Comments on: Bubbles
http://ask.metafilter.com/68724/Bubbles/
Comments on Ask MetaFilter post BubblesTue, 07 Aug 2007 07:02:41 -0800Tue, 07 Aug 2007 07:02:41 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: Bubbles
http://ask.metafilter.com/68724/Bubbles
Maximal volume of an ellipsoid... <br /><br /> The volume of an ellipsoid is: <i>4πabc/3</i><br>
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An ellipsoid is defined with the function: <i>x^2/ a^2 + y^2/ b^2 + z^2/ c^2 = 1</i><br>
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In the case where the sum <i>a + b + c</i> is a fixed value, how might I derive that the ellipsoid of maximal volume is a sphere?<br>
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I'd expect to show that <i>a = b = c</i> but am not certain where to start.<br>
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Note: This is a "homework" question, in that I am working on some math by myself to get back up to speed on some stuff I did many years ago. <b>I'd appreciate hints rather than a full solution.</b> Feel free not to answer if this bothers you.post:ask.metafilter.com,2007:site.68724Tue, 07 Aug 2007 06:55:29 -0800Blazecock PileonmathproblemcalculusvectorcalculushomeworkellipsoidspheremathematicsBy: caek
http://ask.metafilter.com/68724/Bubbles#1027512
You want to extremize some function subject to constraints. The normal way of doing this is to differentiate the function, but the constraints complicate things. One way of dealing with this is Lagrange Multipliers. As requested, I haven't worked this through to confirm the method would work in this case, but I have a hunch it's what you want.comment:ask.metafilter.com,2007:site.68724-1027512Tue, 07 Aug 2007 07:02:41 -0800caekBy: a robot made out of meat
http://ask.metafilter.com/68724/Bubbles#1027515
Method 1<br>
Step 1: use a+b+c=constant to eliminate a variable in the volume<br>
Step 2: use calculus* to check that there exists a local maximum<br>
Step 3: check the "boundaries" of the space. What are the limits to which you could reasonably take a,b, and c subject to your constraint? If there are no boundaries, observe what happens as the variables go to infinity.<br>
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Method 2<br>
Use "lagrange multipliers" to get everything in one equation, repeat steps 2,3<br>
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If you don't remember the * needed, that's easy to fix.comment:ask.metafilter.com,2007:site.68724-1027515Tue, 07 Aug 2007 07:07:24 -0800a robot made out of meatBy: a robot made out of meat
http://ask.metafilter.com/68724/Bubbles#1027517
Also, if you have software available, in method 1 after eliminating a variable you have a 2-d maximization problem. Consider making a plot so that you can see what's going on.comment:ask.metafilter.com,2007:site.68724-1027517Tue, 07 Aug 2007 07:10:01 -0800a robot made out of meatBy: jepler
http://ask.metafilter.com/68724/Bubbles#1027520
If you had only one unknown in '4πabc' rather than 3 of them, you can find minima and maxima by looking for zeroes in the derivitive with respect to the single unknown.comment:ask.metafilter.com,2007:site.68724-1027520Tue, 07 Aug 2007 07:12:18 -0800jeplerBy: andrew cooke
http://ask.metafilter.com/68724/Bubbles#1027534
all the expressions you give are symmetrical (there's nothing to distinguish a, b and c except for the letters used; a+b+c = 1 is identical to b+a+c = 1 for example).<br>
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this means that the result has to be symmetrical too (which is equivalent to saying a = b = c).<br>
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however, i am a physicist not a mathematician, so i don't know how you "prove" the above in a more formal way (although i would love to know how...). i mention it mainly because you can often use this kind of argument to check results, so it's useful to be aware of it.comment:ask.metafilter.com,2007:site.68724-1027534Tue, 07 Aug 2007 07:31:18 -0800andrew cookeBy: a robot made out of meat
http://ask.metafilter.com/68724/Bubbles#1027547
andrew: partially true. If there was a unique answer it would have to be a symmetrical one. However, before you do the math there is no guarantee that a subspace isn't the answer, or that a maximum even exists.comment:ask.metafilter.com,2007:site.68724-1027547Tue, 07 Aug 2007 07:46:19 -0800a robot made out of meatBy: Eringatang
http://ask.metafilter.com/68724/Bubbles#1027555
Hint: <br>
You have three equations with three variables. Simplify. Then maximize.comment:ask.metafilter.com,2007:site.68724-1027555Tue, 07 Aug 2007 08:08:25 -0800EringatangBy: Johnny Assay
http://ask.metafilter.com/68724/Bubbles#1027561
Yeah, if I were you I'd use Lagrange multipliers. The <a href="http://en.wikipedia.org/wiki/Lagrange_Multiplier">Wikipedia article</a> on the subject has a basic overview of how to use them.comment:ask.metafilter.com,2007:site.68724-1027561Tue, 07 Aug 2007 08:15:22 -0800Johnny AssayBy: edd
http://ask.metafilter.com/68724/Bubbles#1027563
I think we should do this without calculus. Where's the fun in it otherwise?<br>
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Ignore the constants 4, π and 3, who cares. We want to maximise abc when a+b+c=1 (who cares what that constant is too?).<br>
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a, b and c must all differ from a third by something. a=1/3+A, b=1/3+B etc.<br>
So a+b+c=1/3+A+1/3+B+1/3+C=1, so A+B+C=0.<br>
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Now abc - spend some time writing that out and get<br>
1/27+(A+B+C)/9+(AB+BC+CA)/3+ABC<br>
Forget the 1/27, who cares about that. A+B+C is 0, so forget about that. We want to maximise<br>
(AB+BC+CA)/3+ABC.<br>
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Lets assume A is positive. Lets say for now, that both B and C are negative (at least one must be so that A+B+C can be 0). If both B and C are negative they must be smaller numbers than A. That means BC is the smallest term, and that's the only positive one. So, (AB+BC+CA)/3 must be negative in this case.<br>
If only one of B and C are negative, lets say B is, then B must be larger than either positive numbers, A or C. Then AB and BC are both negative and CA is the smallest number and the only positive one. So (AB+BC+CA)/3 is still negative.<br>
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Now we need to think back to a=1/3+A. A obviously can't be below -1/3 or a is negative. Which is bad for the health of geometrical shapes. If A were as large as 2/3 then a=1 and b and c both hit zero. So A is between -1/3 and +2/3. Same applies to B and C of course. Since all three always have a modulus less than 1, ABC is less than any of the triplets AB, BC or CA, and therefore has to be less than their average. So no matter what you do, adding it on to that average isn't going to give a sum with the negative (AB+BC+CA)/3 positive. So whatever you do, this difference from 1/27 is going to be negative.<br>
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Therefore the best you can do is to have A=B=C=0, and a=b=c=1/3.comment:ask.metafilter.com,2007:site.68724-1027563Tue, 07 Aug 2007 08:16:00 -0800eddBy: DarkForest
http://ask.metafilter.com/68724/Bubbles#1027564
This is interesting, and I think I want to try it myself. I wonder about approaching it algebraically such that<br>
a = x + d<br>
b = x + e<br>
c = x + f<br>
and then show that any non-zero values for d, e, and f always detract from the volume.<br>
<small>don't laugh if this is completely wrong-headed, I like math, but I've never studied it much and my college calculus is long forgotten</small>comment:ask.metafilter.com,2007:site.68724-1027564Tue, 07 Aug 2007 08:17:17 -0800DarkForestBy: edd
http://ask.metafilter.com/68724/Bubbles#1027565
(yes, there's a smattering of moduluses not explicitly stated in that argument. Adding |s as appropriate is left as an exercise for the reader :-P )comment:ask.metafilter.com,2007:site.68724-1027565Tue, 07 Aug 2007 08:17:49 -0800eddBy: DarkForest
http://ask.metafilter.com/68724/Bubbles#1027567
... or do what edd said...comment:ask.metafilter.com,2007:site.68724-1027567Tue, 07 Aug 2007 08:18:45 -0800DarkForestBy: Blazecock Pileon
http://ask.metafilter.com/68724/Bubbles#1027714
That's very clever, edd, thanks. From reading, I think I'll also try the Lagrange multipliers as a different approach.comment:ask.metafilter.com,2007:site.68724-1027714Tue, 07 Aug 2007 10:22:22 -0800Blazecock PileonBy: jamjam
http://ask.metafilter.com/68724/Bubbles#1027800
It may amuse you to try to achieve your result from a consideration of the relationship between arithmetic and geometric means of arbitrary collections of positive numbers, on the one hand, and, almost without calculation on the other, by assuming the isoperimetric theorem in two dimensions (a circle has maximum area of any closed curve of a given perimeter) and then looking at any arbitrary three dimensional solid as being built up of its cross-sections.comment:ask.metafilter.com,2007:site.68724-1027800Tue, 07 Aug 2007 11:13:09 -0800jamjamBy: edd
http://ask.metafilter.com/68724/Bubbles#1028140
Definitely check the Lagrange Multipliers. If I were actually doing this in practice, I'd do it with Lagrange Multipliers. It's far more general, and I spent longer coming up with that argument than I'd ever have spent doing it the LM approach. But I like to do things the smartarse way now and again (much to the desperation of certain tutors when I did a problem on multiple integrals without so much as integrating once).comment:ask.metafilter.com,2007:site.68724-1028140Tue, 07 Aug 2007 15:21:41 -0800edd