Ok so I have found a bunch of local extrema using the method of Lagrange multipliers.

Now how do I classify them as minimum or maximum?

I cant use the second derivative test because its not a single function - it is a collection of a function along with a bunch of constraints.

Is there any global algorithmic solution for this? Or do I just eyeball it and use human ingenuity that will differ on a case by case basis?

  • $\begingroup$ Do you mean local max vs. local min? A general solution for twice differentiable objective functions is to compute the Hessian matrix $H$, then compute a basis of the directions that are accessible from your point. This is a basis of the vectors which are orthogonal to $\nabla g_k$ where $g_k$ ranges over all your scalar constraints. $\endgroup$ – Ian Apr 10 '16 at 22:10
  • $\begingroup$ Say this basis is $x_1,\dots,x_n$. Then if $x_i^T H x_i>0$ for all $i$, then you have a min; if it is negative for all $i$, then you have a max; if it is strictly positive for some $i$ and strictly positive for some other $i$, then you have a saddle point; otherwise this method fails. $\endgroup$ – Ian Apr 10 '16 at 22:10
  • $\begingroup$ Note that more commonly in multivar calc you are just asking about global max and global min, which can be determined by simply substituting back into the objective function. $\endgroup$ – Ian Apr 10 '16 at 22:11
  • $\begingroup$ @Ian I would appreciate it if you elaborated about your hessian matrix and orthogonality towards constraints in a post $\endgroup$ – AlanSTACK Apr 10 '16 at 22:14

There is a second derivative test for constrained extrema. Here's the statement for the case of one constraint $g(x)=c$. Suppose $a$ is a constrained critical point of $f$ on the constraint set $g(x)=c$. Then we have $\nabla f(a)=\lambda\nabla g(a)$ for some scalar $\lambda$. Consider the Hessian matrix $H$ (matrix of second partial derivatives) of $f(x)-\lambda g(x)$, evaluated at $a$. Then if $v^\top Hv>0$ for all $v$ tangent to the constraint hypersurface at $a$, we conclude that $a$ is a constrained local minimum; if $v^\top Hv<0$ for all $v$ tangent to the constraint hypersurface at $a$, we conclude that $a$ is a constrained local maximum. [Interestingly, it does not work with the Hessian of $f$ alone.]

(One proves this by taking a parametrization $\phi$ of the constraint hypersurface near $a$ (say, with $\phi(0)=a$) and realizing that the function $f\circ\phi$ has a standard critical point at $0$, which can be characterized by looking at the Hessian of $f\circ\phi$ at $0$.)

  • $\begingroup$ To clarify. This extension of the second derivative test (the ordinary unconstrained one) is not usually taught in basic calculus classes, right? $\endgroup$ – AlanSTACK Apr 10 '16 at 22:28
  • $\begingroup$ Very much not taught. ;) But it appears as an exercise in occasional more advanced textbooks (including my own Multivariable Mathematics book) :P I believe I recall seeing it in some edition of Marsden and Tromba's book, as well. $\endgroup$ – Ted Shifrin Apr 10 '16 at 22:31
  • $\begingroup$ Thanks for the help. Also is there an amazon link to this book of yours? $\endgroup$ – AlanSTACK Apr 10 '16 at 22:35
  • $\begingroup$ See my profile on MSE. Thanks. :) $\endgroup$ – Ted Shifrin Apr 10 '16 at 22:37

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