The altitude of a cylinder of the greatest possible volume which can be inscribed in a sphere of radius $3 \sqrt{3}$ is

Open in App

Solution

$v o l u m e o f a c y l i n d e r = π r^{2} h i f w e f i t t h i s c y l i n d e r i n a s p h e r e o f r a d i u s \sqrt{27} f o r g r e a t e s e t p o s s i b l e v o l u m e o f c y l i n d e r t h e n w e c a n s a y t h a t {(r a d i u s o f s p h e r e)}^{2} = {(r a d i u s o f c y l i n d e r)}^{2} + {(\frac{h e i g h t o f c y l i n d e r}{2})}^{2} {(r)}^{2} + \frac{h^{2}}{4} = 27 v = π r^{2} h v = π (27 - \frac{h^{2}}{4}) h f o r g r e a t e s v o l u m e v^{'} = 0 v^{'} = π (27 - \frac{3 h^{2}}{4}) = 0 h = 6$

Suggest Corrections

Similar questions

= \sqrt{3}

5 \sqrt{3}

500 π {c m}^{3}

5 \sqrt{3 c m} i s 500 π {cm}^{3} .

\frac{2 R}{\sqrt{3}}

Join BYJU'S Learning Program