If a right circularcone having maximum volume, is inscribed in a sphere of radius 3 ~cm , then the curved surface area (in cm ^2 ) of this cone is

Step-by-step solution

Sphere of radius r=3 ~cm Let b, h be base radius and height of cone respectively. So, volume of cone = 1 2 b^2 h In right angled A B C by Pythagoras theorem aligned &(h-r)^2+b^2=r^2 & b^2=r^2-(h-r)^2=r^2- (h^2-2 h r+ r ^2 )=2 h r-h^2 & Volume (v) 1 3 h [2 h r-h^2 ]= 1 3 [2 h r^2-h^3 ] & d v d h = 1 3 [4 h r-3 h^2 ]=0 h(4 r-3 h)=0 & d^2 v d h^2 = 1 3 [4 r-6 h] aligned At h= 4 r 3 , d^2 v d h^2 = 1 3 [4 r- 4 r 3 6 ]= 1 3 [4 r-8 r < ] 0 maximum volume ocurs at h = 4 r 3 = 4 3 3=4 ~cm As from (1), (h-r)^2+b^2=r^2 b^2=2