# Thread: maximize volume of cone in sphere

1. ## maximize volume of cone in sphere

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

2. 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
Your calculation look OK. Except I'd give $r = \frac{2 \sqrt{2}}{3}$ units and the maximum volume as $\frac{32 \pi}{81}$ cubic units (you used cm^3 but the radius of the sphere is not given a unit in the question).

3. 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
Of related interest: http://www.mathhelpforum.com/math-he...stion-3-a.html