IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

vedrulz

show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3

srujana

Refer to the attached figure,

$r \rightarrow Radius\ of\ the\ sphere\\<br/>R \rightarrow Radius\ of\ the\ cone\\<br/>H \rightarrow Altitude\ of\ the\ cone\\$

From Geometry,

(H-r)^2+R^2=r^2

$\Rightarrow$ R^2=H(2r-H)

$Volume\ of\ the\ cone,\ V=\frac{1}{3}\pi R^2H$

Differentiating wrt H and equating the first derivative to 0,we get

-H^2+(2r-H)(2H)=0

$\Rightarrow H=0\ or\ H=\frac{4r}{3}$

H=0 is surely not the ans as the volume of the cone becomes 0.

Hence the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3

Greatdreams

Excellent work, however post being removed to calculus section .