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Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle \alpha is one-third that of the cone and the greatest volume of the cylinder is   \frac{4}{27}\pi h^3\tan^2\alpha.

 

 

 

 
 
 
 
 

Answers (1)

Let height and radius of the cylinder inscribed in the cone H and R resp.

Let \angle OAC=\alpha.  \because \triangle AOC\sim \triangle AO'B

\therefore \frac{AO}{AO'}=\frac{OC}{O'B}\Rightarrow \frac{h}{h-H}=\frac{r}{R} \; \; \; \&\; \; \; H=\left ( 1-\frac{R}{r} \right )h

Also in  \triangle AOC  , \tan \alpha =\frac{OC}{OA}\Rightarrow r=h\tan \alpha

\therefore H=h\left ( 1-\frac{R}{h\tan \alpha } \right )\: \: \: \: -(i)

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

and  \frac{\mathrm{d}^2V }{\mathrm{d} R^2}=\pi h\left ( 2-\frac{6R}{h\tan\alpha } \right )

For local points of maxima and or minima

\frac{\mathrm{d} V}{\mathrm{d} R}=\pi h\left ( 2R-\frac{3R^2}{h\tan \alpha } \right )=0

\Rightarrow 2R-\frac{3R^2}{h\tan \alpha }=0\Rightarrow R=\frac{2h\tan\alpha }{3}

\therefore \left [\frac{\mathrm{d}^2V }{\mathrm{d} R^2} \right ]_{at\; R=\frac{2h\tan \alpha }{3}}=\pi h\left ( 2- 4 \right )=-2h\pi <0

\therefore V is maximum at   R=\frac{2h\tan \alpha }{3}

Now height of the cylinder H=h\left ( 1-\frac{R}{h\tan \alpha } \right )=h\left ( 1-\frac{2}{3} \right )

\therefore H=\frac{h}{3}

Also volume of the cylinder 

V=\pi \left ( \frac{2h\tan \alpha }{3} \right )^2h\left ( 1-\frac{2}{3} \right )

\Rightarrow V=\frac{\pi h}{3}\left ( \frac{4h^2\tan^2\alpha }{9} \right )

\therefore V=\frac{4\pi h^3\tan^2\alpha }{27}

 

Posted by

Ravindra Pindel

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