# right circular cone

• Oct 27th 2009, 06:20 AM
right circular cone
Hello,
Show that semi - vertical angle of right circular cone of given surface area and maximum volume is sin-1(1/3)?
• Oct 28th 2009, 06:45 AM
Quote:

Hello,
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 $\displaystyle = \theta$, the radius $\displaystyle = r$, the surface area $\displaystyle = S$ and the volume $\displaystyle = V$:

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

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

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

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

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

$\displaystyle =\tfrac13(S^2r^2-2\pi Sr^4)^{\frac12}$
$\displaystyle \Rightarrow 3\frac{dV}{dr}=\tfrac12(S^2r^2-2\pi Sr^4)^{-\frac12}.(2S^2r-8\pi Sr^3)$
$\displaystyle = 0$ when $\displaystyle 2S^2r-8\pi Sr^3=0$
$\displaystyle \Rightarrow r = 0$ or $\displaystyle S - 4\pi r^2 = 0$

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

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