Inequality. (1 Viewer)

[Fus Ro Dah]

 

[Fus Ro Dah]

Oh and I know it's tempting to use Lagrangian Multipliers for max(P(x,y,z)) but please restrain from doing so.

 

[barbernator]

P(x,y,z)=2. if it is correct, i will post solution. hmmm my algebra came out as that, but subbing in stuff it doesn't work...must have subben in wrong lol

 

[seanieg89]

The expression is identically 2 for (x,y,z) satisfying the given constraint. It falls out from the substitution x=u/v, y=v/w, z=w/u. Such a triple (u,v,w) exists (and is unique up to non-zero real multiples) for any given triple (x,y,z).

[The point of the substitution is it incorporates the contraint into its definition.]

 

[Fus Ro Dah]

Cool result isn't it? Was trying this question and when I found out that P(x,y,z)=2, I thought it was pretty funny.

 

[seanieg89]

It is pretty surprising haha.

 

[asianese]

Is it a problem that we don't learn functions of 3 variables?

 

[Fus Ro Dah]

Not really. Most Extension 2 inequality questions are of 3 variables and can actually be defined to be P(a,b,c) or P(x,y,z). It doesn't affect the question at all, it's just a much shorter and easy way to refer to the expression instead of typing it all over again.