Question

Prove that the altitude of the right circular cone of maximum volume that can be inscribed 8 in a sphere of radius r is $\frac{4r}{3}$. Also show that the maximum volume of the cone is $\frac{8}{27}$ of the volume of the sphere.

Answer 1

Let VAB be a cone of maximum volume inscribed in a sphere of radius r.
Let OC = x. Then $AC=\sqrt{r^2-x^2}$ = Radius of cone, $VC=OC+VO=x+r$= Height of cone.
Then volume of cone, $V=\frac{1}{3}\pi \left(AC{)}^2\right(VC)$
$\therefore V=\frac{1}{3}\pi (r^2-x^2)(r+x)$
$\Rightarrow \frac{dV}{dx}=\frac{1}{3}\pi [r^2-x^2-2x(r+x\left)\right]=\frac{1}{3}\pi [r^2-3x^2-2rx]$
And, $\frac{d^{2}V}{d{x}^3}=\frac{1}{3}\pi [-6x-2r]$.
For points of local maxima & local minima, we have $\frac{dV}{dx}=0$
i.e., $\frac{1}{3}\pi [r^2-3x^2-2rx]=0\Rightarrow (r-3x)(r+x)=0\Rightarrow x=\frac{r}{3},x=-r$
We shall reject x=-r. So, ${\begin{array}{c}\frac{d^{2}V}{d{x}^2}\end{array}|}_{atx=\frac{r}{3}}=\frac{1}{3}\pi [-2r-2r]=-\frac{4r\pi }{3}<0$
So, V is maximum at $x=\frac{r}{3}$.
Now, height of cone $VC=x+r=\frac{r}{3}+r=\frac{4r}{3}$.
Now, volume of the cone =$\frac{1}{3}\pi \left(r^2{\left(\frac{r}{3}\right)}^2\right)(r+\frac{r}{3})=\frac{1}{3}\pi \left(\frac{32}{27}{r}^3\right)=\frac{8}{27}\left(\frac{4}{3}\pi r^3\right)$
$\therefore$ Maximum volume of cone =$\frac{8}{27}$ (volume of sphere) .
Hence proved.