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Show that semi-vertical angle of right circular cone of given surface area and maximum volume is \( \sin^{-1} \left(\frac{1}{3}\right)\)

1 Answer

  • 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}-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)
$4\pi r^2=\pi r l+\pi r^2$
$4\pi r^2-\pi r^2=\pi r l$
$3\pi r^2=\pi r l$
$\sin \alpha=\large\frac{1}{3}$
Thus $V$ is maximum when $S=$constant
answered Aug 9, 2013 by sreemathi.v

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