Finding the minimum and maximum distances using Lagrange Multipliers

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Maximizing the squared distance, d², is valid because while d² and d are not the same values, they reach their extrema at the same points in the context of optimization. The derivative of d² with respect to time shows that both d² and d will have stationary points at the same coordinates, provided d is not zero. However, if the original function can take on both negative and positive values, the relationships between maxima and minima can differ, as illustrated by examples like f(x) = x³ and g(x) = f(x)². In the specific case of distance, where values are non-negative, these complications do not arise. Understanding this relationship is crucial for applying Lagrange multipliers effectively in distance optimization problems.
theBEAST
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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.
 

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theBEAST said:

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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Letting distance be "D", so we can distinguish it from the differential, "d", d(D^2)/dt= 2D (dD/dt)= 0. If D itself is not 0, d(D^2)/dt will be 0 if and only if dD/dt is 0.
 
Ray Vickson said:
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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