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Math Help - LaGrange Multipliers

  1. #1
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    LaGrange Multipliers

    Grrr... This problem has been stressing me out for the past half hour


    Find the extreme values of f(x,y) = xy on the ellipse ((x^2)/8) + ((y^2)/2) = 1


    i've gone ahead and figured out the gradient f and g and ended up with (we'll call L lambda since I don't know how to insert that

    y = (Lx)/4

    x = Ly

    ((x^2)/8) + ((y^2)/2) = 1

    I've gotten it down to this system of equations but i'm having trouble solving for this. Any help is appreciate.
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  2. #2
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    Hello, Jonathan!

    Sorry, your set-up is all wrong . . .


    Find the extreme values of f(x,y) = xy on the ellipse \frac{x^2}{8} + \frac{y^2}{2} \: =\:1
    We want the extreme values of f(x,y) with the constraint: . x^2+4y^2-8\:=\:0

    Our function is: . F(x,y) \;=\;xy + \lambda(x^2+4y^2-8)


    Then: . \begin{array}{ccccc}\frac{\partial F}{\partial x} &=& y + 2x\lambda &=& 0 \\ \\[-3mm]<br />
\frac{\partial F}{\partial y} &=& x + 8y\lambda &=&0 \\  \\[-3mm]<br />
\frac{\partial F}{\partial\lambda} &=& x^2+4y^2-8 &=& 0 \end{array}


    Now solve that system . . .

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  3. #3
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    Wow... just wow... you made it so much easier.



    I just figured it out using my same setup but this would of been so much easier. This is what I did


    Eq. 1 = y = (Lx)/4

    Eq. 2 = x = Ly

    Eq. 3 = ((x^2)/8) + ((y^2)/2) = 1


    I substituted Eq.2 into Eq.1 for the x value and ended up with L = +2 and -2

    Taking the value of lambda of +2 I put it in Eq.2 and got x = 2y

    I then subbed 2y for the x value in Eq.3 and ended up with a y value of +1 and -1. Finding x is simple and got +2 and -2

    I ended with extreme values of 2 and -2 for my final answer. All in all though, your way seems WAYYY easier and i'm gonna try it right now. Thanks!
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