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Math Help - Lagrange Multipliers

  1. #1
    Member iPod's Avatar
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    Lagrange Multipliers

    Find the maxima and the minima of the given function given the constraint.

    f(x,y)=x^3-y^2

    Constraint:

    x+y=1

    Using the Lagrange Multiplier method, I managed to obtain

    3x^2=\lambda

    -2y=\lambda

    Solving for the variables gave me:

    x=\pm \sqrt{\frac{\lambda}{3}}

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

    So then I plug these values into my constraint to end up with

    \pm \sqrt{\frac{\lambda}{3}}-\frac{\lambda}{2}=1

    Then I end up with the quadratic, after simplifying;

    3\lambda^2+8\lambda+12=0

    However, this quadratic can't be solved using real numbers. Have I done anything wrong or am I using the wrong approach?

    Thank you
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Lagrange Multipliers

    Quote Originally Posted by iPod View Post
    However, this quadratic can't be solved using real numbers. Have I done anything wrong or am I using the wrong approach?
    You are right. There are no critical points. Another way: substituting y=1-x, we obtain

    \varphi (x)=f(x,1-x)=x^3-(1-x)^2=x^3-x^2+2x-1

    So, \varphi'(x)=3x^2-2x+2>0\quad \forall x\in\mathbb{R} (there are no extrema).
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  3. #3
    MHF Contributor

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    Re: Lagrange Multipliers

    Because a specific value \lambda is NOT a part of the solution, I find it simplest to eliminate \lambda first. Here, instead of solving for x and y in terms of \lambda, I would use 3x^2= \lambda and -2y= \lambda to say that 3x^2= -2y so that y= -(3/2)x^2. Put that into the condition x+ y= 1 to get x- (3/2)x^2= 1 which is equivalent to 3x^2- 2x+ 2= 0. The discriminant for that quadratic equation is (-2)^2- 4(3)(2)= 4- 24= -20 so, yes, there is no real value of x satisfying this problem.
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