Posted by Srinivas Garimella on September 07, 1998 at 10:34:33:
In Reply to: Some clarifications posted by Michael Trick on September 06, 1998 at 15:44:26:
I believe you mean "minimum" (not "maximum") in your comment 1.
Srinivas
: Some questions are coming up a few times (and I at least was
: quite on clear on some aspects), so here are a few clarifications:
: 1) If a function is convex, then any critical point (i.e. gradient
: equals 0) is a maximum: you do not need the Hessian test at that
: point.
: 2) If the functions are continuous and differentiable (and we will
: assume that), and the function has a maximum (i.e. it does not go
: off to infinity in the feasible boundary), then the best
: of all points that satisfy K-T conditions is the optimal solution.
: Again, no second derivative test is needed.