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Question:
You have a circular piece of paper with radius 1 and wish to form a cone shaped cup. You may cut out any wedge you like from the paper, call the angle of the wedge x. The point of the wedge must be at the center of the circle. After cutting out the wedge you then attach the two straight edges remaining to form a cone. Assuming the paper cone can hold water, what should x be to maximize the water holding capacity of the cone?
Answer:
>When we perform the construction described in the brain teaser we get a right-angled circular cone of slant height s=1 and circumference c = 2pi - x, where x is in radians. Therefore this problem is equivalent to finding the value of c in [0,2pi] for which
(1) V(c) = ( c^2*sqrt( 4*pi^2 - c^2 ) )/( 24*pi^2 )
is maximized. Since V is 0 at both endpoints and always >= 0, a maximum can only occur in (0,2pi), unless V = 0 over [0,2pi] which is not the case as can be seen by plugging any value in (0,2pi) into (1). Therefore we must find a value for which
(2) dV/dc = ( 8*pi^2*c - 3*c^3 ) / ( 24*pi^2*sqrt( 4*pi^2 - c^2 ) ) = 0
The denominator is unimportant since it cannot cause a fraction to be zero and it will never itself be zero because of the nature of the problem. Therefore this problem is equivalent to finding a c in (0,2pi) for which
(3) 8*pi^2*c - 3*c^3 = 0
The only c in this range that fulfills the requirement is 2*sqrt(2/3)*pi. Because the function is continuous, >= 0 in [0,2pi], and is equal to 0 at 0 and 2pi, this must be a maximum. Therefore the volume is maximized when
(4) x = 2*pi - 2*sqrt(2/3)*pi
x = ~ 1.153 radians = ~66.1 degrees
-- Daniel Giaimo
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