#### Arden1528

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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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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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.

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

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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.Originally posted by Arden1528

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

the partial of X: 2X+(lag. symbol)2X=0

thus getting (Lag. Symbol)= -1

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Either that, or x=0.Originally posted by Arden1528

the partial of X: 2X+(lag. symbol)2X=0

thus getting (Lag. Symbol)= -1

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?

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

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?

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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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