Question

Show that semi vertical angle of a cone of maximum volume and of given slant height is tan^-1√2

Answer 1

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

Let assume that

Radius of cone be r units.

Height of cone be h units.

Slant height of cone be l units.

Semi-vertical angle of cone be x.

Now, we know, Slant height, vertical height and radius of a cone are connected by the relationship

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

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

Now, Volume of cone (V) of radius r and height h is given by

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

can be rewritten as using equation (1), we get

$\sf \: V = \dfrac{\pi}{3} ({l}^{2} - {h}^{2}) h$

$\sf \: V = \dfrac{\pi}{3} (h{l}^{2} - {h}^{3})$

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

$\sf \: \dfrac{dV}{dh} = \dfrac{d}{dh}\left(\dfrac{\pi}{3} (h{l}^{2} - {h}^{3})\right)$

$\sf \: \dfrac{dV}{dh} = \dfrac{\pi}{3} ({l}^{2} - 3{h}^{2})$

For maxima or minima, we have

$\sf \: \dfrac{dV}{dh} =0$

$\sf \: \dfrac{\pi}{3} ({l}^{2} - 3{h}^{2}) = 0$

$\sf \: {l}^{2} = {3h}^{2}$

$\implies\sf \: l = \sqrt{3}h$

Now, on substituting the value of l in equation (1), we get

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

$\sf \: {r}^{2} = {2h}^{2}$

$\implies\sf \: r = \sqrt{2} \: h$

Now, In right-angle triangle AOB

$\sf \: tanx = \dfrac{AO}{OB}$

$\sf \: tanx = \dfrac{r}{h}$

$\sf \: tanx = \dfrac{ \sqrt{2} h}{h}$

$\sf \: tanx = \sqrt{2}$

$\implies\sf \: x = {tan}^{ - 1} \sqrt{2}$

$\rule{190pt}{2pt}$

Additional Information

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