Fus Ro Dah
Member
- Joined
- Dec 16, 2011
- Messages
- 248
- Gender
- Male
- HSC
- 2013
-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Inequality. (1 Viewer)
- Thread starter Fus Ro Dah
- Start date
Fus Ro Dah
Member
- Joined
- Dec 16, 2011
- Messages
- 248
- Gender
- Male
- HSC
- 2013
Oh and I know it's tempting to use Lagrangian Multipliers for max(P(x,y,z)) but please restrain from doing so.
barbernator
Active Member
- Joined
- Sep 13, 2010
- Messages
- 1,439
- Gender
- Male
- HSC
- 2012
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 
Last edited:
seanieg89
Well-Known Member
- Joined
- Aug 8, 2006
- Messages
- 2,662
- Gender
- Male
- HSC
- 2007
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.]
[The point of the substitution is it incorporates the contraint into its definition.]
Last edited:
Fus Ro Dah
Member
- Joined
- Dec 16, 2011
- Messages
- 248
- Gender
- Male
- HSC
- 2013
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.
Fus Ro Dah
Member
- Joined
- Dec 16, 2011
- Messages
- 248
- Gender
- Male
- HSC
- 2013
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.