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Math Help - solve using Lagrange multiplier

  1. #1
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    solve using Lagrange multiplier

    find the point on the plane ax + by + cz = d that is nearest the origin.

    I tried minimizing x^2 + y^2 + Z^2 with ax + by + cz - d = 0
    I don't get the correct answer. Please hep
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  2. #2
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    Quote Originally Posted by tennis14321 View Post
    find the point on the plane ax + by + cz = d that is nearest the origin.

    I tried minimizing x^2 + y^2 + Z^2 with ax + by + cz - d = 0
    I don't get the correct answer. Please hep
    Please show the details of your calculations.
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  3. #3
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    f = x^2 + y^2 + Z^2 + lamda (ax + by + cz -d) = 0
    F (patial y) = 2y + lamda * b
    F (patial z) = 2z + lamda * c
    Euler: d/dx F (partial y') - F (partial y) = 0 => 2y + lamda * b = 0
    d/dx F (partial z') - F (partial z) = 0 => 2z + lamda * c = 0
    Then I'd get y and z...

    I know this is incorrect . Please help.
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  4. #4
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    Quote Originally Posted by tennis14321 View Post
    f = x^2 + y^2 + Z^2 + lamda (ax + by + cz -d) = 0
    F (patial y) = 2y + lamda * b
    F (patial z) = 2z + lamda * c
    Euler: d/dx F (partial y') - F (partial y) = 0 => 2y + lamda * b = 0
    d/dx F (partial z') - F (partial z) = 0 => 2z + lamda * c = 0
    Then I'd get y and z...

    I know this is incorrect . Please help.
    L = x^2 + y^2 + z^2 + \lambda (ax + by + cz - d)

    \frac{\partial L}{\partial x} = 0 \Rightarrow 2x + \lambda a = 0 .... (1)

    \frac{\partial L}{\partial y} = 0 \Rightarrow 2y + \lambda b = 0 .... (2)

    \frac{\partial L}{\partial z} = 0 \Rightarrow 2z + \lambda c = 0 .... (3)

    \frac{\partial L}{\partial \lambda} = 0 \Rightarrow ax + by + cz - d = 0 .... (4)

    Substitute equations (1), (2) and (3) into equation (4) and solve for \lambda.

    Then substitute \lambda into equations (1), (2) and (3) to get x, y and z.
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  5. #5
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    aghh
    I was so close. I just couldn't think anymore.
    Thanks so much.
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