# BubblesAugust 7, 2007 6:55 AM   Subscribe

Maximal volume of an ellipsoid...

The volume of an ellipsoid is: 4πabc/3

An ellipsoid is defined with the function: x^2/ a^2 + y^2/ b^2 + z^2/ c^2 = 1

In the case where the sum a + b + c is a fixed value, how might I derive that the ellipsoid of maximal volume is a sphere?

I'd expect to show that a = b = c but am not certain where to start.

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. I'd appreciate hints rather than a full solution. Feel free not to answer if this bothers you.
posted by Blazecock Pileon to Education (15 answers total) 2 users marked this as a favorite

Best answer: 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.
posted by caek at 7:02 AM on August 7, 2007

Method 1
Step 1: use a+b+c=constant to eliminate a variable in the volume
Step 2: use calculus* to check that there exists a local maximum
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.

Method 2
Use "lagrange multipliers" to get everything in one equation, repeat steps 2,3

If you don't remember the * needed, that's easy to fix.
posted by a robot made out of meat at 7:07 AM on August 7, 2007

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.
posted by a robot made out of meat at 7:10 AM on August 7, 2007

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.
posted by jepler at 7:12 AM on August 7, 2007

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).

this means that the result has to be symmetrical too (which is equivalent to saying a = b = c).

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.
posted by andrew cooke at 7:31 AM on August 7, 2007

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.
posted by a robot made out of meat at 7:46 AM on August 7, 2007

Hint:
You have three equations with three variables. Simplify. Then maximize.
posted by Eringatang at 8:08 AM on August 7, 2007

Yeah, if I were you I'd use Lagrange multipliers. The Wikipedia article on the subject has a basic overview of how to use them.
posted by Johnny Assay at 8:15 AM on August 7, 2007

Best answer: I think we should do this without calculus. Where's the fun in it otherwise?

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?).

a, b and c must all differ from a third by something. a=1/3+A, b=1/3+B etc.
So a+b+c=1/3+A+1/3+B+1/3+C=1, so A+B+C=0.

Now abc - spend some time writing that out and get
1/27+(A+B+C)/9+(AB+BC+CA)/3+ABC
Forget the 1/27, who cares about that. A+B+C is 0, so forget about that. We want to maximise
(AB+BC+CA)/3+ABC.

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.
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.

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.

Therefore the best you can do is to have A=B=C=0, and a=b=c=1/3.
posted by edd at 8:16 AM on August 7, 2007

This is interesting, and I think I want to try it myself. I wonder about approaching it algebraically such that
a = x + d
b = x + e
c = x + f
and then show that any non-zero values for d, e, and f always detract from the volume.
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
posted by DarkForest at 8:17 AM on August 7, 2007

(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 )
posted by edd at 8:17 AM on August 7, 2007

... or do what edd said...
posted by DarkForest at 8:18 AM on August 7, 2007

Response by poster: That's very clever, edd, thanks. From reading, I think I'll also try the Lagrange multipliers as a different approach.
posted by Blazecock Pileon at 10:22 AM on August 7, 2007

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.
posted by jamjam at 11:13 AM on August 7, 2007

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).
posted by edd at 3:21 PM on August 7, 2007

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