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Lagrange multipliers, guidance needed

  1. Mar 6, 2015 #1
    1. The problem statement, all variables and given/known data

    f(x,y) is function who's mixed 2nd order PDE's are equal.

    consider k_f:

    Untitled.png

    determine the points on the graph of the parabloid f(x,y) = x^2 + y^2 above the ellipse 3x^2 + 2y^2 = 1 at which k_f is maximised and minimised.






    3. The attempt at a solution




    is this the langrange equation that I'd use?

    [tex]L(x,y,\lambda )=k_{f}(x,y)-\lambda k_{g}(x,y)[/tex]

    or maybe it's just:
    [tex]L(x,y,\lambda )=k_{f}(x,y)-\lambda g(x,y)[/tex]

    im not sure if I should put the g function (3x^2 + 2y^2 = 1) into the k_f. If I do need to do this, then the algebra is very ugly.
     
  2. jcsd
  3. Mar 7, 2015 #2

    Ray Vickson

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    Have you made an attempt to simplify ##k_f(x,y)## as much as possible before using it in a Lagrangian or anywhere else? Far from ugly, the algebra will be very simple and straightforward.
     
  4. Mar 7, 2015 #3
    my k_f(x,y) is:

    [tex]\frac{4}{(1+4x^{2}+4y^{2})^2}[/tex]

    How would you simplify this/the problem? I tried simplifying with wolfram alpha, it doesn't return anything useful.

    my k_g(x,y) is:

    [tex]\frac{24}{(1+36x^{2}+16y^{2})^2}[/tex]

    please, someone help. It's a bit of an emergency.
     
    Last edited: Mar 7, 2015
  5. Mar 7, 2015 #4

    Ray Vickson

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    Why do you care about ##k_g##? It has nothing at all to do with the question as originally posed. Also: knowing the form of ##k_f## should allow you to simplify the problem significantly, to get a problem that is much easier than the original one. I cannot say more, without essentially doing the problem for you
     
  6. Mar 7, 2015 #5

    ok, then the Lagrange equation we are interested in is:

    [tex]L(x,y, \lambda)=\frac{4}{(1+4x^{2}+4y^{2})^2}- \lambda(3x^{2} + 2y^{2}-1)[/tex]
    what do you mean by form of k_f(x,y)? could you come up with a mini-alternative example to what you mean, I just don't see how you can simplify it, unless this has something unique to do with the problem as opposed to general algebraic manipulation.
     
  7. Mar 7, 2015 #6

    Ray Vickson

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    OK, so don't simplify it; just solve it as-is. It does work, but takes a bit more effort.
     
  8. Mar 8, 2015 #7
    [itex]\frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0[/itex]

    [itex]\frac{\partial L}{\partial y}= (-64y(1+4x^{2}+4y^{2})^{-3}) - 4y\lambda = 0[/itex]

    [itex]\frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0[/itex]

    this leads to a contradiction though, using the first and second equation we get:

    [itex](\frac{-32}{3}(1+4x^{2}+4y^{2})^{-3}) = \lambda = ((-24)(1+4x^{2}+4y^{2})^{-3})[/itex]
     
  9. Mar 8, 2015 #8

    Ray Vickson

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    No contradiction; you have just made some fatal errors.

    You can write
    [tex] 0= L_x = -Rx - 6x \lambda , \; \text{and} \; 0 = L_y = -Ry - 4 y \lambda [/tex]
    where ##R = 64/(1 + 4x^2+4 y^2)^3##. You have attempted to convert ##-Rx - 6 x \lambda = 0## into ##-R - 6 \lambda = 0## (similarly for ##y##), and you are NOT allowed to do that without some extra conditions.
     
  10. Mar 8, 2015 #9
    1. [itex]\frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0[/itex]

    2. [itex]\frac{\partial L}{\partial y}= (-64y(1+4x^{2}+4y^{2})^{-3}) - 4y\lambda = 0[/itex]

    3. [itex]\frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0[/itex]

    if you get an equation for lambda from the first and second equation without dividing through by any variables you get that (using [itex]\lambda[/itex]):

    [itex]\frac{32}{3}xy(1+4x^{2}+4y^{2})^{3} = -24xy(1+4x^{2}+4y^{2})^{3}[/itex]

    how am I meant to get a relationship between x and y from the above equation that I can utilize in the third equation?
     
  11. Mar 8, 2015 #10

    Ray Vickson

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    For any real ##x,y## we have ##1 + 4 x^2 + 4 y^2 \geq 1 > 0##, so you are allowed to divide it out on both sides, to end up with
    [tex] \frac{32}{3} xy = - 24 xy [/tex]
    What does that tell you?
     
    Last edited: Mar 8, 2015
  12. Mar 8, 2015 #11
    that x =y = 0, however this cannot be true due to the third equation.

    therefore x and y has to take multiple values that satisfy the following relationships:

    [itex]\frac{32}{3}xy = -24xy[/itex]

    [itex]1 - 3x^{2}-2y^{2}=0[/itex]

    going along the right lines?

    and thanks for the help so far.
     
  13. Mar 8, 2015 #12

    Ray Vickson

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  14. Mar 8, 2015 #13
    ok, so then x must equal y which must equal zero.

    Or perhaps my calculations are wrong...
     
  15. Mar 8, 2015 #14
    ...............
     
  16. Mar 8, 2015 #15
    someone please help me!
     
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