Finding the minimum and maximum distances using Lagrange Multipliers

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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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Ray Vickson
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Homework Statement


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.
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
 
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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.
 
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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
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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