For 0 < r < 1 trivially the area A(r) increases towards A(1) = 4 pi > 12 . For 1 < r < sqrt(2) the area of a spherical cap cut off exceeds the area of the circle where the sphere meets the cube, so A(1 + x) - A(1) < 4 pi r^2 - 6 pi (r^2 - 1) - 4 pi = 4 pi (1 + 2 x + x^2) - 6 pi (2 x + x^2) - 4 pi = - 2 pi (2 x + x^2) < 0 , and A(r) inceeds A(1) . For sqrt(2) < r < sqrt(3) the area of a spherical trigon nestling in a cube corner inceeds the area of the trihedron enclosing it; at r = sqrt(2) the trihedra have max total area 6*4 - 6*2 = 12 , which inceeds A(1) . For r > sqrt(3) trivially A = 0 . Hence A has maximum at r = 1 . WFL On 8/10/13, Dan Asimov <dasimov@earthlink.net> wrote:
A simple question in 3-space, but I'm not sure it can be solved in closed form:
Let C denote a cube of side 2 centered at the origin. Let S(r) denote the sphere of radius r centered at the origin.
QUESTION: For which r does the area of S(r) lying inside of C reach its absolute maximum?
(Of course any maximizing r satisfies 1 <= r <= sqrt(3).)
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun