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Math Help - [SOLVED] Lagrange multipliers to find max/min

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    [SOLVED] Lagrange multipliers to find max/min

    Use Lagrange multipliers to find the max and min values of the function subject to the given constraints.

    I get multiple answers for \lambda, which is my problem.

    function:
    f(x,y,z)=x+2y

    subject to:
    x+y+z=1 and  y^2+z^2=4

    \nabla f <1,2,0> = \lambda<1,1,1>
    \nabla f <1,2,0> = \lambda<0,2y,2z>

    Now I set them equal to solve for \lambda using all of the possible combinations, here is where my answers get messed up. They are all different, and 1= \lambda 0 makes no sense.
    1= \lambda
    2= \lambda
    0= \lambda
    1= \lambda 0
    2= \lambda 2y
    0= \lambda 2z

    From here any solution i pick does not give me the solution of 1,\sqrt{2},-\sqrt{2} as a max. Also 1,-\sqrt{2},\sqrt{2} is the min.

    Any help is appreciated as always.
    Thanks!
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  2. #2
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    Quote Originally Posted by snaes View Post
    Use Lagrange multipliers to find the max and min values of the function subject to the given constraints.

    I get multiple answers for \lambda, which is my problem.

    function:
    f(x,y,z)=x+2y

    subject to:
    x+y+z=1 and  y^2+z^2=4

    \nabla f <1,2,0> = \lambda<1,1,1>
    \nabla f <1,2,0> = \lambda<0,2y,2z>

    Now I set them equal to solve for \lambda using all of the possible combinations, here is where my answers get messed up. They are all different, and 1= \lambda 0 makes no sense.
    1= \lambda
    2= \lambda
    0= \lambda
    1= \lambda 0
    2= \lambda 2y
    0= \lambda 2z

    From here any solution i pick does not give me the solution of 1,\sqrt{2},-\sqrt{2} as a max. Also 1,-\sqrt{2},\sqrt{2} is the min.

    Any help is appreciated as always.
    Thanks!
    Lagrange multipliers with two constraints - you'll need two multipliers.

    Form

     <br />
F = f + \lambda g(x,y,z) + \mu h(x,y,z) = 0<br />

    Then \nabla F = 0, with g(x,y,z) = 0 and h(x,y,z) = 0 (the constaints).

    However, looking at your problem, since f has no z, then eliminate z between your constraints and only use one multiplier.
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  3. #3
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    Ahh thats how. Thanks alot!
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