# Show that the cone of the greatest volume which can be inscribed in a given sphere has an altitude equal to 2/3 of the diameter of the sphere.

Arun Kumar IIT Delhi
7 years ago
Hi

$\\h_{cone}=r+rsin\theta \\r_{cone}=rcos\theta \\=>v=1(r+rsin\theta)(rcos\theta)^2/3 \\=>lnv=2ln(rcos\theta)+ln(r+rsin\theta)+lnc \\=>0=2(-sin\theta)/cos\theta+cos\theta/(1+sin\theta)=0 \\=>2sin\theta(1+sin\theta)=cos^2\theta \\=>1-sin^2\theta=2sin\theta+2sin^2\theta \\=>3sin^2\theta+2sin\theta-1=0 \\=>sin\theta={-2 \pm 4 \over 6}=1/3 \\=>h=r4/3=d*2/3$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty