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

1. Oct 3, 2014

### freshman2013

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. Oct 4, 2014

### HallsofIvy

Then look at values close to that extremum.

3. Oct 4, 2014

### economicsnerd

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.