Closed-Ended Problem Algorithm:
Jelly Beans

Problem Statement: How many jelly beans will fit in a cylindrical jar 10 inches high and 8 inches in diameter?

ATTITUDE

This is a fun problem that may prove useful in the future for winning a guessing contest.

ACTIONS

Re-describe the problem:

What is the volume of the jar and what is the volume of a jelly bean and what is the packing factor (i.e., what is the volume fraction of the jelly beans in the filled jar… remembering there will be some empty space between the beans)?

Assume the jelly beans are spheres.

SOLUTION PROCEDURE

Identify:

Knowns

Jar height (H= 10 inches) and jar diameter (D=8 inches).
Jelly bean diameter (d=0.5inches)

Unknowns

Solid Fraction (the fraction of the jar volume occupied by beans, not including the void space between the beans).

Volume of the cylinder	=	π(D2/4)H in3
	=	3.1416*((82)/4)*10
	=	502.6
Volume of one jelly bean	=	π(D3/6) in3
	=	3.1416*(0.53/6)
	=	0.0654
Look up the solid fraction of spheres packed together. It is ~0.65.
Volume of Beans	=	(Jar Volume)*(Solid Fraction)
	=	502.6*0.65
	=	326.7 in3
Number of Beans	=	(Volume of Beans)/(Volume of One Bean)
	=	(326.7 in3)/(0.0654 in3 per bean)
	=	4,995 beans in the jar

ACCURACY

Check and Recheck:

Explore the effects of cylindrical beans rather than spherical beans. Check the jelly bean dimensions.

If beans are cylindrical (D = 0.5in, H = 0.5in), redo the calculations.

Volume of one jelly bean	=	π(D2/4)*H
	=	3.1416*(0.52/4)*0.5
	=	0.0982 in3
Look up the solid fraction of cylinders packed together in jar. It is ~0.67.
Volume of Beans	=	(Jar Volume)*(Solid Fraction)
	=	502.6*0.67
	=	336.7 in3
Number of Beans	=	(Volume of Beans)/(Volume of One Bean)
	=	(336.7 in3)/(0.0982 in3 per bean)
	=	3429 beans in the jar

The true number of jelly beans in the jar is probably between the two figures, since the beans are not truly spheres or cylinders, but this technique gives us a bracket that the true answer lies between.

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