Lagrange multi.


by Arden1528
Tags: lagrange, multi
Arden1528
#1
Nov9-03, 12:56 PM
P: n/a
We have started to do Lagrange Multi. in my class and my book has a very short section on how to solve these. I was wondering if someone couls help.
The problem is f(x,y)=x^2-y^2 with the constraint x^2+y^2=1.
I have found the partial derv. but I am not sure on what else to do. Any help would be sweet, later.
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Ambitwistor
Ambitwistor is offline
#2
Nov9-03, 01:21 PM
P: 837
If you have a function f(x,y), and a constraint g(x,y) = 0, then to find the constrained extrema you set the partial derivatives of f + λg to zero, and solve for x and y. Remember to take partial derivatives not only with respect to x and y, but also with respect to λ; otherwise, you won't impose the constraint.
Arden1528
#3
Nov9-03, 01:37 PM
P: n/a
In my problem I have the constraint equal to 1, should I adjust it and have it equal to 0?

Ambitwistor
Ambitwistor is offline
#4
Nov9-03, 01:38 PM
P: 837

Lagrange multi.


Originally posted by Arden1528
In my problem I have the constraint equal to 1, should I adjust it and have it equal to 0?
It's traditional to write the constant as zero, but it really doesn't matter; it will work with any constraint g(x,y) = constant, because the constant goes away after you take the derivatives.
Arden1528
#5
Nov9-03, 01:48 PM
P: n/a
So then for my problem I would get something like

the partial of X: 2X+(lag. symbol)2X=0
thus getting (Lag. Symbol)= -1
Ambitwistor
Ambitwistor is offline
#6
Nov9-03, 01:52 PM
P: 837
Originally posted by Arden1528
the partial of X: 2X+(lag. symbol)2X=0
thus getting (Lag. Symbol)= -1
Either that, or x=0.
Arden1528
#7
Nov9-03, 01:58 PM
P: n/a
Then take that 1 and 0 and plug them into the orig. equation of F(X)
and get (+-1,0) or (0,+-1) or (0,0). Then I have the conditions to make this either a max or min. Value. Is this correct?
Ambitwistor
Ambitwistor is offline
#8
Nov9-03, 02:02 PM
P: 837
Originally posted by Arden1528
Then take that 1 and 0 and plug them into the orig. equation of F(X)
and get (+-1,0) or (0,+-1) or (0,0).
Almost; (0,0) is not a solution of the constraint. Otherwise, you're right.
Arden1528
#9
Nov9-03, 02:08 PM
P: n/a
Alright, then for my next problem I have
F(X);x^2y and G(X);x^2+2y^2=6

The partials of x I get
2xy+(Lang.)2x=0, giving me (Lang.)=y, can this be true?
Ambitwistor
Ambitwistor is offline
#10
Nov9-03, 02:15 PM
P: 837
That's one solution (again, another is x=0), but you'll have to impose the other derivative constraints too, and you'll find that, in that case, they restrict what y can be.
Arden1528
#11
Nov9-03, 02:23 PM
P: n/a
I have thought about the three equations
Partial x
2xy=(Lang.)2x
Partial y
x^2=(Lang.)4y
and
x^2+2y^2=6
and am looking for numbers that satisfy all equations. So I would get something like...


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