Results 1 to 3 of 3

Thread: Maxima and Minima using Lagrange Multipliers

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    7

    Maxima and Minima using Lagrange Multipliers

    Find the local maxima and minima of the following problem by introducing two Lagrange multipliers:

    $\displaystyle f(x_1,x_2,x_3) = x_1 + 2x_2 + 2x_3$

    subject to

    $\displaystyle x_1^2 + x_2^2 + 4x_3^2 = 2$ and $\displaystyle x_1^2 + (x_2 - 1)^2 + 4_3^2 = 3$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17
    Alright. There are some parts of this problem that should be pretty straightforward. Have you introduced said multipliers? Have you written the equations. It'd be awesome if you can write down the vector equations. (you can use \nabla for $\displaystyle \nabla$, unles you know it of course)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2010
    Posts
    7
    Okay, I am stuck. Here is what I have now though....


    Define Constraints:
    $\displaystyle g_1(x_1, x_2, x_3) = x_1^2 + x_2^2 + 4x_3^2 = 1 $ , $\displaystyle g_2(x_1, x_2, x_3) = x_1^2 + (x_2 - 1)^2 + 4x_3^2 = 3 $

    Then

    $\displaystyle L(x) = x_1 + 2x_2 + 2x_3 + \lambda_1(x_1^2 + x_2^2 + 4x_3^2) + \lambda_2(x_1^2 + (x_2 - 1)^2 + 4x_3^2) $

    Critical Points satisfy

    $\displaystyle \frac{\partial L}{\partial x_1} = 1 + 2 \lambda_1 x_1 + 2 \lambda_2 x_1 = 0 $
    $\displaystyle \frac{\partial L}{\partial x_2} = 2 + 2 \lambda_1 x_2 + 2 \lambda_2 (2x_2 - 2) = 0 $
    $\displaystyle \frac{\partial L}{\partial x_3} = 2 + 8 \lambda_1 x_3 + 8 \lambda_2 x_3 = 0 $

    But im having trouble finding the values for $\displaystyle x_1, x_2, x_3 $ and then finding $\displaystyle \lambda_1 , \lambda_2$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Maxima and Minima
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 23rd 2009, 12:03 AM
  2. Maxima and Minima
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 30th 2009, 04:06 AM
  3. Maxima and minima 3
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 28th 2009, 08:30 AM
  4. Maxima/Minima
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Dec 2nd 2008, 07:11 AM
  5. maxima and minima
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Nov 13th 2008, 04:12 PM

Search Tags


/mathhelpforum @mathhelpforum