How do I solve a Lagrange Multiplier problem with a given constraint?

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To solve a Lagrange Multiplier problem with the function f(x,y)=x^2-y^2 and the constraint x^2+y^2=1, the method involves setting up the equations using partial derivatives. The partial derivative of x yields 2x + (λ)(2x) = 0, leading to λ = -1. The next step involves substituting values into the original function to find potential maximum or minimum points. For the second problem with f(x,y)=x^2y and g(x,y)=x^2+2y^2=6, the equations derived from the partial derivatives need to be solved simultaneously to find valid solutions. Ultimately, the goal is to identify values that satisfy all derived equations to determine extrema.
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.
 
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In my problem I have the constraint equal to 1, should I adjust it and have it equal to 0?
 
So then for my problem I would get something like

the partial of X: 2X+(lag. symbol)2X=0
thus getting (Lag. Symbol)= -1
 
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?
 
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?
 
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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