Question

The altitude of a right circular cone of minimum volume circumscribed about a sphere of radius $r$ is

• A. $2r$
• B. $3r$
• C. $5r$
• D. $4r$

Answer 1

Answer: (D)

$4r$

Let $R$ be the radius of the cone, $l$ its slant height and h be the height.
$V=\frac{1}{3}\pi R^{2}h$
We have to make $V$ a function of single variable.
$\frac{r}{h-r}=\frac{R}{l}=sin\alpha$ ...... (1)
or $\frac{r}{h-r}=\frac{R}{\sqrt{R^2+h^2}}$
$\therefore r^2(R^2+h^2)=R^2(h^2-2hr+r^2)$
or $r^2{h}^2=R^{2}h(h-2r)$
$\therefore R^{2}h=\frac{r^2{h}^2}{h-2r}$ ..... (2)
$\therefore V=\frac{1}{3}\pi \frac{r^2{h}^2}{h-2r}$. where r is given
$\therefore V=\frac{1}{3}\pi \frac{r^2}{\frac{1}{h}-\frac{2r}{h^2}}$
Now V will be minimum if $z=\frac{1}{h}-\frac{2r}{h^2}$ is max.
$\frac{dz}{dh}=\frac{-1}{h^2}+\frac{4r}{h^3}=0$ $\therefore h=4r$.
$\frac{d^{2}z}{d{h}^2}=\frac{2}{h^3}-\frac{12r}{h^4}=\frac{2}{h^3}[1-\frac{6r}{h}]=\frac{2}{h^3}(1-\frac{6}{4})=-$ ive
$\therefore$ z is max. and hence V is minimum when h $=$ 4r.
$\therefore sin\alpha =\frac{r}{h-r}=\frac{1}{3}\because h=4r.$