A paper drinking cup filled with water has the shape of a co

A paper drinking cup filled with water has the shape of a cone with height h and semivertical angle ?. A ball is placed carefully in the cup, thereby displacing some of the water and making it overflow. What is the radius of the ball that causes the greatest volume of water to spill out of the cup?

Solution

It maybe simpler to think of a sphere of radius 1, and look at a cone of semi vertical angle a, and then see where the cone should be cut so that the proportion of the sphere inside the cone be maximized. Then you just have to rescale. So S = S(O,1) and A, the vertex fof the cone is at (0,0,Z) with Z = 1/sin a. You are going to cut the cone somewhere between -1 and +1 at a height t. It\'s volume will then be 1/3 pi * (Z-t)*(sin a)^2(Z-t)^2. The volume of the sphere above the plane z = t will be V = pi(1-t)(3(1-t^2) +(1-t)^2)/6 = pi(1-t)^2*(2+t)/3 . Hence you want to maximize F(t) = (1-t)^2(2+t) / (Z - t)^3. on [-1,1] F\'(t) = 0 iff (-2(1-t)(2+t)+(1-t)^2)(Z-t)^3 +3(Z-t)^2 (1-t)^2(2+t) = 0 or (-2(2+t)+(1-t))(Z-t) +3 (1-t)(2+t) = 0 that is (-3-3t)(Z-t) +3 (1-t)(2+t) = 0 or (Z - t)(1 + t) = (1 - t)(2 + t). luckily t^2 terms cancel and Z + tZ - t = 2 - t, so finally t = 2/Z - 1 Let\'s do the rescaling r / h = 1 / (Z - t) = 1 / (1 + Z - 2/Z) = 1 / (1 + (1/sin a) - 2 sin a). So r / h = (sin a) / ( 1 + sin a - 2 (sin a )^2 ) = (sin a)/(sin a + cos(2a))