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23 April, 20:21

160 marbles fit inside a tin box measuring 4.5 x 4.5 x 7.5 if they are transferred to a tin cylinder that is 4.5 in diameter how tall must the cylinder be so all the marbles fit inside comfortably

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  1. 24 April, 00:20
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    (box volume/ball volume) * packing density = # mables that fit in box max packing density is pi/3*sqrt (2) for spheres, according to the Kepler conjecture, so if ball volume is b, (4.5 * 4.5 * 7.5 / b) * (pi/3*sqrt (2)) = 160 151.875 / b * (pi/3*sqrt (2)) = 160 151.875 * pi / b * 3 * sqrt (2) = 160 50.625 * pi / b * sqrt (2) = 160 10.125 * pi / 32 * sqrt (2) = b b ~ 0.7 For a tin cylinder of 4.5 diameter, the volume is (4.5/2) ^2 * pi * h. Using the formula again, (box volume/ball volume) * packing density = # mables (5.0625 * pi * h / (10.125 * pi / 32 * sqrt (2)) * (pi/3*sqrt (2)) = 160 (5.0625 * pi * h * 32 * sqrt (2) / 10.125 * pi) * (pi/3*sqrt (2)) = 160 5.0625 * h * 32 * sqrt (2) / 10.125 * (pi/3*sqrt (2)) = 160 162 * h * sqrt (2) * pi / 10.125 * 3 * sqrt (2) = 160 54 * h * pi / 10.125 = 160 54 * h * pi = 1620 h * pi = 30 h = 30/pi h ~ 9.55 So if the marbles are of equal size, and both the box and cylinder are packed as tightly as possible, the cylinder would have to be 9.55 units tall.
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