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Thread: Need help with Constrained Optimization

  1. #1
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    Smile Need help with Constrained Optimization

    Need help to continue with the following solution: Find the stationary points of the function, $\displaystyle f(x,y) = x^2 + y^2$

    subject to the constraint, $\displaystyle 3x^2 + 7y^2 + 12x - 28y + 1 = 0 $

    $\displaystyle L_{xy \lambda} = x^2 + y^2 + \lambda (3x^2 + 7y^2 + 12x - 28y + 1)$

    $\displaystyle L_x = 2x + \lambda (6x + 12)$

    $\displaystyle L_y = 2y + \lambda (14y - 28)$

    $\displaystyle L_ \lambda = 3x^2 + 7y^2 + 12x - 28y + 1 $

    $\displaystyle L_x = L_y = L_ \lambda = 0$

    $\displaystyle 2x + \lambda (6x + 12) = 0$

    $\displaystyle 2y + \lambda (14y - 28) = 0$

    $\displaystyle 3x^2 + 7y^2 + 12x - 28y + 1 = 0 ---------[1]$

    $\displaystyle x + \lambda (3x + 6) = 0$

    $\displaystyle y + \lambda (7y - 14) = 0$

    $\displaystyle \lambda = \frac{-x}{3x + 6}$

    $\displaystyle \lambda = - \frac{y}{7y - 14}$

    equating, $\displaystyle - \frac{x}{3x + 6} = - \frac{y}{7y - 14}$

    $\displaystyle x(7y -14) = y(3x + 6)$

    $\displaystyle 7xy - 14x = 3xy + 6y$

    $\displaystyle 7xy - 3xy - 14x = 6y$

    $\displaystyle 4xy - 14x = 6y$

    $\displaystyle x(4y - 14) = 6y$

    $\displaystyle x = \frac {6y}{4y - 14}$

    Substituting for x in [1]

    $\displaystyle 3 \left ( \frac{6y}{4y - 14} \right )^2 + 7y^2 + 12 \left ( \frac{6y}{4y - 14} \right ) - 28y + 1 = 0$

    $\displaystyle ................$

    $\displaystyle ................$
    Last edited by ashura; Dec 1st 2006 at 05:41 AM. Reason: make correction
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ashura View Post
    Need help to continue with the following solution: Find the stationary points of the function, $\displaystyle f(x,y) = x^2 + y^2$

    subject to the constraint, $\displaystyle 3x^2 + 7y^2 + 12x - 28y + 1 = 0 $

    $\displaystyle L_{xy \lambda} = x^2 + y^2 + \lambda (3x^2 + 7y^2 + 12x - 28y + 1)$

    $\displaystyle L_x = 2x + \lambda (6x + 12)$

    $\displaystyle L_y = 2y + \lambda (14y - 28)$

    $\displaystyle L_ \lambda = 3x^2 + 7y^2 + 12x - 28y + 1$

    $\displaystyle L_x = L_y = L_ \lambda = 0$

    $\displaystyle 2x + \lambda (6x + 12) = 0$

    $\displaystyle 2y + \lambda (14y - 28) = 0$

    $\displaystyle 3x^2 + 7y^2 + 12x - 28y + 1 = 0$

    $\displaystyle x + \lambda (3x + 6) = 0$

    $\displaystyle y + \lambda (7y - 14) = 0$

    $\displaystyle \lambda = \frac{-x}{3x + 6}$

    $\displaystyle \lambda = - \frac{y}{7y - 14}$

    equating, $\displaystyle - \frac{x}{3x + 6} = - \frac{y}{7y - 14}$

    $\displaystyle x(7y -14) = y(3x + 6)$

    $\displaystyle 7xy - 14x = 3xy + 6y$

    $\displaystyle 7xy - 3xy - 14x = 6y$

    $\displaystyle 4xy - 14x = 6y$

    $\displaystyle x(4y - 14) = 6y$

    $\displaystyle x = \frac {6y}{4y - 14}$

    $\displaystyle 3 \left ( \frac{6y}{4y - 14} \right ) + 7y^2 + 12 \left ( \frac{6y}{4y - 14} \right )$
    You've got the method. A hint. You've got the system of equations:

    $\displaystyle 2x + \lambda (6x + 12) = 0$

    $\displaystyle 2y + \lambda (14y - 28) = 0$

    $\displaystyle 3x^2 + 7y^2 + 12x - 28y + 1 = 0$

    Pick one to solve for $\displaystyle \lambda$ and sub this in to get two equations in x and y. It will go slightly more smoothly.

    -Dan
    Last edited by Jameson; Dec 1st 2006 at 07:26 AM. Reason: fix math tag
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  3. #3
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    Unhappy

    That's what i'm trying to do here,

    $\displaystyle 3x^2 + 7y^2 + 12x - 28y + 1 = 0 ---------[1]$

    $\displaystyle x + \lambda (3x + 6) = 0$

    $\displaystyle y + \lambda (7y - 14) = 0$

    $\displaystyle \lambda = \frac{-x}{3x + 6}$

    $\displaystyle \lambda = - \frac{y}{7y - 14}$

    equating, $\displaystyle - \frac{x}{3x + 6} = - \frac{y}{7y - 14}$

    $\displaystyle x(7y -14) = y(3x + 6)$

    $\displaystyle 7xy - 14x = 3xy + 6y$

    $\displaystyle 7xy - 3xy - 14x = 6y$

    $\displaystyle 4xy - 14x = 6y$

    $\displaystyle x(4y - 14) = 6y$

    $\displaystyle x = \frac {6y}{4y - 14}$

    Substituting for x in [1]

    $\displaystyle 3 \left ( \frac{6y}{4y - 14} \right )^2 + 7y^2 + 12 \left ( \frac{6y}{4y - 14} \right ) - 28y + 1 = 0$

    But I would like a simpler expression for,

    $\displaystyle x = \frac {6y}{4y - 14}$
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ashura View Post
    That's what i'm trying to do here,

    $\displaystyle 3x^2 + 7y^2 + 12x - 28y + 1 = 0 ---------[1]$

    $\displaystyle x + \lambda (3x + 6) = 0$

    $\displaystyle y + \lambda (7y - 14) = 0$

    $\displaystyle \lambda = \frac{-x}{3x + 6}$

    $\displaystyle \lambda = - \frac{y}{7y - 14}$

    equating, $\displaystyle - \frac{x}{3x + 6} = - \frac{y}{7y - 14}$

    $\displaystyle x(7y -14) = y(3x + 6)$

    $\displaystyle 7xy - 14x = 3xy + 6y$

    $\displaystyle 7xy - 3xy - 14x = 6y$

    $\displaystyle 4xy - 14x = 6y$

    $\displaystyle x(4y - 14) = 6y$

    $\displaystyle x = \frac {6y}{4y - 14}$

    Substituting for x in [1]

    $\displaystyle 3 \left ( \frac{6y}{4y - 14} \right )^2 + 7y^2 + 12 \left ( \frac{6y}{4y - 14} \right ) - 28y + 1 = 0$

    But I would like a simpler expression for,

    $\displaystyle x = \frac {6y}{4y - 14}$
    Sorry. I had a brain fart this morning. I see what you are doing now.

    I've played with this system for a while and can't see any way to make it less of a nightmare. Whether or not you get the equation in x or in y you wind up with a quartic equation which just doesn't look like it's got nice solutions. You'll have to do it numerically. Looking at the graph of the x equation it looks like you've got 2 real solutions lurking. One is near x = 1/2 and the other is large and negative.

    -Dan
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