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Math Help - Lagrange Multipiers

  1. #1
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    Lagrange Multipiers

    The teacher gave us some problems on the board for us to jot down. He said it would prepare us with the final.

    Problem:

    Optimize the function f(x,y,z)=8x-4z subject to the constraint x^2 + 10y^2 + z^2 = 5

    I am not 100% sure what optimize means and if Lagrange multiplier applies but this is what i have so far.

    f_{x}= 8
    f_{y}= 0
    f_{z}= -4

    g_{x}=2x
    g_{y}= 20y
    g_{z}= 2z

    Combining we have:

    (1)   8 = 2x\lambda
    (2) 0 = 20y\lambda
    (3) -4 = 2z\lambda

    Solving (1) for x:

    (4)  x = \frac{4}{\lambda}
    (5) y = \lambda
    (6) z = \frac{-2}{\lambda}

    Now, add in (4),(5), and (6) for the constraint

    \left(\frac{4}{\lambda}\right)^2 + 10\lambda^2+\left(\frac{-2}{\lambda}\right)^2 = 5

    then

    \frac{16}{\lambda^2} + 10\lambda^2+\frac{4}{\lambda^2} = 5

    So, at this point i am stuck. I don't know how to finish.
    I don't even know if i was suppose to use Lagrange multiplier or if its correct up to that point?

    If someone could let me know if i did the problem correctly and help me finish?
    Thank you so much.
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  2. #2
    MHF Contributor
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    Re: Lagrange Multipiers

    Hey icelated.

    Hint: Multiply both sides by lambda^2 and you should get an equation after re-arranging: a*lambda^4 + b*lambda^2 + c = 0.

    Make a substitution u = lambda^2 and solve the quadratic a*u^2 + b*u + c = 0 and then solve for lambda.
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  3. #3
    MHF Contributor

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    Re: Lagrange Multipiers

    Quote Originally Posted by icelated View Post
    The teacher gave us some problems on the board for us to jot down. He said it would prepare us with the final.

    Problem:

    Optimize the function f(x,y,z)=8x-4z subject to the constraint x^2 + 10y^2 + z^2 = 5

    I am not 100% sure what optimize means and if Lagrange multiplier applies but this is what i have so far.

    f_{x}= 8
    f_{y}= 0
    f_{z}= -4

    g_{x}=2x
    g_{y}= 20y
    g_{z}= 2z

    Combining we have:

    (1)   8 = 2x\lambda
    (2) 0 = 20y\lambda
    (3) -4 = 2z\lambda

    Solving (1) for x:

    (4)  x = \frac{4}{\lambda}
    (5) y = \lambda
    No, 0= 20y\lambda gives y= 0 NOT " y= \lambda"

    (6) z = \frac{-2}{\lambda}

    Now, add in (4),(5), and (6) for the constraint

    \left(\frac{4}{\lambda}\right)^2 + 10\lambda^2+\left(\frac{-2}{\lambda}\right)^2 = 5

    then

    \frac{16}{\lambda^2} + 10\lambda^2+\frac{4}{\lambda^2} = 5

    So, at this point i am stuck. I don't know how to finish.
    I don't even know if i was suppose to use Lagrange multiplier or if its correct up to that point?

    If someone could let me know if i did the problem correctly and help me finish?
    Thank you so much.
    Follow Math Help Forum on Facebook and Google+

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