Math, asked by madanyadavjee689, 4 months ago

25. Show that the semi-vertical angle of the cone of the maximum volume and of
given slant height is tan-12.​

Answers

Answered by mathdude500
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, from equation (2), we have

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

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

\sf \: \dfrac{d^{2} V}{d {h}^{2} } = \dfrac{\pi}{3} (0 -  6{h}) \\

\sf \: \dfrac{d^{2} V}{d {h}^{2} } =  - 2\pi \: h \\

\implies\sf \: \sf \: \dfrac{d^{2} V}{d {h}^{2} }  < 0 \\

\implies\sf \: V \: is \: maximum \: \\

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}

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