Question

$\huge{ \underline{ \underline{ \textsf{ \textbf{ \red{Question:-}}}}}}$
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan⁻¹ √2.​

Answer 1

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

Let assume that

Height of cone be h units.

Radius of cone be r units.

Slant height of cone be l units.

Semi - vertical angle of cone be x.

Now,

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

$\rm \implies\:\boxed{ \tt{ \: {r}^{2} = {l}^{2} - {h}^{2} \: }} - - - (1) \:$

Now, Consider Volume of cone,

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

On substituting the value from equation (1), we get

$\rm :\longmapsto\:V = \dfrac{\pi}{3} \:[{l}^{2} - {h}^{2}] h$

$\rm :\longmapsto\:V = \dfrac{\pi}{3} \:[{l}^{2}h - {h}^{3}]$

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

$\rm :\longmapsto\:\dfrac{d}{dh}V =\dfrac{d}{dh} \dfrac{\pi}{3} \:[{l}^{2}h - {h}^{3}]$

$\rm :\longmapsto\:\dfrac{dV}{dh} = \dfrac{\pi}{3} \dfrac{d}{dh} \:[{l}^{2}h - {h}^{3}]$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dV}{dh} = \dfrac{\pi}{3} \: [ {l}^{2} - {3h}^{2}]$

For maxima or minima,

$\rm :\longmapsto\:\dfrac{dV}{dh} = 0$

$\rm :\longmapsto\:\dfrac{\pi}{3}[ {l}^{2} - {3h}^{2}] = 0$

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

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

$\bf\implies \:l = \sqrt{3}h - - - (2)$

On substituting (2) in equation (1), we get

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

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

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

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

Now, we have

$\rm :\longmapsto\:\dfrac{dV}{dh} = \dfrac{\pi}{3} \: [ {l}^{2} - {3h}^{2}]$

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

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dh}^{2} } = \dfrac{\pi}{3} \: [0 - 6h] = - 2\pi \: h \: < \: 0$

$\bf\implies \:V \: is \: maximum$

Now, In triangle ABC,

$\rm :\longmapsto\:tanx = \dfrac{r}{h}$

$\rm :\longmapsto\:tanx = \dfrac{ \sqrt{2} \: h}{h}$

$\rm :\longmapsto\:tanx = \sqrt{2}$

$\rm \implies\:\boxed{ \tt{ \: x \: = \: {tan}^{ - 1} \: \sqrt{2} \: \: }}$

More to know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$