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Thread: LaGrange Multipliers

  1. March 21st 2009, 06:09 PM #1
    TheBerkeleyBoss
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    LaGrange Multipliers

    Hey guys! I just took a Calculus III midterm, and I believe I did fine. However, the last problem was the most difficult, and I am unsure if I will get full credit on it.

    It involves LaGrangian Multipliers...I'm going to start the problem with two objective equations both equaling "lambda" (after I've taken partial derivatives of both the objective and constraint equations). I cannot recall the original objective equation.

    The constraint equation is:

    x^2 + y^2 \leq 9


    Equation of all partials calculated and equal to each other through the value \lambda :

    x(1-x) = \lambda = y(1-y)



    therefore....

    x(1-x) = y(1-y)

    -x^2 + x = -y^2 + 1

    multiply both sides by -1....

    x^2 - x = y^2 -y

    I believe that I need to complete the squares...however, I'm not sure...I used the quadratic formula a few steps further, but the values I got do not make any sense. Can anyone help? Thanks!
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  2. March 22nd 2009, 02:59 AM #2
    Opalg
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    Quote Originally Posted by TheBerkeleyBoss View Post
    ....

    x(1-x) = y(1-y)

    -x^2 + x = -y^2 + 1

    multiply both sides by -1....

    x^2 - x = y^2 -y

    I believe that I need to complete the squares...however, I'm not sure...
    It's hard to see where this is going without knowing the whole problem. But I would continue the calculation like this:

    If x^2 - x = y^2 -y then x^2-y^2-(x-y) = 0. Factorise this as (x-y)(x+y-1) = 0. Therefore either x=y or x+y=1.
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  3. March 22nd 2009, 04:32 AM #3
    HallsofIvy
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    Quote Originally Posted by Opalg View Post
    It's hard to see where this is going without knowing the whole problem. But I would continue the calculation like this:

    If x^2 - x = y^2 -y then x^2-y^2-(x-y) = 0. Factorise this as (x-y)(x+y-1) = 0. Therefore either x=y or x+y=1.
    Or simply go from x^2- y^- (x-y)= 0 to x^2- y^2= x- y so (x-y)(x+ y)= (x-y).

    If x- y\ne 0 divide both sides by it to get x+y= 1. Of course, if x- y= 0, then x= y.
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