Browse Questions

Show that the semi-vertical angle of the right circular cone of given total surface area and maximum volume is $sin^{-1} \frac{1}{3}$.

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Toolbox:
• Surface Area =$\pi r l+\pi r^2$
• $V=\large\frac{1}{3}$$\pi r^2h Step 1: Let r be the radius l be the slant height and h be the vertical height of a cone of semi-vertical angle \alpha Surface area S=\pi r l+\pi r^2------(1) l=\large\frac{S-\pi r^2}{\pi r} The volume of the cone V=\large\frac{1}{3}$$\pi r^2h$
$\qquad\qquad\qquad\qquad\quad=\large\frac{1}{3}$$\pi r^2\sqrt{l^2-r^2} \qquad\qquad\qquad\qquad\quad=\large\frac{\pi r^2}{3}$$\sqrt{\large\frac{(S-\pi r^2)^2}{\pi^2r^2}-\normalsize r^2}$
$\qquad\qquad\qquad\qquad\quad=\large\frac{\pi r^2}{3}$$\sqrt{\large\frac{(S-\pi r^2)^2-\pi^2r^4}{\pi^2r^2}} \qquad\qquad\qquad\qquad\quad=\large\frac{\pi r^2}{3}$$\large\frac{\sqrt{S^2-2\pi Sr^2+\pi^2r^4-\pi^2r^4}}{\pi r}$
$\qquad\qquad\qquad\qquad\quad=\large\frac{r}{3}$$\sqrt{S^2-2\pi Sr^2+\pi^2r^4-\pi^2 r^4} \qquad\qquad\qquad\qquad\quad=\large\frac{r}{3}$$\sqrt{S(S-2\pi r^2)}$
Step 2:
$V^2=\large\frac{r^2}{9}$$S(S-2\pi r^2) V^2=\large\frac{S}{9}$$(Sr^2-2\pi r^4)$
$\large\frac{dV^2}{dr}=\frac{S}{9}$$[2Sr-8\pi r^3] \large\frac{d^2V^2}{dr^2}=\frac{S}{9}$$[2S-24\pi r^2]$------(2)
Now $\large\frac{dV^2}{dr}$$=0 \Rightarrow \large\frac{S}{9}$$(2Sr-8\pi r^3)=0$
$\Rightarrow (S-4\pi r^2)=0$
Putting $S=4\pi r^2$ in (2)
$\large\frac{d^2V}{dr^2}=\frac{S}{9}$$[8\pi r^2-24\pi r^2]=-ve$
$\Rightarrow V$ is maximum when $S=4\pi r^2$
Step 3:
Putting the value in equ(1)
$4\pi r^2=\pi r l+\pi r^2$
$4\pi r^2-\pi r^2=\pi r l$
$3\pi r^2=\pi r l$
$3r^2=rl$
$\large\frac{r}{l}=\frac{1}{3}$
$\sin \alpha=\large\frac{1}{3}$
$l=\sin^{-1}\big(\large\frac{1}{3}\big)$
Thus $V$ is maximum when $S=$constant
$\alpha=\sin^{-1}\large\frac{1}{3}$