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mehtank
12-26-2001, 11:21 PM
to keep your brain un-dead over the holiday. actually, the more i think about it, the advanced-er it becomes...

the thought came to me with the talk of possible objects that's been going around: what happens if you overinflate an inflatable cube? it seems to be an interesting mathematical problem, so i'll phrase it [a bit] more rigorously.

what is the maximum volume of a region completely bounded by 6 smooth surfaces, subject to the following conditions:
~ each surface is of unit area, bounded by 4 continuous curves of unit length intersecting at right angles
~ two surfaces intersect completely along a common edge, and are orthogonal at every point along their intersection
~ three surfaces come together at a corner

i think those are the only conditions... basically you have a unit cube made of rubber walls, but where the edges and vertices maintain fixed relations. what shape will give it the maximum enclosed volume, and what is that volume?

mehtank
12-26-2001, 11:49 PM
and now i'll continue by giving my thoughts so far.

i have no idea how to even begin solving that.

so how about taking it down a dimension? now, you have 4 edges of unit length, intersecting at right angles, what's the maximum area you can have, and what's the shape that gives it?

well, its obviously symmetrical, so we can look at a quarter of the resulting shape. lets center the shape at the origin, with each corner lying on an axis. then, each edge lies within one quadrant, so we'll only look at the first quadrant.

let the curve representing the edge be described [in polar coordinates] by r = f(theta). we know:
~ the length of the curve is 1
~ the slope of the curve at theta = 0, pi/2 is -1
~ we want to maximize the area under the curve

sounds like calculus of variations or something like that to me. got some fun integrals in there. a cookie to whoever can solve this problem. and if you can solve the original one, i'll be deeply in awe.

-ANkuR

Matt Attallah
12-27-2001, 05:15 PM
This is from a 1 minute thought, so if it seems too easy and wrong, that is why. But i thought, wouldn't it turn into a circle? Or some "deformed" circle. The sides will keep going until they break and the corners just won't go too far, so you have 6 sides of "curveniess", thus your "deformed" circle. How to get the exact spacings? Trial and error. If it breaks, DUCT TAPE IT!! :p Well, that is my very simple, and probaly wrong, answer...

Tom Fairchild
12-27-2001, 05:43 PM
Big kudos to Matt. I thought that it would be a limit of the cube approaching a sphere, but figgured that was too elementary of an answer to give mehtank all the trouble. This is coming from the guy who drove his calculus teacher up the wall by using common sense instead of integrals to do problems (and honestly, she was right ;)). Even so though, I'm pretty sure that it would be a sphere, assuming that the corners wouldn't break by then.

~Tom Fairchild~, who'll bring it up with his math prof. once school starts back up again.

Anthony X.
12-27-2001, 09:27 PM
Although a sphere/circle will probably maximize volume/area, Mehtank's specification that the edges are orthogonal does not allow this since all edge intersections must be at right angles.

If we consider only in 2D as his second message asks, perhaps the edges would be a cosine curve (between -pi and pi) or a gaussian curve (bell curve) rotated 45, 135, 215, and 305 degrees respectively? This would allow for the edges to intersect at right angles at least. I don't know if it will maximize area though. Unfortunately, I don't have a clue how to find the length of a curve (and limit it to one) so I'm stuck. Any suggestions? :confused:

Just my two cents.

Anthony. :)

Matt Attallah
12-28-2001, 12:09 AM
Although a sphere/circle will probably maximize volume/area, Mehtank's specification that the edges are orthogonal does not allow this since all edge intersections must be at right angles.

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I have no Idea what Anthony X. ment/said, but i think i know. (you all know what i mean!) But isn't that where the "curviness" and strech will over run the corners?? The corners won't be able to move due to the amount of force it will take to move it, thus the sides would explode before it happens. So... the sides would eather buldge, enclose its self, or just plain explode. Think about a soccer ball... (i know that is a pentagon, ok ok about 50 pentagons togther, but the basic fundamentals, i think, can form the basic idea i am trying to get togther! :p)