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Math Help - Lagrange Multipliers :D

  1. #1
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    Lagrange Multipliers :D

    ts a simple question i think?.. But im not too sure what the answer is myself, which is obviously why im asking...

    Ok, the question goes like this

    Maximise and Minimise  x^2 +3y^2 subject to  x^2 +xy +2y^2 = 56 .

    I understand that i have to differentiate them in respect to x, y, and then lambda.. Rearrange them accordingly to find x and y.

     2x + \lambda(2x+y) = 0

             6y+ \lambda(x+4y) = 0

     x^2 + xy + 2y^2 -56 = 0

    Thats what i get.. Not sure where to go from there, i realise that i have to substitute bits in.. but ive tried everything, and cant really seem to get any further. thanks in advance
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  2. #2
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    Quote Originally Posted by brd_7 View Post
    ts a simple question i think?.. But im not too sure what the answer is myself, which is obviously why im asking...

    Ok, the question goes like this

    Maximise and Minimise  x^2 +3y^2 subject to  x^2 +xy +2y^2 = 56 .

    I understand that i have to differentiate them in respect to x, y, and then lambda.. Rearrange them accordingly to find x and y.

     2x + \lambda(2x+y) = 0 \, Edit: ..... (1)

             6y+ \lambda(x+4y) = 0 \, Edit: .... (2)
     x^2 + xy + 2y^2 -56 = 0 \, Edit: .... (3)

    Thats what i get.. Not sure where to go from there, i realise that i have to substitute bits in.. but ive tried everything, and cant really seem to get any further. thanks in advance
    From (1): \lambda = -\frac{2x}{2x + y}, y \neq 2x ..... (1')


    Sub (1') into (2): 6y - \frac{2x(x + 4y)}{2x + y} = 0.

    Multiply through by 2x + y: 6y(2x + y) - 2x(x+4y) = 0 \therefore x^2 - 2xy + 3y^2 = 0 .... (2')


    Does it help if I now suggest trying to solve (2') and (3) simultaneously? Then test the nature of the resulting solutions.
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  3. #3
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    Ok, i tried substituting but i just got a big mess again.. A good sign is that i can get equations only in terms of x or only in terms of y.. But still cant seem to get anything.. could you maybe show me one more step? Thanks for the help
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  4. #4
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    Quote Originally Posted by brd_7 View Post
    Ok, i tried substituting but i just got a big mess again.. A good sign is that i can get equations only in terms of x or only in terms of y.. But still cant seem to get anything.. could you maybe show me one more step? Thanks for the help
    Did you check my working or just copy it. Don't answer, I know it's the latter. Tsk tsk. Because there's a typo in equation (2'). It should read:

    x^2 - 2xy - 3y^2 = 0. That makes a big big difference ......

    Solutions are x = 2 \sqrt{7}, \, y = -2 \sqrt{7} or x =- 2 \sqrt{7}, \, y = 2 \sqrt{7} or x = 6, y = 2 or x = - 6, y =-2.

    Moral of the story: The method is usually OK but you should ALWAYS check the details.
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