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Math Help - Lagrange Multipliers

  1. #1
    Senior Member chella182's Avatar
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    Lagrange Multipliers

    I always get stuck half way through these things

    Find the maximum value of

    f(x,y,z)=x^2+9y^2+100z^2,

    subject to the constraint that

    x^6+y^6+z^6=1

    [NB: your answer should be given as a surd, not a decimal.]

    So I formed

    L(x,y,z|\lambda) (not sure if that's the correct notation)

    to get

    x^2+9y^2+100z^2-\lambda(x^6+y^6+z^6-1)

    and took partial derivatives to get

    \frac{\partial L}{\partial x}=2x-6\lambda x^5

    \frac{\partial L}{\partial y}=18y-6\lambda y^5

    \frac{\partial L}{\partial z}=200z-6\lambda z^5

    \frac{\partial L}{\partial \lambda}=-(x^6+y^6+z^6-1)

    ...but it's at this point I get stuck. I've tried looking in a few other threads, but none of it's seemed to help
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  2. #2
    Junior Member slider142's Avatar
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    Let x = (x, y ,z). Just to be sure you get the methodology of using Lagrange multipliers, the general theorem is that if you are studying values of a function f(x) restricted to a level curve g(x) = C, we note that when a value x on g is an extremum for f parametrized over the level curve, the gradient of g at x is parallel to the gradient of f at x. Mathematically, we say
    \nabla f(x) = \lambda\nabla g(x)
    where \lambda is some scalar. So the value of \lambda is not actually important, it is just a way of saying the gradients are parallel.
    Rearranging the equation above gets you
    \nabla f(x) - \lambda\nabla g(x) = 0
    which is the form you seem to be familiar with. Can you work out what you need to do from here to find x?
    Last edited by slider142; August 4th 2009 at 03:50 PM.
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  3. #3
    Senior Member chella182's Avatar
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    Honestly? That makes no sense. I just wanna know what I do with all of those partial differentials, 'cause the methods used in other examples I've looked at don't appear to work in my particular question. One example rearranged to make x, y and z the subject or something, but I can't do that here.
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  4. #4
    Junior Member slider142's Avatar
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    Quote Originally Posted by chella182 View Post
    Honestly? That makes no sense. I just wanna know what I do with all of those partial differentials, 'cause the methods used in other examples I've looked at don't appear to work in my particular question. One example rearranged to make x, y and z the subject or something, but I can't do that here.
    Did you note the required condition
    \nabla f(x) - \lambda\nabla g(x) = 0?
    The above is the same as
    \nabla (f(x) - \lambda g(x)) = 0
    Your text calls the function above L(x|\lambda) = f(x) - \lambda g(x). Taking the partials and setting them equal to zero is what is then required to solve the above vector equation.
    You have in your post the equations that correspond to
    \nabla L(x, y, z, \lambda ) = \left(\frac{\partial L}{\partial x}, \frac{\partial L}{\partial y}, \frac{\partial L}{\partial z}, \frac{\partial L}{\partial \lambda }\right)
    All you have to do is equate it to the 0 vector (just equate each partial derivative to 0) and solve the resulting list of simultaneous equations.
    If you do not see the connection, I believe it would be a good idea to review this section of your textbook.
    Last edited by slider142; August 4th 2009 at 04:42 PM.
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  5. #5
    Senior Member chella182's Avatar
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    I do see it, I know it's simultaneous equations but I just can't do it for this example, which is why I'm asking. We don't get textbooks, I've looked through my notes and I still can't fathom how to do this example.
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  6. #6
    Junior Member slider142's Avatar
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    Quote Originally Posted by chella182 View Post
    I do see it, I know it's simultaneous equations but I just can't do it for this example, which is why I'm asking. We don't get textbooks, I've looked through my notes and I still can't fathom how to do this example.
    It just gets tedious. You get 4 polynomials, which can be factored so that you have the option of x = 0, y=0 or z = 0:
    x(1 - (3\lambda)x^4) = 0
    y(9 - (3\lambda)y^4) = 0
    z(100 - (3\lambda)z^4) = 0
    x^6 + y^6 + z^6 -1 = 0
    They cannot all be simultaneously 0 because of the last equation. So choose each to be 0 for separate cases and solve the resulting 3 polynomials for each case. There is also the case where none are 0. I will do one to show that it is not difficult, just tedious. Suppose x = 0. Then we have the system:
    y(9 - (3\lambda)y^4) = 0
    z(100 - (3\lambda)z^4) = 0
    y^6 + z^6 -1 = 0
    The second factors are differences of squares, so we actually have:
    y(3 + \sqrt{3\lambda}y^2)(3 - \sqrt{3\lambda}y^2)= 0
    z(10 + \sqrt{3\lambda}z^2)(10 - \sqrt{3\lambda}z^2) = 0
    y^6 + z^6 -1 = 0
    The second factors are never 0, so we again have the option of either the variables y or z are individually 0, or the third factors are 0. If neither y nor z is 0, then we have:
    3 - \sqrt{3\lambda}y^2= 0
    10 - \sqrt{3\lambda}z^2 = 0
    y^6 + z^6 -1 = 0
    which is straightforward to solve for y and z using \sqrt{3\lambda}, although tedious.
    In the case that none of the variables are 0, we have only the last factors to consider, since the others cannot be 0:
    1 - \sqrt{3\lambda}x^2 = 0
    3 - \sqrt{3\lambda}y^2= 0
    10 - \sqrt{3\lambda}z^2 = 0
    x^6 + y^6 + z^6 -1 = 0
    Like the others, simply solve the first three equations for each variable as a function of \sqrt{3\lambda} and then use those values in the last equation which will become a single variable equation in \sqrt{3\lambda}.
    Last edited by slider142; August 4th 2009 at 05:31 PM.
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  7. #7
    Senior Member chella182's Avatar
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    Nevermind, I did it I got...

    x=\sqrt[4]{\frac{1}{3\lambda}}

    y=\sqrt[4]{\frac{3}{\lambda}}

    z=\sqrt[4]{\frac{100}{3\lambda}}

    then I put that into \frac{\partial L}{\partial\lambda}=0 to get

    \lambda=\sqrt[2/3]{\frac{1028}{3^{3/2}}}

    Putting that back into the expressions for x, y and z I got

    x=\frac{1}{\sqrt[6]{1028}}

    y=\frac{\sqrt[4]{9}}{\sqrt[6]{1028}}

    z=\frac{\sqrt[4]{100}}{\sqrt[6]{1028}}

    When put into x^6+y^6+z^6 you get 1

    No idea if that's what I'm asked for (think I need to put those values into the f(x,y,z) actually) but that should get me enough marks in an exam
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  8. #8
    Junior Member slider142's Avatar
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    Very good! However, there is more than one solution! I have edited my previous reply.
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  9. #9
    Senior Member chella182's Avatar
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    How can there be more than one solution when you're maximising a function? & I still can't properly get my head around what you've written
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  10. #10
    Junior Member slider142's Avatar
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    Quote Originally Posted by chella182 View Post
    How can there be more than one solution when you're maximising a function? & I still can't properly get my head around what you've written
    The Lagrange multiplier criterion does not guarantee that you get a maximum, it only gets you constrained local extrema. They could be minima or maxima and they could be local or global. Only comparing their values and/or using the second derivative test will tell you what they are.
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  11. #11
    Senior Member chella182's Avatar
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    Okay, well in all of my questions they're definitely maxima, 'cause we've not been taught how to test if they actually are maxima yet.
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