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Lagrange multipliers: How do I know if its a max of min

  1. Oct 3, 2014 #1
    in the problem f(x,y)=x^2+y^2 and xy=1, I get 2 as a local extrema and it is a min
    in the problem f(x1,x2...xn) = x1+x2..+xn (x1)^2+...(xn)^2=1 I get sqrt(n) and its a max. How do I know if these are max or min values? If I get more than two extrema, I just compare them and one's a max and the other's min. What if there's only one?
     
  2. jcsd
  3. Oct 4, 2014 #2

    HallsofIvy

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    Science Advisor

    Then look at values close to that extremum.
     
  4. Oct 4, 2014 #3
    1) The objective ##f## is bounded below (since ##f\geq 0##) but not above (since ##f\left(x,\frac1x\right) = x^2 +
    \frac1{x^2}\geq x^2 \to \infty## as ##x\to \infty##). So it may be possible to have a global minimum in the constraint set, but you'll never find a global maximum.

    2) The objective is convex (which follows either from using the triangle inequality or from computing the Hessian), so that any local minimum in the convex set ##\{(x,y):\enspace x,y\geq0, \enspace xy\geq 1\}## is a global minimum.
     
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