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Thread: 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.

    $\displaystyle f(x,y)=x^3-y^2$

    Constraint:

    $\displaystyle x+y=1$

    Using the Lagrange Multiplier method, I managed to obtain

    $\displaystyle 3x^2=\lambda$

    $\displaystyle -2y=\lambda$

    Solving for the variables gave me:

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

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

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

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

    Then I end up with the quadratic, after simplifying;

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

    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 $\displaystyle y=1-x$, we obtain

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

    So, $\displaystyle \varphi'(x)=3x^2-2x+2>0\quad \forall x\in\mathbb{R}$ (there are no extrema).
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  3. #3
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    Re: Lagrange Multipliers

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