Answer to Question #311403 in Calculus for j j

Question

Find the dimensions of a right circular cone of minimum volume V that can be

circumscribed about a sphere of radius 8 inches.

Answer 1

The volume of a cone = $​ \tfrac{1}{3}*π *r^2*h$

Let the radius of the sphere = a

the radius of the cone =r

by similar triangles $sin (θ)=\tfrac{a}{h-a}$ = $\tfrac{r}{\sqrt{h^2+r^2}} ​ ​ ​$

$=a*(\sqrt{h^2+r^2) }=r(h-a)$

on squaring both sides

$a^2(h^2+r^2)=r^2(h-a)^2$

$=a^2h^2+a^2r^2= h^2r^2-2ahr^2+a^2r^2$

making $r^2$ to be the subject of the formulae we have

$r^2= \frac{a^2h}{h-2a}$

The volume V = $​ \tfrac{1}{3}*π *\frac{h^2}{h-2a}*h$

at minimum volume ;

$\frac{dV}{dh}= \frac{\pi a^2}{3}[\frac{(h-2a)2h-h^2(1)}{(h-2a)^2}]=0$

$2(h-2a)h-h^2=0$

$2h-4a-h=0$

$h-4a=0$

$h=4a$

but a =8 inches

therefore $h=4*8=32$

The height of the cone is 32 inches.

$r^2= \frac{8^2*32}{32-2*8}$

$r^2=\frac{2048}{16}$

$r^2=128$

$r= \sqrt{128}$

$r= 11.3137$

The height of the cone is 32 inches and the radius is 11.3137inches.