Question

A cone of maximum volume is inscribed in a given sphere, then ratio of the height of the cone to diameter of the sphere is

• A. $\frac{2}{3}$
• B. $\frac{3}{4}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{2}{3}$

Answer 2

Let the diameter of the sphere,

AE = 2r

Let the radius of cone is x and height is y.

\ AD = y Since BD² = AD . DE

̃ x² = y (2r – y) …(i)

Volume of cone

V = $\frac{1}{3}$ px² y = $\frac{1}{3}$py (2r – y) y

= $\frac{1}{3}$ p (2ry² – y³)

On differentiating, we get

̃ $\frac{dV}{dy}$ = $\frac{1}{3}$p (4ry – 3y²)

For maxima and minima, put $\frac{dV}{dy}$ = 0

̃ $\frac{1}{3}$p (4ry – 3y²) = 0 ̃ y (4r – 3y) = 0

̃ y = $\frac{4}{3}$ r, 0

Again differentiating, we get

$\frac{d^{2}V}{dy^{2}}$ = $\frac{1}{3}$p (4r – 6y)

At y = $\frac{4}{3}$r, $\frac{d^{2}V}{dy^{2}}$ = $\frac{1}{3}$p (4r – 8r) = –ive

\ Volume of cone is maximum at y = $\frac{4}{3}$r

Now, Ratio = $\frac{Height\mspace{6mu} ofcone}{Diameterofsphere}$

= $\frac{y}{2r}$ = $\frac{\frac{4r}{3}}{2r}$ = $\frac{2}{3}$.