Posted by Subhotosh Khan on October 10, 2002 at 13:29:18:
In Reply to: check me, Mr K (going your way!) posted by Denis Borris on October 10, 2002 at 12:03:56:
: [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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