In the figure, ∆PQR represents the 2D view of the cone and the circle represents the sphere. PA is perpendicular to QR and PS is diameter of circle. C will lie on PA due to symmetry.

Let the radius and height of cone be r and h and the radius of sphere be R. Also, the semi vertical angle of cone is α.

In ∆PAR

-(1)

∠APR=α

∠PAR=90°

Hence, ∠ PRA=180° -90° -α =90° -α

Also ∠PRS=90° (Angle in a semicircle)

Hence ∠ARS=∠PRS-∠PRA=α

In ∆RAS

AS=PS-PA=2R-h

–(2)

From (1) and (2), we get

r²=2Rh-h²

The volume of cone will be

Differentiating V with respect to h, we get

Differentiating V’ with respect to h, we get

For maxima at h=c, V’(c)=0 and v’’(c)<0

V’=0 ⇒

Hence, the ratio of height of cone to diameter of sphere is .