Question

A right circular cone is inscribed inside a larger right circular cone with a volume of 190 cm3. The axes of the cones coincide and the vertex of the inner cone touches the center of the base of the outer cone. Find the ratio of the heights of the cones that maximizes the inner cone volume.

Answer 1

Answer:$\frac{1}{3}$

Step-by-step explanation:

From Similar triangles

$\frac{r}{R}=\frac{H-h}{H}$

$r=R\times \frac{H-h}{H}$

$Volume of small cone$

$V=\frac{\pi }{3}r^2h$

$V=\frac{\pi }{3}R^2\left [ 1-\frac{h}{H}\right ]^2h$

$V=\frac{\pi }{3}R^2H\left [ 1+\frac{h^2}{H^2}+-2\cdot \frac{h}{H}\right ]^2\frac{h}{H}$

Let $\frac{h}{H}$ be x

$V=V_{big}\left ( 1+x^2-2x\right )\left ( x\right )$

Differentiate to get V maximum

$\frac{\mathrm{d}V}{\mathrm{d}x}=1+x^2-2x +\left ( 2x-2\right )x$

$\left ( 3x-1\right )\left ( x-1\right )=0$

x=1 is not possible thus $x=\frac{1}{3}$

$\frac{h}{H}=\frac{1}{3}$