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Lagrange multipliers Q

  1. Feb 20, 2006 #1
    I'm stuck on the following question

    "Find the maximum and minimum values of f(x,y,z) = [tex] x^2y^2-y^2z^2 + z^2x^2[/tex] subject to the constraint of [tex] x^2 + y^2 + z^2 = 1[/tex] by using the method of lagrange multipliers.

    Write the 4 points where the minimum value is achieved and the 8 points where the max is achieved"

    So far, I've managed to calculate the gradient for both formulae and have the resulting equations:

    [tex]2xy^2 + 2z^2x = \lambda(2x)[/tex]

    [tex]2x^2y - 2z^2y = \lambda(2y)[/tex]

    [tex]2x^2z - 2y^2z = \lambda(2z)[/tex]

    Do I just substitute these equations in turn into the constraint equation and determine what the points are that satisfy it?

    Also, how do you find out if a point is a local max/min when you know the critical point when looking at a case like this?
    Last edited: Feb 20, 2006
  2. jcsd
  3. Feb 20, 2006 #2


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    You should be able to simplify those. Obviously, x= 0 makes the first of those equations true. If x is not 0, then the first equation is [itex]y^2+ z^2= \lamba[/itex].

    Similarly, y= 0 satisfies the second equation. If y is not 0, then the second equation is [itex]x^2- z^2= \lamda[/itex]

    z= 0 satisfies the third equation. If z is not 0, then the third equation is [itex]x^2- y^2= \lambda[/itex].

    Setting [itex]x^2- z^2= x^2- y^2[/itex] gives [itex] z^2= y^2[/itex] or z= y. Similarly, setting [itex]y^2+ z^2= x^2- y^2[/itex] gives z= x. Can you finish from here?
  4. Feb 20, 2006 #3
    Ok, I can follow you up to a certain point.

    For whenever x,y or z = 0, I get the following critical points:

    I get [itex] z^2 = \frac{x^2} {3} [/itex] here.

    I then get a bit lost from now on. By substitution and a number of steps I get [itex] z = \frac {1} {\sqrt{5}} [/itex] and [itex] z = -\frac {1} {\sqrt{5}} [/itex]. Do I just substitute that in and get the critical points?
  5. Feb 20, 2006 #4


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    hmm..you've missed out quite a few points:
    Rewrite your equations as follows:
    If you set x=0, you get in addition to your own points the following 4:
    [tex](0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}), (\lambda=-\frac{1}{2})[/tex]

    Having dealt with those, note that from combining the first and fourth equation, we must have:
    (Replacing the first and eliminating x from further consideration)
    You can now proceed further with these three equations.

    If I've counted correctly, there are a total of 18 (x,y,z) solutions here, in addition to the 8 previous ones.
    Last edited: Feb 20, 2006
  6. Feb 20, 2006 #5


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    Oops, you are right! Now you know y= z and x2= 3z2. Now put those into your constraint:
    [itex]x^2+ y^2+ z^2= 3z^2+ z^2+ z^2= 5z^2= 1[/itex]
    So, yex, [itex]z= \pm\frac{1}{\sqrt{5}}[/itex].

    Now use y= z and x2= 3z2 to find corresponding values of x and y.
  7. Feb 21, 2006 #6


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    FYI, the max/min is [itex]\pm\frac{1}{4}[/itex]
  8. Feb 21, 2006 #7


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    Since [tex]f(x,z,y) = x^2z^2-z^2y^2 + y^2x^2 = x^2y^2-y^2z^2 + z^2x^2 = f(x,y,z)[/tex]

    which is to say that f is symmetric in the variables y and z, doesn't it follow that the extrema will have y=z?
  9. Feb 21, 2006 #8


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    Nope; what follows is that for any particular extremum with one choice of y and z, there must also be another extremum with their values commuted.
    The 26 critical points are:
    [tex](\pm{1},0,0), (0,\pm{1},0), (0,0,\pm{1}),(\pm\sqrt{\frac{3}{5}},\pm\sqrt{\frac{1}{5}},\pm\sqrt{\frac{1}{5}}),(0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}),(\pm\frac{1}{\sqrt{2}},0,\pm\frac{1}{\sqrt{2}}),(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}},0)[/tex]
    The last 12 are the extrema.
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