Originally Posted by
Gul A cone is inscribed in a sphere of radius 1, find the dimensions of the cone, and the maximum volume of the cone?
I get a right right triangle:
(h-1)^2 + r^2 = 1^2
h^2 - 2h +1 + r^2 = 1
r^2 = -h^2 + 2h
r^2 = 2h - h^2
Volume is: 1/3*pi*r^2*h
V = 1/3*pi*(2h - h^2 )*h
V = 1/3*pi*(2h^2-h^3)
V' = 1/3*pi(4h - 3h^2) = 0
V' = 4h - 3h^2 = 0 , h(4-3h) = 0 , h = 4/3
r^2 = 2h - h^2
r^2 = 2(4/3) - (4/3)^2
r^2 = 8/3 - 16/9
r^2 = 8/9
r = sqrt(8/9)
r = .94
Therefore Vmax = 1/3*pi*(8/9)*(4/3) = 1.241 cm^3
Have i gone wrong somwhere