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Math Help - Lagrange Multiplier problem

  1. #1
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    Lagrange Multiplier problem

    Take home exam problem. Someone want to tell me if I'm missing something with this Lagrange multiplier problem?

    Problem: Find the maximum and minimum values of f(x,y)=xy on the curve (x+2)^2 + y^2 =1

    This is what I've done-
    gradient of f=(y,x)
    gradient of g=(2x+4,2y)

    According to Lagrange: gradient f = lambda*(gradient g)
    So we have the following equations.
    y=lambda*(2x+4)
    x=lambda*2y
    (x+2)^2 + y^2 =1

    That's 3 equations and 3 unknowns but I can't seem to find a solution. I feel like I'm missing something really stupid. Or my brain is hitting a wall to solve those equations. Can someone help?
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  2. #2
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    Quote Originally Posted by splash View Post
    Take home exam problem. Someone want to tell me if I'm missing something with this Lagrange multiplier problem?

    Problem: Find the maximum and minimum values of f(x,y)=xy on the curve (x+2)^2 + y^2 =1

    This is what I've done-
    gradient of f=(y,x)
    gradient of g=(2x+4,2y)

    According to Lagrange: gradient f = lambda*(gradient g)
    So we have the following equations.
    y=lambda*(2x+4)
    x=lambda*2y
    (x+2)^2 + y^2 =1

    That's 3 equations and 3 unknowns but I can't seem to find a solution. I feel like I'm missing something really stupid. Or my brain is hitting a wall to solve those equations. Can someone help?
    Okay, we have...
    f(x,y)=xy
    Thus,
    \nabla f(x,y)=<y,x>
    And,
    c(x,y)=(x+2)^2+y^2
    Thus,
    \nabla c(x,y)=<2x+4,2y>
    This agrees with what you have.

    By Lagrange multipliers we have,
    <y,x>=k<2x+4,2y>
    Thus,
    y=(2x+4)k
    x=2ky
    (x+2)^2+y^2=1
    From the second equation we have,
    k=\frac{x}{2y}
    Substitute that into the first equation,
    y=\frac{x(2x+4)}{2y}
    Thus,
    y^2=x(x+2)
    Substitute that into third equation,
    (x+2)^2+x(x+2)=1
    Thus,
    x^2+4x+4+x^2+2x=1
    Thus,
    2x^2+6x+3=0
    Now solve.
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