right circular cone

Hello anilpadavil

Quote:

Originally Posted by anilpadavil

Hello,
please help me ,
Show that semi - vertical angle of right circular cone of given surface area and maximum volume is sin-1(1/3)?

If the semi-vertical angle $= \theta$ , the radius $= r$ , the surface area $= S$ and the volume $= V$ :

The vertical height, $h = \frac{r}{\tan\theta}$ and the slant-height, $l = \frac{r}{\sin\theta}$ .

$\Rightarrow S = \pi r^2 + \pi r l$
$= \pi r^2 + \frac{\pi r^2}{\sin \theta}$
$\Rightarrow \sin\theta = \frac{\pi r^2}{S-\pi r^2}$

$\Rightarrow \tan\theta = \frac{\pi r^2}{\sqrt{S^2 - 2\pi S r^2}}$

So $V = \tfrac13\pi r^2 h$
$= \frac{\pi r^3}{3\tan\theta}$

$= \frac{\pi r^3\sqrt{S^2-2\pi S r^2}}{3\pi r^2}$

$=\tfrac13(S^2r^2-2\pi Sr^4)^{\frac12}$

$\Rightarrow 3\frac{dV}{dr}=\tfrac12(S^2r^2-2\pi Sr^4)^{-\frac12}.(2S^2r-8\pi Sr^3)$
$= 0$ when $2S^2r-8\pi Sr^3=0$
$\Rightarrow r = 0$ or $S - 4\pi r^2 = 0$

$r = 0$ gives a minimum volume; and the volume is a maximum when $\pi r^2 = \tfrac14S$

For this value of $r, \sin\theta = \frac{\pi r^2}{S-\pi r^2}= \frac{\tfrac14S}{S-\tfrac14S}=\frac13$

Grandad