Calculus Help~~~~~

Calculation is quite tedious. Here just list out the key points.
Let x be the angle of the sector to be cut out from the paper circle. So arc length = Rx in radians, which will becomes the base circumference of the cone(cup).
Let radius of the circular base be y, therefore, 2py = Rx where p stands for pi.
So y = Rx/2p.
Let height of the cone be h, the slant side of the cone = R. then by Pythagoras theorem, we get
R^2 = h^2 + y^2 = h^2 + (Rx/2p)^2.....................(1)
Volume of cone(cup), V = py^2h/3.......................(2)
From (2), dV/dx can be found. From (1), dh/dx can be found. Sub dh/dx into the expression of dV/dx and put it to zero. We get x = sqrt(8/3)(pi).
That means, volume of cup is maximum when x = sqrt(8/3)(pi) radians which is about 294 degree. From (1), you can find h and from (2) you can find the value of max. capacity.

2008-09-08 11:48:06 補充：
Volume of cone, V = py^2h/3 = p(Rx/2p)^2h/3 = R^2x^2h/12p, then dV/dx = R^2/12p[2xh + x^2(dh/dx)], dh/dx can be found from (1).