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Math Help - Lagrange multiplier method

  1. #1
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    Lagrange multiplier method

    For the function f(x,y) = 2x^3 + 3xy + 2y^3,
    if the values of c and y are constrained to lie on the straight line y+x=1 use the lagrange multiplier method to find the value of (x,y) where f(x,y) has a turning point.

    So far I have computed f sub x, f sub y and f sub lambda to produce 3 equations which imply that y=1-x,
    6x^2 + 3 -3x + lambda =0 and
    6x^2 - 9x + 6 + lamda = 0

    at this point I get stuck
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  2. #2
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    Quote Originally Posted by linearalgebradunce View Post
    For the function f(x,y) = 2x^3 + 3xy + 2y^3,
    if the values of c and y are constrained to lie on the straight line y+x=1 use the lagrange multiplier method to find the value of (x,y) where f(x,y) has a turning point.

    So far I have computed f sub x, f sub y and f sub lambda to produce 3 equations which imply that y=1-x,
    6x^2 + 3 -3x + lambda =0 and
    6x^2 - 9x + 6 + lamda = 0

    at this point I get stuck
    I cannot follow what you've done at all.


    L = 2x^3 + 3xy + 2y^3 + \lambda (y + x - 1).

    \frac{\partial L}{\partial x} = 0 \Rightarrow 6x^2 + 3y + \lambda = 0 .... (1)

    \frac{\partial L}{\partial y} = 0 \Rightarrow 6y^2 + 3x + \lambda = 0 .... (2)

    \frac{\partial L}{\partial \lambda} = 0 \Rightarrow y + x = 1 .... (3)


    Note:

    From (1) and (2) it follows that

    6x^2 + 3y = 6y^2 + 3x \Rightarrow 2x^2 - x - 2y^2 + y = 0 .... (4)

    From (3) it follows that

    x = 1 -y .... (5)


    Solve equations (4) and (5) simultaneously, test nature of extrema etc.
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  3. #3
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    My apologies in my post I inputted the wrong equations corresponding to f sub x, etc.
    I found the same 3 equations initially and have now found the value of (x,y) by solving the equations simultaneously, I have never seen or used the Lagrange multiplier before however and I do not know what the next steps are?! I have an example I am trying to follow but I cannot see what they have done after they have their (x,y) values.
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