LOAD('OPTMIZ);
/*
We have already seen how STAP finds poles of the objective or gradient
only by accident.  The following examples are included as reminders
about other limitations of using calculus to find extrema:*/
stapoints(x, [], x**3-y**2) ;
/* Lagrange multipliers require the constraints to have con-
tinuous tangents, and this is violated for the above example at X=0,
Y=0, where the objective is a minimum.  A direct use of elimination
would fail too, because it results in an undefined derivative at the
minimum.  In such cases, each piece between discontinuities must be
considered a separate constraint.

The next example has a minimum at X=0, Y=0, where the two
active constraints are parallel, making them linearly dependent.
Such cusps or "wiskers" on the feasible region violate the so-called
"constraint-qualification" requirement; so the extremum is not found:*/
stapoints(x, [-y, y-x**3]) ;
/* The following example doesn't have a strictly feasible point;
so it violates Slater's condition; and the extremum at X=0 is not
found: */
stapoints(x, x**2) ;
/* Finally, it is important to remember that unbounded regions
may have nonstationary or asymptotically stationary points
at infinity, which STAP will not find.  Such situations are usually
obvious from qualitative considerations, provided the objective and
constraints are not too complicated and numerous, but the macsyma LIMIT
function can be of help.  However, it is important to remember that a
multivariate limit depends upon the way it is taken.  This is
illustrated by the following example, where you will have to type
NONZERO; in response to two interrupts: */
peano:  (y - c - (a*x)**2)*(y - c - (b*x)**2) ;
limit(peano, y, inf);
limit(peano, x, inf);
radcan(ev(peano, y=c+(a**2+b**2)*x**2/2));
limit(%, x, inf);