Posted by Denis Borris on October 10, 2002 at 12:03:56:
In Reply to: continuation of the "Kurt cones"; Mr K vs Mr B ! posted by Denis Borris on October 09, 2002 at 22:41:17:
[where are you, Kurt, you @#^%&$@!!]
OK, Mr K, is this all ok (got some help with this):
Let t be the angle (in RADIANS!) and assume circle has radius 1. (there!!)
Let b = radius of the bigger cone (BC).
So the base of BC has circumference = 2pi - t, so b = 1 - t / 2pi
The circumference of the smaller cone (SC) is t, so it's radius = t / 2pi
So letting x = t / 2pi :
the height of BC = sqrt(1 - (1 - x)^2) = sqrt (2x - x^2)
Height of SC = sqrt(1 - x^2)
Volume of BC = 1/3 * pi * (1 - x)^2 * sqrt (2x - x^2)
Volume of SC = 1/3 * pi * x^2 * sqrt (1 - x^2)
V = K[(1 - x)^2 * sqrt (2x - x^2) + x^2 * sqrt (1 - x^2)] : where K = pi/3
So take the derivative, set it = 0, and solve for x .....EASY?!
We have V = K[(1 - x)^2sqrt(1 - (1 - x)^2) + x^2sqrt(1 - x^2)]
Let u = 1-x; du/dx=-1
V = K[u^2sqrt(1 - u^2) + x^2sqrt(1 - x^2)]
Taking dV/dx and rearranging terms (and do the du/dx part):
(u - 2u^3) / sqrt(1 - u^2) = (x - 2x^3) / sqrt(1 - x^2)
I don't know how to solve that, but since u+x = 1,
clearly x = u = 1/2 will be a solution.
If the second derivative of V at x=1/2 is negative, we'll know
we have a local maximum anyway, right?
But wether it's the global maximum for x from 0 to 1 is another matter :(
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