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Math Help - Lagrange Multiplier Problem

  1. #1
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    Lagrange Multiplier Problem

    Solve the following problem using Lagrange multipliers. 1.) Find the critical points of f(x,y)= \frac{x}{y} on the curve y=x^2+1.

     \nabla{f}=<\frac{1}{y},x>

     \nabla{g}=<2x,-1>

    Thus, yeilding my three equations with three unknowns:

    \frac{1}{y}=\lambda*2x

    x=\lambda*(-1)

    x^2-y+1=0

    Solving the equations, I get the following:

    y=\frac{-1}{2\lambda^2}

    x=-\lambda

    Plugging these into the third equation yeilds:

    (-\lambda)^2-\frac{-1}{2\lambda^2}=-1

    \lambda^2+\frac{1}{2\lambda^2}=-1

    This is where I'm stuck, how can \lambda^2 (a positive number) plus \frac{1}{2\lambda^2} (also a positive number) = -1 when added together. Can someone help me find my mistake so I can finish the problem? Thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Lagrange Multiplier Problem

    Quote Originally Posted by dbakeg00 View Post
    \frac{1}{y}=\lambda*2x
    Right.

    x=\lambda*(-1)
    It should be x/y^2=\lambda

    x^2-y+1=0
    Right.
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  3. #3
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    Re: Lagrange Multiplier Problem

    The derivative of \frac{x}{y} with respect to y, as FernandoRevilla suggests, is NOT "x" but -\frac{y}{x^2}. Write \frac{y}{x} as x^{-1}y and it should be easy to see why.
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  4. #4
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    Re: Lagrange Multiplier Problem

    Hello, dbakeg00!

    I was taught a different method for Lagrange multipliers.


    Solve, using Lagrange multipliers.
    Find the critical points of: . f(x,y)\:=\:\frac{x}{y} on the curve y\,=\,x^2+1

    \text{We have: }\:g(x,y,\lambda) \;=\;\frac{x}{y} + \lambda(x^2+1-y)


    Set the partial derivatives equal to zero and solve.

    . . \begin{array}{ccccccc}\dfrac{\partial g}{\partial x} &=& \dfrac{1}{y} + 2x\lambda &=& 0 & [1] \\ \\[-4mm] \dfrac{\partial g}{\partial y} &=& \text{-}\dfrac{x}{y^2} - \lambda &=& 0 & [2] \\ \\[-3mm] \dfrac{\partial g}{\partial\lambda} &=& x^2+1-y &=& 0 & [3] \end{array}


    From [2]: . \text{-}\frac{x}{y^2} - \lambda \:=\:0 \quad\Rightarrow\quad \lambda \,=\,\text{-}\frac{x}{y^2}

    Substitute into [1]: . \frac{1}{y} + 2x\left(\text{-}\frac{x}{y^2}\right) \:=\:0
    . . \frac{y-2x^2}{y^2} \:=\:0 \quad\Rightarrow\quad y-2x^2 \:=\:0 \quad\Rightarrow\quad y \:=\:2x^2\;\;[4]


    Substitute into [3]: . x^2 + 1 - 2x^2 \:=\:0
    . . x^2\:=\:1 \quad\Rightarrow\quad x \:=\:\pm1


    Sustitute into [4]: . y \:=\:2(\pm1)^2 \quad\Rightarrow\quad y \:=\:2


    The critical points are: . \left(1,2,\tfrac{1}{2}\right),\;\left(\text{-}1,2,\text{-}\tfrac{1}{2}\right)

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  5. #5
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    Re: Lagrange Multiplier Problem

    Thanks Fernando & Halls...that was a dumb mistake which I should have caught. Soroban, I'll have to study your way for a bit and see how I like it. Thanks again to all for the help.
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