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Finding the minimum and maximum distances using Lagrange Multipliers

  1. May 29, 2012 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=47774&stc=1&d=1338346735.png

    What I don't understand is why you can maximize the distances squared - d2. Isn't d2 different from d? I don't see how they can get you the same value.
     

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    Last edited: May 29, 2012
  2. jcsd
  3. May 29, 2012 #2

    Ray Vickson

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    They don't have the same values (unless they happen to be 0 or 1), but they are maximized or minimized at the same points (x,y,z). Think about it: how could it be otherwise?

    RGV
     
    Last edited: May 29, 2012
  4. May 30, 2012 #3

    HallsofIvy

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    Letting distance be "D", so we can distinguish it from the differential, "d", [itex]d(D^2)/dt= 2D (dD/dt)= 0[/itex]. If D itself is not 0, [itex]d(D^2)/dt[/itex] will be 0 if and only if [itex]dD/dt[/itex] is 0.
     
  5. May 30, 2012 #4

    Ray Vickson

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    As HallsofIvy has pointed out, the stationary points are the same in both problems. However, if the original f can take both negative and positive values, a min in the squared problem can be a saddle point in the original problem (eg: f(x) = x^3 has a saddle point at x = 0 but g(x) = f(x)^2 = x^6 has a global minimum at x = 0) and a min in the original problem can be a max in the squared problem, etc. None of these issues arise if the original f is >= 0, as it is in the distance problem you cite.

    RGV
     
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