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# Think outside the can - a challenge for Y-san

### #1

Posted 05 April 2013 - 05:55 PM

Knowing how it is second nature for my friend Y-san to solve extremal problems by equating derivatives of functions to zero, I was delighted to find a problem that can be solved with only a pencil, some drawing paper and a little thought. Here it is.

Imagine an empty can that weighs 1.5 ounces. It is a perfect cylinder with weightless top and bottom, so that any asymmetry introduced by punching holes in the top can be disregarded. The can holds 12 ounces of beer; so its total weight, when filled, is 13.5 ounces. The can is 8 inches high. Without using calculus, determine the level of the beer at which the center of gravity is at its lowest point.

Consider that, as beer is removed from a full can, the center of gravity is lowered. But empty and full, the center of gravity is in the same place. So there must be a filling factor where it is lowest.

*Vidi vici veni.*

### #2

Posted 05 April 2013 - 07:43 PM Best Answer

**Edited by jhawk, 05 April 2013 - 07:53 PM.**

### #3

Posted 09 April 2013 - 11:39 AM

Spoiler for let me rethink this]

**jhawk**, sorry for not replying sooner.

That is the correct answer. Care to share your method?

I have a clever thought process (it's not mine) that leads to it that I will share if you want to go first.

It involves freezing.

*Vidi vici veni.*

### #4

Posted 10 April 2013 - 05:20 AM

### #5

Posted 10 April 2013 - 05:26 PM

Nice.Not sure if this was the same process that jhawk used, but it's what I would have done if it weren't already answered. And it doesn't seem to involve freezing.

Spoiler for

Freezing would allow balancing the can sideways on a fulcrum at its beer height. Then exactly what you said is obvious. Either adding beer or subtracting beer will make the can fall toward its empty side.

*Vidi vici veni.*

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