Lagrange multi.

Arden1528

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.

Ambitwistor

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

In my problem I have the constraint equal to 1, should I adjust it and have it equal to 0?

Ambitwistor

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

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

Either that, or x=0.

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). Then I have the conditions to make this either a max or min. Value. Is this correct?

Ambitwistor

Almost; (0,0) is not a solution of the constraint. Otherwise, you're right.

Arden1528

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

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

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...