For the proof of lagrange multipliers, it is based on the assumption that the function you are optimizing, f(x,y,z), takes on an extreme value at the point (x0,y0,z0), and that any curve that passes through this point has the tangent vector perpendicular to the gradient vector. That seems fair enough. But it is possible to have more than one point be the extreme value. In which case you could make a curve that passes through both extreme value points. Even if this is the case, I dont think it posses any problems, because when you work it out, you still get the derivative equal to zero at the first extreme value your concerned with. But what about critical points that are meaningless. Like in the case of x^3, x=0 would be an extreme value, but its not a max or a min. Likewise, there might be an extreme value on your constrained optimization function that acts like x^3 near zero. The function can increase, slow down, stop, then decrease on the surface near that point. So if you use lagranges method, you would find that point, and plug it into the constrained equation, and find a value for f(x,y,z) at that point. Is there something to this that im missing that allows for lagranges method to avoid this pitfall?(adsbygoogle = window.adsbygoogle || []).push({});

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# Lagrange multipliers proof

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