Results 1 to 3 of 3

Math Help - LaGrange Multipliers

  1. #1
    Junior Member
    Joined
    Mar 2009
    From
    Berkeley, California
    Posts
    26

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. lagrange multipliers
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 10th 2009, 03:24 AM
  2. Lagrange Multipliers
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 1st 2009, 12:49 PM
  3. Lagrange Multipliers in 4D
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 23rd 2009, 10:37 AM
  4. Lagrange multipliers
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 22nd 2009, 02:17 PM
  5. Lagrange Multipliers
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 2nd 2009, 10:26 PM

Search Tags


/mathhelpforum @mathhelpforum