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**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