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Lagrange Function

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the optimal value of the function
    f (x,y) = 3.5x^2+y^2-42x-28y+5xy+190
    subject to
    6x+5y = 37

    2. Relevant equations
    Use the second order condition to determine if the optimal point is maximum
    or minimum


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 9, 2010 #2
    You need to at least attempt a solution if you want help. You can start by using Lagrange multipliers.
     
  4. Mar 9, 2010 #3
    df/dx=12.25x -42 +5y
    df/dy=2y-28+5x
    df/dlagrange = -(6x+5y-37)

    You make them all equal to 0 so
    df/dx=12.25x+5y-42=0
    df/dy=5x+2y-28=0
    df/dlangrange=6x+5y-37=0
    Is this right? then you find a common factor between the first two equations in order to solve for x and y? how do you find lagrange?
     
  5. Mar 9, 2010 #4

    ideasrule

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    Homework Helper

    No, that's not right. What's the Lagrangian function?
     
  6. Mar 9, 2010 #5
    I'm not sure. I am having difficulty solving this equation.
     
  7. Mar 9, 2010 #6

    ideasrule

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    Homework Helper

  8. Mar 10, 2010 #7
    dL/dx= 12.25x-42+5y-lagrange
    dL/dy= 5x-28+2y-lagrange
    dL/dL=-(6x+5y-37)

    So 12.25x-42+5y-lagrange=0
    5x-28+2y-lagrange=0
    -(6x+5y-37)=0

    What do you do next?
    Multiple them by a common factor.what do you do with lagrange?
     
  9. Mar 10, 2010 #8
    Why did you just type out the word "lagrange"? As ideasrule said before, first write out the lagrange function before you start taking derivatives.
     
  10. Mar 10, 2010 #9
    I typed out the word lagrange because i do not have the backward L function on my keyboard. It means -backward l . I know how to get the derivative now . I just don't know how to solve for x, y and lagrange(backward l)
     
  11. Mar 10, 2010 #10
    You solve for the lagrange function first then take the derivative. Can we at least see your lagrange function first.
     
  12. Mar 10, 2010 #11
    Sure.
    The function would be:f (x,y) = 3.5x^2+y^2-42x-28y+5xy+190 - backward L(6x+5y-37)
     
  13. Mar 10, 2010 #12
    You should definitely learn to use tex, but you can just use "L" for lambda and leave out "backward". That looks fine to me. So now you can take the partial derivatives, but this time use "L" and not "lagrange" since that is hard to understand.
     
  14. Mar 10, 2010 #13
    Ya so the derivative are:
    dL/dx= 12.25x-42+5y-L
    dL/dy= 5x-28+2y-L
    dL/dL=-(6x+5y-37)

    You then set them to 0 so they are
    dL/dx= 12.25x-42+5y-L=0
    dL/dy= 5x-28+2y-L=0
    dL/dL=-(6x+5y-37)=0
    Then, could you times the 2nd equation there which is dL/dy by 2 to give you common y terms of 10y?
     
  15. Mar 10, 2010 #14
    What happened to the terms multiplied by L? There was a (6x+5y-37) multiplied by L.
     
  16. Mar 10, 2010 #15
    so your saying you would put the dl/dl as -L(6X+5Y-37)=0
     
  17. Mar 10, 2010 #16
    I am going by an example that the teacher gave us and he said that the dl/dl would just go as -(6x+5y-37)
     
  18. Mar 10, 2010 #17
    i just do not know how to solve for x, y and l
     
  19. Mar 10, 2010 #18
    No, the first 2 partial derivatives you did. There are x and y terms multiplied by the L. What happened to them? For example:

    dL/dx = 12.25x-42+5y-6L = 0

    It should be 6L, because you have a 6x.
     
  20. Mar 10, 2010 #19
    I see what you mean now.
    So there is a 5 in front of the y for the constraint so that means that
    dy/dl would have to be : 5x-28 +2y-5L=0
     
  21. Mar 10, 2010 #20
    dl/dy that is
     
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