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From: Ray Vickson on 6 Aug 2010 18:13
On Aug 6, 11:56 am, "Robert H. Lewis" <rle...(a)fordham.edu> wrote:
> I don't see any simple, easy algebraic manipulations. Maybe someone will prove me wrong.

As explained in my first posting, we can manipulate the optimality
equations to *prove* that two of x, y and z must be equal; thus, if we
set x = y = s and z = t we have 2*s+t = 12 and s^2*t = 54. You get a
cubic for s with s1 = 3 as one root, s2 = (3/2) + (3/2)*sqrt(5) as
another positive root, plus one negative root. The point x = y = s1, z
= 12 - 2*s1 is a global max, while x = y = s2, z = 12 - 2*s2 is a
global min. Other max/min are obtained by permuting the components of
these two points.

R.G. Vickson

>
> Purely algebraically, you have a system of 5 polynomial equations in 5 variables.  You could compute the resultants. Granted, the text book probably has not gone into that topic.
>
> Geometrically, the first equation is of course just a slanting plane. The second is fairly well known, the z = c/(xy) hyperbolic "curtains".  A graph is here:
>
> http://www.math.wisc.edu/~miller/old/m223-96/progs/rat0.gif
>
> There are 4 sheets to the image, one above/below each xy-quadrant. If you look at that image, picture the slanting plane z = 12 - x - y.  When x=y, that is z = 12 - 2x, so that will help visualize it.  The intersection with the third quadrant sheet (x<0, y<0) is a sort of slanting hyperbola. It is unbounded. Therefore, there are points on this intersection arbitrarily far from the origin.  To get one, let y, say, be some large negative number, like y = -10.  Then z = 22-x and also z = -5.4/x. Find an x that makes these equal.  It's roughly x = -.24. Repeat for -100, -1000, etc.
>
> The intersection with the first quadrant is kind of cool.  (I think the 12 and 54 were chosen so that there is an intersection.)  It's going to be one of those cubics, a sort of squashed ellipse.
>
> So what's the answer?  Don't know, but if you let x=y, you get (3,3,6).  I'd bet on that, and its permutations.
>
> Robert H. Lewis
> Fordham University

From: achille on 6 Aug 2010 20:18
On Aug 7, 6:13 am, Ray Vickson <RGVick...(a)shaw.ca> wrote:
> On Aug 6, 11:56 am, "Robert H. Lewis" <rle...(a)fordham.edu> wrote:
>
> > I don't see any simple, easy algebraic manipulations. Maybe someone will prove me wrong.
>
> As explained in my first posting, we can manipulate the optimality
> equations to *prove* that two of x, y and z must be equal; thus, if we
> set x = y = s and z = t we have 2*s+t = 12 and s^2*t = 54. You get a
> cubic for s with s1 = 3 as one root, s2 = (3/2) + (3/2)*sqrt(5) as
> another positive root, plus one negative root. The point x = y = s1, z
> = 12 - 2*s1 is a global max, while x = y = s2, z = 12 - 2*s2 is a
> global min. Other max/min are obtained by permuting the components of
> these two points.
>
> R.G. Vickson
>
A minor point, the point (x,y,z) = (3,3,6) is not a global
maximum. It is a 'max' only when you restrict (x,y,z) to
the first octant( x, y, z > 0).

The set of solutions for x+y+z = 12, xyz = 54 has 3 other
connected components, lying in the octants [--+], [-+-] and
[+--] respectively. On these connected components,

x^2+y^2+z^2 >= 153+45 sqrt(5) ~ 253.6230589874905

and the remaining negative root s3 = 3/2 (1-sqrt(5))
corresponds to one of the local minima (s3,s3,12-2*s3)
in the octant [--+].
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