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Homework Help: Solving a system in five unknowns for lagrange multiplier

  1. Jul 12, 2014 #1
    1. The problem statement, all variables and given/known data

    I have to find the extrema of a given function with two constraints

    [tex]f(x,y,z) = x+y+z;x^2-y^2=1;2x+z=1[/tex]

    3. The attempt at a solution
    If I create a new function F

    then I have

    [tex]F(x,y,z,\lambda,\mu)=x+y+z-(x^2\lambda - y^2\lambda -\lambda) -(2x\mu + z\mu - \mu) [/tex]

    and taking the partials

    [tex]\left\{\begin{array}{cc} F_{x} = 1-2x\lambda - 2\mu =0\\ F_{y} = 1+2y\lambda = 0 \\F_{z} = 1 - \mu = 0 \\ F_{\lambda} = -x^2 + y^2 + 1 = 0 \\ F_{\mu} = -2x - z +1 = 0 \end{array}\right. [/tex]

    so now,

    [tex]\mu =1[/tex]
    solving for lambda

    [tex]\lambda = -\frac{1}{2y} = \frac{1-2}{2x}[/tex]

    now solving for x [or y] [tex]x=y[/tex]

    but this causes an issue with [tex]-x^2+y^2+1=0[/tex] because 1 does not equal zero
  2. jcsd
  3. Jul 13, 2014 #2


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    There are no local extrema then.

  4. Jul 13, 2014 #3
    okay, that's what I had figured. Making sure Idid not miss something
  5. Jul 13, 2014 #4


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    But can be extrema on the boundary.

  6. Jul 13, 2014 #5
    I can't find the zeros for the constraints though so what would I test in the function to see if there is.
  7. Jul 13, 2014 #6

    Ray Vickson

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    No. The fact that the Lagrangian equations plus the constraints have no solution precludes any extrema on the boundary.

    This can also be verified directly: the two constraints can be used to find x and z in terms of y, then the results can be substituted into f(x,y,z); there are two solutions for x and z in terms of y, hence two versions of f. The two resulting problems are each one-dimensional problems, which can be examined graphically to check for the absence of local extrema. In other words, the "max" is +∞ and the "min" is -∞, but without any finite local constrained extrema at all.
    Last edited: Jul 13, 2014
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