given two functions f(x,y) and g(x,y) [assume both are "well-behaved"
: continuous and differentiable] If we are solving a min/max problem
on f subject to the constraint that g(x,y)=0 is it possible (knowing
nothing more about f or g) that there could be 0 solutions?  3
solutions?  infinitely many solutions?  (I think the answer will be
found by thinking about LaGrange Multipliers -- but I'm not sure) 
Thanks for your help

Since you ask about minima and maxima (I think), I'll assume these are
real-valued functions.

You have an equation g(x,y) = 0 that presumably restricts the
"feasible" points (x,y) to a curve of some kind [you did say to assume
g is "well-behaved"].

That is, the solutions to g(x,y) = 0 form a 1-dimensional smooth
manifold, with the possible exception of finitely many points of
self-intersection and/or cusps (kinks) in the curve.

As far as the question of whether maxima/minima exist, or whether
there can be multiple occurrences, the issues of self-intersection or
cusps are not really critical (no pun intended).  Sure, you can
artificially introduce an obstacle to the existence of a maximum or
minimum by declaring g(x,y) to be mysteriously undefined at the points
where the extrema would otherwise occur, but again we will assume
"nice" behavior.

In that case the big issue is whether the curve has bounded or
unbounded extent.  A closed and bounded subset of the Euclidean plane
is compact, and a continuous function f(x,y) on a compact set will
have both (at least one) maxima and minima.  Think, for example, of
g(x,y) = 0 defining a circle in the plane.  The range of values as you
go around must have a high and a low point (though it may have more
than one of each).

On the other hand the curve could be open ended (even if it crosses
itself).  A simple case would be a straight line, and since then y is
explicitly a function of x, this is really a one-dimensional
optimization problem over (-oo,+oo).  In consequence there may not be
a maximum or a minimum point, or there may be one kind of extremum but
not the other, or there could be both a maximum AND a minimum.

So the easiest case to discuss in theory is the compact (closed and
bounded) curve, since there we are guaranteed the continuous function
f will have (at least) one maximum and one minimum (and maybe more).

If it's a question about computing the maximum and minimum values,
this is part of the subject of numerical analysis.  Googling for
one-dimensional numerical optimization will give you some ideas of the
diverse methods that are available.

regards, mathtalk-ga

Let me add that the phrase "solving a min/max problem" means something
to a mathematician that seems inconsistent with the data of the
original Question here.

Given a function of two variables, say f(x,y), one poses a "min/max"
problem by asking (for example) how to choose a value of variable x,
such that the range of values y gives the least possible (minimum)
value of the maximum value of f(x,y):

        Min Max f(x,y)
         x   y

Hence the name "min/max problem".  The additional constraint g(x,y) =
0, which tends to reduce the problem from two variables to one, and
the mention of Lagrange multipliers (a method for imposing a
constraint using techniques for otherwise unconstrained optimization)
lead me to think, however, this is _not_ the kind of problem you have
in mind.

To briefly return to the specific questions you raised (could there be
no solutions? finitely many solutions? infinitely many?), these are
issues of existence and uniqueness, two more search keywords that
might help you find the information you are looking for.

regards, mathtalk-ga

Isn't your question just related to this page? -
http://www.dpgraph.com/MCExamples/MCExamples.html
"[g is tangent at one or more points to a level curve of f,] the
gradients of f and g are scalar multiples of each other and the
coordinates of tangency points can be determined by solving a system
of 3 (usually nonlinear) equations. "

I'm getting the notion that your question is conditional.

Search : "calculus subject to constraint"