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Math Help - Extrema with constraints

  1. #1
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    Extrema with constraints

    Here is the question:

    Find the extrema of f(x, y) = x + y^2, with constraint : 2x^2 + y^2 = 1

    My attempts at solution:

    I tried to solve it using Lagrange multiplier and I got:

    1 = 4x \lambda ............................. I

    2y = 2y \lambda ............................. II

    2x^2 + y^2 = 1    ............................. III

    Now here is where I am stuck. How can I use the first two equations to get the value of either x or y?
    Last edited by Keep; May 19th 2010 at 04:27 PM.
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  2. #2
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    Doesn't equation II tell you that \lambda =1 (assuming y\ne 0 )?
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Keep View Post
    Here is the question:

    Find the extrema of f(x, y) = x + y^2, with constraint : 2x^2 + y^2 = 1

    My attempts at solution:

    I tried to solve it using Lagrange multiplier and I got:

    1 = 4x lambda ............................. I

    2y = 2y lambda ............................. II

    2x^2 + y^2 = 1 ............................. III

    Now here is where I am stuck. How can I use the first two equations to get the value of either x or y?

    2y-2y\lambda = 0

    2y(1 - \lambda) = 0

    y = 0, \lambda = 1, x = \pm \frac{1}{\sqrt{2}}
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  4. #4
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    Your system of equations is correct.

    From eq. II, \lambda=1
    Plug that into eq. I, x=\frac{1}{4}
    Plug x=\frac{1}{4} into eq. III, y=\pm\frac{1}{2}\sqrt\frac{7}{2}

    The other critical point occurs when y=0,
    Use eq.III , x=\frac{1}{\sqrt2}


    SO, your 4 critical points are...
    x=\frac{1}{4} , y=\pm\frac{1}{2}\sqrt\frac{7}{2}
    y=0 , x=\pm\frac{1}{\sqrt2}
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  5. #5
    MHF Contributor matheagle's Avatar
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    I'd use Calc 1.

    Since y^2=1-2x^2 you only need to get the max's and min's of

    f(x)=x+1-2x^2
    Last edited by matheagle; May 19th 2010 at 06:51 PM.
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  6. #6
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    Quote Originally Posted by ojones View Post
    Doesn't equation II tell you that \lambda =1 (assuming y\ne 0 )?
    Ooh OK I didn't look at it like that, I was trying to divide equation 1 by 2 and it wan't working. Yeah equation 2 says \lambda is 1. Substitute that in equation 1 and we get x as 1/4. Substitute for x in equation III we get y = +/- \sqrt {7/8}
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Keep View Post
    Ooh OK I didn't look at it like that, I was trying to divide equation 1 by 2 and it wan't working. Yeah equation 2 says \lambda is 1. Substitute that in equation 1 and we get x as 1/4. Substitute for x in equation III we get y = +/- \sqrt {7/8}
    Sorry forgot to give you your other x point to try
    x = \frac{1}{4}
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  8. #8
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    So the maximum is x=\frac{1}{4} , y=\sqrt \frac{7}{8} and the minimum is x=\frac{-1}{\sqrt2}, y=0
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