Out of a circular piece of paper, you wish to form a cone cup, so you cut out a circle wedge (with its extreme at the circle center) and join the resulting straight sides, forming a conical cup.

What size should the wedge be, to maximize the capacity of the cone?

Suppose that the circle has radius 1 and we cut out a wedge of angle (and circumference) 2*pi*x where 0<x<1. Then the circumference of the base of the cone is 2*pi*(1-x) so its radius is r = 1-x.  A plane vertically through the apex and perpendicular to the base intersects the cone in an isosceles triangle with sides 1 and base 2*r. The height of this triangle is therefore h = sqrt(1-r^2) = sqrt((1-(1-x)^2) = sqrt (2*x - x^2). The volume of the cone is a constant times r*h so if we maximize r*h we maximize the volume of the cone. Now

    r*h = (1-x)*sqrt(2*x - x^2)

and setting to 0 its derivative with respect to x gives

   (1-x)* (1/2)*(1/sqrt(2*x - x^2))*(2 - 2*x) - sqrt(2*x - x^2) = 0

which after simplfying and rearranging gives

   2*x^2 - 4*x + 1 = 0

which has roots 1 +/- sqrt(1/2) and taking the root between 0 and 1 gives us

   x = 1 -  sqrt(1/2) ~= .293.

So the wedge removed is about .293*360 ~= 105.5 degrees, which, however, disagrees with previous results. Pehaps I have made an error, but I have double checked all my steps. When I experiment with a circle cut out of a piece of paper, 105.5 degrees seems very plausible to me.

Posted by Richard on 2005-04-02 02:39:17