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Math Help - LaGrange Multiplier Problem

  1. #1
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    LaGrange Multiplier Problem

    Hi. I have the equation x^2 + 4y^2, with the constraint xy=1. (with m being used to indicate theta)
    Hence, F(x,y,m)= x^2 + 4y^2 + m(xy-1).

    Derivatives:

    Fx= 2x + my=0
    Fy= 8y+ mx=0
    Fm= xy-1=0.

    What is the correct way to solve for y and x; I keep getting y= the sq. root of 1/2, but the book says this is wrong.

    Many thanks
    Jay
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  2. #2
    Ted
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    Quote Originally Posted by jaijay32 View Post
    Hi. I have the equation x^2 + 4y^2, with the constraint xy=1. (with m being used to indicate theta)
    Hence, F(x,y,m)= x^2 + 4y^2 + m(xy-1).

    Derivatives:

    Fx= 2x + my=0
    Fy= 8y+ mx=0
    Fm= xy-1=0.

    What is the correct way to solve for y and x; I keep getting y= the sq. root of 1/2, but the book says this is wrong.

    Many thanks
    Jay
    x^2 + 4y^2 is not an equation.
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  3. #3
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    f_{x} = 2x + my = 0 \implies m = -\frac{2x}{y}

    Substitute m = -\frac{2x}{y} into f_{y} = 8y + mx = 0 to produce:

    f_{y} = 8y + (-\frac{2x}{y})x = 8y - \frac{2x^2}{y} = 0.

    Solving for y produces:

    y = \pm\frac{x}{2}.

    Substitute y = \pm\frac{x}{2} into f_{m} = xy - 1 = 0 to produce:

    f_{m} = x(\pm\frac{x}{2}) - 1 = 0.

    Solving for x and disregarding the imaginary solutions produces:

    x = \pm\sqrt{2}

    Substituting x = \pm\sqrt{2} into y = \pm\frac{x}{2} produces:

    y = \pm\frac{\pm\sqrt{2}}{2} = \frac{\sqrt{2}}{2}
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  4. #4
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    Quote Originally Posted by NOX Andrew View Post
    f_{x} = 2x + my = 0 \implies m = -\frac{2x}{y}

    Substitute m = -\frac{2x}{y} into f_{y} = 8y + mx = 0 to produce:

    f_{y} = 8y + (-\frac{2x}{y})x = 8y - \frac{2x^2}{y} = 0.

    Solving for y produces:

    y = \pm\frac{x}{2}.

    Substitute y = \pm\frac{x}{2} into f_{m} = xy - 1 = 0 to produce:

    f_{m} = x(\pm\frac{x}{2}) - 1 = 0.

    Solving for x and disregarding the imaginary solutions produces:

    x = \pm\sqrt{2}

    Substituting x = \pm\sqrt{2} into y = \pm\frac{x}{2} produces:

    y = \pm\frac{\pm\sqrt{2}}{2} = \frac{\sqrt{2}}{2}
    After substituting x you should get two sets of solutions:

    x=-\sqrt {2}, y=-1/2\,\sqrt {2}, m=-4
    x=\sqrt {2}, y=1/2\,\sqrt {2}, m=-4
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