Question 305002
Four solid metal balls fit snugly inside a cylindrical can. 
A geometry student claims that two extra balls of the same size can be put into the can, provided all six balls can be melted and the molten liquid poured into the can. 
Is the student correct? (hint: let the radius of each ball be r.)
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Volume of the can:
Vc = (pi)r^2*h
Note: h = 8r
Vc = (pi)r^2*(8r)
Vc = (8pi)r^3
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Volume of the 4 balls:
4Vb = 4(4/3)(pi)r^3
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Volume remaining when only 4 balls are in the can:
(8pi)r^3 - (16/3)(pi)r^3
= (8-(16)3))(pi)r^3
= (8/3)(pi)r^3
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Can volume of 2 balls fit into that remaining space?
[(8/3)(pi)r^3]/[(4/3)(pi)r^3] = 2
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Answer: Yes
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Cheers,
Stan H.
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