pi h^2}{4}\: \: \: \: \: \: (iii)\\ \Rightarrow 0 = \pi R^2- \frac{3\pi h^2}{4} \: \: \: \: \left [ \because \frac{\mathrm{d} v}{\mathrm{d} }=0 \right ]\\ \Rightarrow h = \frac{2R}{\sqrt{3}}

Again differentiating equation (iii) w.r.t. h we get,

\frac{\mathrm{d}^2V }{\mathrm{d} h^2}=\frac{\mathrm{d} }{\mathrm{d} h}\left ( \pi R^2-\frac{3\pi h^2}{4} \right )

\Rightarrow \frac{\mathrm{d}^2 V}{\mathrm{d} h^2}=0-\frac{3}{4}\pi \times 2h\\ \Rightarrow \frac{\mathrm{d}^2 V}{\mathrm{d} h^2} = \frac{-3\pi h}{2}\\ At\: \: h=\frac{2R}{\sqrt3}\: \: \: we\: \: have\: \: \\ \frac{\mathrm{d}^2 V}{\mathrm{d} h^2}=\frac{-3\pi}{2}\left ( \frac{2R}{\sqrt3} \right )=-\sqrt3\pi R\\ \frac{\mathrm{d}^2 V}{\mathrm{d} h^2}<0,\: \: hence \: \: h= \frac{2R}{\sqrt3}is \: a\: point \: of\: maximum

So, V is maximum when h=\frac{2R}{\sqrt3}

hence, the highest of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \frac{2R}{\sqrt3}

Hence proved 

From (i) , we have 

x^2=4R^2-h^2\\ \Rightarrow x^2=4R^2- \left ( \frac{2R}{\sqrt3} \right )^2\\ \Rightarrow x^2=\frac{8}{3} R^2\\ \therefore Maximum\: \: volume\: \: of\: \: \: cylinder\\= \pi \left ( \frac{x}{2} \right )^2\times h \\ \Rightarrow \pi \times \frac{x^2}{4}\times h \Rightarrow \frac{\pi}{4}x^2h\\ \Rightarrow \frac{\pi}{4}\times \frac{8R^2}{3}\times \frac{2R}{\sqrt3} \Rightarrow \frac{4\pi R^3}{3\sqrt3} 

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