Question

area.
7. If the curved surface of right circular cylinder inscribed in a sphere of radius r is maximum, show
that the height of the cylinder is V2 r.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let radius of cylinder be R and height of cylinder be 'h' that can be inscribed in a sphere of radius 'r'.

So,

By pythagoras theorem,

$\rm :\longmapsto\: {r}^{2} = {R}^{2} + {\bigg(\dfrac{h}{2}\bigg) }^{2}$

$\rm :\longmapsto\: {r}^{2} = {R}^{2} + \dfrac{ {h}^{2} }{4}$

$\rm :\implies\:{R}^{2} = {r}^{2} - \dfrac{ {h}^{2} }{4} - - - (1)$

Now,

Curved Surface Area of cylinder is

$\rm :\longmapsto\:Curved \: Surface_{(cylinder)} =2 \pi \: {R}h$

$\rm :\longmapsto\:C =2 \pi \: {R}h$

On squaring both sides, we get

$\rm :\longmapsto\: {C}^{2} = 4 {\pi}^{2} {R}^{2} {h}^{2}$

$\rm :\longmapsto\: {C}^{2} = 4 {\pi}^{2} {R}^{2} {h}^{2}$

Put the value of R, we get

$\rm :\longmapsto\: {C}^{2} = 4 {\pi}^{2} \: \bigg( {r}^{2} - \dfrac{ {h}^{2} }{4} \bigg) {h}^{2}$

$\rm :\longmapsto\:f(h) = \pi \: \bigg( {r}^{2} {h}^{2} - \dfrac{ {h}^{4} }{4} \bigg)$

On differentiating both sides w. r. t. h, we get

$\rm :\longmapsto\:\dfrac{d}{dh}f(h) =\dfrac{d}{dh} \pi \: \bigg( {r}^{2} {h}^{2} - \dfrac{ {h}^{4} }{4}\bigg)$

$\rm :\longmapsto\: f'(h)= \pi \: \bigg( 2h{r}^{2} - \dfrac{ {4h}^{3} }{4} \bigg)$

$\rm :\longmapsto\: f'(h)= \pi \: \bigg( 2h{r}^{2}-{h}^{3}\bigg) - - - (2)$

For maxima or minima,

$\rm :\longmapsto\:f'(h) = 0$

$\rm :\longmapsto\:2h{r}^{2} - {h}^{3} = 0$

$\rm :\longmapsto\:2h{r}^{2} = {h}^{3}$

$\rm :\longmapsto\:2{r}^{2} = {h}^{2}$

$\rm :\implies\:h = \sqrt{2} \: r - - - (3)$

Again, On Differentiating both sides w. r. t. h of equation(2), we get

$\rm :\longmapsto\: f''(h)= \pi \: \bigg( 4h{r}-{3h}^{2}\bigg)$

$\rm :\longmapsto\: f''(h)_{h = \sqrt{2}r} = \pi \: \bigg( 4 (\sqrt{2}r) {r}-{3( \sqrt{2}r) }^{2}\bigg)$

$\rm :\longmapsto\: f''(h)_{h = \sqrt{2}r} = \pi \: \bigg( 4\sqrt{2} {r}^{2} \: - \: {6r}^{2}\bigg) < 0$

$\bf\implies \:C \: is \: maximum \: when \: h = \sqrt{2}r$