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Math Help - Using Lagrange Multiploiers locate any extrema of the function

  1. #1
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    Using Lagrange Multiploiers locate any extrema of the function

    Using Lagrange Multiploiers locate any extrema of the function
    x^2 + y^2 + z^2
    subject to x + z = 6 and y + z = 8

    I got (4/3, 10/3, 14/3)
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  2. #2
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    No you didn't. Check your arithmetic. You are SO CLOSE.

    Hint: The function is symmetric, but the constraints are not.

    If you're still struggling, show your work and guidance can be offered.
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    No you didn't. Check your arithmetic. You are SO CLOSE.

    Hint: The function is symmetric, but the constraints are not.

    If you're still struggling, show your work and guidance can be offered.
    Mmmm... I don't know about the OP, but I am having trouble interpreting this comment.

    This problem can be transformed into a 1-D unconstrained optimisation problem in the variable z. This transformed problem has a unique minimum at z=14/3, which then gives the x and y for the minimum by substitution into the constraints.

    CB
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  4. #4
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    Quote Originally Posted by TKHunny View Post
    No you didn't. Check your arithmetic. You are SO CLOSE.

    Hint: The function is symmetric, but the constraints are not.

    If you're still struggling, show your work and guidance can be offered.
    I see nothing wrong with the arithmetic.

    As Captain Black says, you can use the two constraints to reduce this to a quadratic function of the single variable, z, then show that z= 14/3 is a minimum for that function and solve for x and y from the constraints.

    But since this says to use "Lagrange Multipliers", we have 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}= \lambda(\vec{i}+ \vec{k})+ \mu(\vec{j}+ \vec{k}) where \lambda and \mu are the multipliers.

    That gives the three equations 2x= \lambda, 2y= \mu, and 2z= \lambda+ \mu. Since 2x+ 2y= \lambda+ \mu, that is the same as 2z= 2x+ 2y or z= x+ y. Then x= z- y and the first constraint is 2z- y= 6 while the second is y+ z= 8. Adding those equations 3z= 14 so z= 14/3. Then y= 8- 14/3= 10/3 and x= 4/3 .
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    I see nothing wrong with the arithmetic.

    As Captain Black says, you can use the two constraints to reduce this to a quadratic function of the single variable, z, then show that z= 14/3 is a minimum for that function and solve for x and y from the constraints.

    But since this says to use "Lagrange Multipliers", we have 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}= \lambda(\vec{i}+ \vec{k})+ \mu(\vec{j}+ \vec{k}) where \lambda and \mu are the multipliers.

    That gives the three equations 2x= \lambda, 2y= \mu, and 2z= \lambda+ \mu. Since 2x+ 2y= \lambda+ \mu, that is the same as 2z= 2x+ 2y or z= x+ y. Then x= z- y and the first constraint is 2z- y= 6 while the second is y+ z= 8. Adding those equations 3z= 14 so z= 14/3. Then y= 8- 14/3= 10/3 and x= 4/3 .
    That is not a notation for Lagrange multiplers that I recognise, but my Lagrange multipler result is the same. I only brought up the unconstrained method as a cross check that there was not something going on that I had missed in the Largangian method.

    CB
    Last edited by CaptainBlack; June 3rd 2010 at 10:44 AM.
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  6. #6
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    That's odd. I could have sworn the 14 and the 10 were reversed. Has it been edited or was I seeing cross-eyed last night? No matter.
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