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Lagrange multipliers and two constraints 
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#1
Jul1911, 11:58 PM

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So I need to find the min and max values of f(x,y,z) = x^2 + 2y^2 + 3z^2 given the constraints x + y + z = 1 and x  y + 2z =2. I've gotten as far as (2x, 4y, 6z) = (u,u,u) + (m,m,2m). I'm stuck trying to solve this system of equations. Any hints?



#2
Jul2011, 12:59 AM

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You have the constraints as two more equations: x+y+z=1 and xy+2z=2.
ehild 


#3
Jul2011, 02:07 AM

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RGV 


#4
Jul2011, 03:59 AM

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Lagrange multipliers and two constraints
whilst lagrange is a good idea, and the comments above from ehild & Ray are good ideas about where to head  i think a geometric method would be a tiny bit quicker here... though its upto preference i guess, and all the same at the end of the day
the intersection of two planes is a line. If you find the equation of that line it reduces to optimising a single variable function. 


#5
Mar3012, 03:41 AM

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how would you go about solving for x,y,z as functions of u and m?



#6
Mar3012, 11:45 AM

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So, rather than being an unnecessary distraction, Lagrange multipliers play an extremely important part in modern optimization theory and practice. They should be known by every serious student who will ever do optimization in his/her professional life. That said, it is often the case that the material is not welltaught, and the actual geometric content of the method is perhaps not spelled out the way it ought to be, but that may be more the fault of textbook choice rather than of the material itself. Of course, if the method just appears as a small section in a Calculus course, the crucial role of the method in optimization practice may be missed or not sufficiently emphasized. RGV 


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