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^2)*h so if we maximize (r^2)*h we maximize the volume of the cone. Now

    (r^2)*h = [(1-x)^2]*sqrt(2*x - x^2)

and setting to 0 its derivative with respect to x gives

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

which after simplfying and rearranging gives

   3*x^2 - 6*x + 1 = 0

which has roots 1 +/- sqrt(2/3) and taking the root between 0 and 1 gives us

   x = 1 -  sqrt(2/3) ~= .1835

So the wedge removed is about .1835*360 ~= 66.06 degrees, which agrees with previous results.

Posted by Richard on 2005-04-02 03:55:28