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November 24, 2014

November 24, 2014

Posted by **DJ** on Sunday, October 16, 2011 at 5:41pm.

- Calculus -
**Steve**, Monday, October 17, 2011 at 11:42amDraw a diagram, with the top of the cone at the top of the sphere. If the cone extends past the diameter of the sphere of radius R, its base will have a radius r, and height h > R.

From the center of the sphere, draw a radius to the base of the cone. Then

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

r^2 + h^2 - 2rR = 0

Now, the cone has a volume of V = 1/3 πr^2 h

To avoid a bunch of square roots, let's square things:

V^2 = 1/9 π^2 r^4 h^2

= π^2/9 r^4 (2Rr - r^2)^2

2VV' = 4π/9 r^3 (2Rr - r^2)^2 + 2π^2/9 r^4 (2Rr-r^2)(2R-2r)

Now, V' is a bunch of stuff all divided by V. V is not zero, so V' will be zero when the numerator is zero. Toss out all that π^2 stuff, and we are left with a bunch of algebra that ends up giving h = 4/3 R.

That will give you the value of r.

do a google for maximum volume cone inscribed in sphere to find more details.

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