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Math Help - Lagrange Multipliers Confusion

  1. #1
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    Question Lagrange Multipliers Confusion

    I have this question that I am lost on:

    Use lagrange multipliers to find the points on the surface z^2 - xy = 49
    that are closest to the origin.

    This is my thinking so far:

    d = sqrt ((x-0)^2 + (y-0)^2 + (z-0)^2

    or
    d^2 = (x-0)^2 + (y-0)^2 + (z-0)^2

    so
    del f = 2(x-0)i + 2(y-0)j + 2(z-0)k

    The constraint is
    g(x,y,z) = z^2 - xy = 49


    so lambda del g(x,y,z) = lambda(k^2 - jk)

    so
    2(x-0)i + 2(y-0)j + 2(z-0)k = lambda(k^2 - jk), which leads to

    2(z - 0) = lambda^2 and 2(y-0)*2(x-0) = lambda^2

    therefore,
    lambda = sqrt(2(z-0)) = sqrt(2(y-0)*2(x-0))

    so sqrt(2*z) = sqrt(4*x*y)
    so 2*z = 4*x*y

    Im not even sure if this is correct, but if it is where do I go from here?

    I have made many attempts, and with one of those I came up with 3 points:

    x=7, y =7
    x=1 , y=49
    x=49, x=1

    but again, I dont think this is correct.
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  2. #2
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    Quote Originally Posted by woody198403 View Post
    I have this question that I am lost on:

    Use lagrange multipliers to find the points on the surface z^2 - xy = 49
    that are closest to the origin.

    This is my thinking so far:

    d = sqrt ((x-0)^2 + (y-0)^2 + (z-0)^2

    or
    d^2 = (x-0)^2 + (y-0)^2 + (z-0)^2

    so
    del f = 2(x-0)i + 2(y-0)j + 2(z-0)k

    The constraint is
    g(x,y,z) = z^2 - xy = 49


    so lambda del g(x,y,z) = lambda(k^2 - jk)

    so
    2(x-0)i + 2(y-0)j + 2(z-0)k = lambda(k^2 - jk), which leads to

    2(z - 0) = lambda^2 and 2(y-0)*2(x-0) = lambda^2

    therefore,
    lambda = sqrt(2(z-0)) = sqrt(2(y-0)*2(x-0))

    so sqrt(2*z) = sqrt(4*x*y)
    so 2*z = 4*x*y

    Im not even sure if this is correct, but if it is where do I go from here?

    I have made many attempts, and with one of those I came up with 3 points:

    x=7, y =7
    x=1 , y=49
    x=49, x=1

    but again, I dont think this is correct.
    Here is how I do them. You should get the same equations that need solving.

    L = x^2 + y^2 + z^2 + \lambda (z^2 - xy - 49).


    \frac{\partial L}{\partial x} = 2x - \lambda y = 0 .... (1)


    \frac{\partial L}{\partial y} = 2y - \lambda x = 0 .... (2)


    \frac{\partial L}{\partial z} = 2z - 2 \lambda z = 0 \Rightarrow z(1 - \lambda) = 0 \Rightarrow z = 0 ~ \text{or} ~ \lambda = 1 .... (3)


    \frac{\partial L}{\partial \lambda} = z^2 - xy - 49 = 0 .... (4)


    Solve these equations simultaneously.


    Substitute \lambda = 1 into (1) and (2): x = y = 0. From (4): z = \pm 7.

    So you have (0, 0, 7) and (0, 0, -7). They require testing.


    Substitute z = 0 into (4): xy = -49 \Rightarrow y = - \frac{49}{x}. Substitute into (1) and (2) and solve for x. Then substitute back to get y. More solutions to test .....
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  3. #3
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    Lagrange Multipliers

    Hey
    Im currently trying to use lagrange multipliers to find the points on the surface z^2-x*y=49 that are closest to the origin.

    im saying d^2=x^2+y^2+z^2 for the distance for the origin

    but i just cant grasp the whole lagrange multipliers concept...

    all help much appreciated
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  4. #4
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    We want to extremize f(x,y,z)=x^{2}+y^{2}+z^{2}

    subject to the constraint g(x,y,z)=z^{2}-xy=49

    Therefore,

    {\nabla}f(x,y,z)=x^{2}+y^{2}+z^{2}={\lambda}{\nabl  a}g(x,y,z)

    at a constrained relative extremum; that is,

    2xi+2yj+2zk={\lambda}(-yi-xj+2zk)

    Which leads to the equations:

    2x=-y{\lambda}, \;\ 2y=-x{\lambda}, \;\ 2z={\lambda}2z

    Now continue?.

    What a lagrange multiplier tells us is that extrema occur where the gradients are parallel. {\lambda} is the lagrange multiplier that represents a multiple such that
    {\nabla}f(x,y,z)={\lambda}g(x,y,z).
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  5. #5
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    Hey thanks
    that all makes sense but do I now sub:


    2x=-y{\lambda}, \;\ 2y=-x{\lambda}, \;\ 2z={\lambda}2z

    into the constraint?

    and if I do how do i then solve for the points with the minimum distance??
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  6. #6
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    Solve them for lambda, then equate and solve for x, y or z, then sub into the constraint.
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  7. #7
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    hey just with the final bit how can you Substitute into (1) and (2) and solve for x when there is a third unknown of \lambda??

    it seems you have 2 equations and still 3 unknowns?

    or am i just not thinking clearly?

    Thanks again
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  8. #8
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    Quote Originally Posted by edeffect View Post
    hey just with the final bit how can you Substitute into (1) and (2) and solve for x when there is a third unknown of \lambda??

    it seems you have 2 equations and still 3 unknowns?

    or am i just not thinking clearly?

    Thanks again
    After the substitution of y = - \frac{49}{x} into (1) and (2), the unknowns are x and \lambda. Two equations, two unknowns .......
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