Does Lagrange Multipliers method provide a way to classify relative minimum and maximum points? In all the problems I've seen about Lagrange's multipliers method they always ask for absolute minimum and maximum points, but what about relative ones?
In this example: 
$f(x,y)=x^2+3y$ subject to the restriction $x^2+y^2=5$
the method provides the following points $ P=(0, \sqrt{5})$
$ Q=(0, -\sqrt{5})$, $ R=(\frac{\sqrt{11}}{2}, \frac{3}{2})$
$ S=(-\frac{\sqrt{11}}{2}, \frac{3}{2})$
so
$f(P)= 3\sqrt{5} \approx6.69$
$f(Q)= -3\sqrt{5}\approx-6.69$
$f(R)= \frac{29}{4}=7.25$
$f(S)= \frac{29}{4}=7.25$
So one concludes that Q is a minimum point and R and S are maximum points
but how do I determine the nature of P: relative maximum, relative minimum ?
A graph of the function tells me it should be a relative minimum (the intersection curve bends down in a neighbordhood of point P):

Besides I tried solving it by studying the single-variable function $h(y)=
5-y^2+3y$  which I got by plugging in the restriction into f. By making $h'(y)=0$ I found $y=3/2$  and considering $-\sqrt{5}\leq y \leq \sqrt{5}$
to be it's domain to stay in the restriction,  $y=\sqrt{5}$ and $y=-\sqrt{5}$ are also possible extrema points.
Plugging in this values into the restriction $x^2+y^2=5$, yields the corresponding x-components, so the same P,Q ,R and S points found with the first method. 
Then, comparing the values of h(y) in the y-components of these  points:
$h(\sqrt{5})=3\sqrt{5}=f(P)$
$h(-\sqrt{5})=-3\sqrt{5}=f(Q)$
$h(\frac{3}{2})= \frac{29}{4}=f(R)=f(S)$
Q is found again to be a minimum point and R ans S to be maximum points.
But now I could use second-derivative test for P:
 $h''(y)=-2 < 0$ ,  for any $y$ and so $P$ should be a local maximum, in contradiction with graph that shows it to be local minimum, What is this second method not working and how do both  methods determine P is a  relative minimum?
 A: The second derivative test tells you about the type of critical point on the interior of the search interval.  Let's temporarily set aside the source of the equation and just try to solve

Extremize $h(y) = 5 - y^2 + 3y$ on $[-\sqrt{5},\sqrt{5}]$.

The second derivative tells you whether an interior critical point, the one at $y = 3/2$ is a local minimum, local maximum, neither, or unresolved.  It cannot speak to the endpoints because the first derivative is not (generally) zero at the endpoints.  As far as the first derivative is concerned we could find even lower points by pushing just outside the search interval.
Additionally, notice that you are looking at the first and second derivatives in the $y$ direction.  The $y$-axis in your diagram is the green one.  At the points $P$ and $Q$, infinitesimal displacements in the $y$ direction leave the surface of the cylinder.  It may well be that the second derivative in a direction that is disallowed by the constraint is negative, but the intersection curve you see on the constraint cylinder is running in the $x$ direction at $P$ and $Q$.  To understand their behaviour, you need to be taking derivatives in the $x$-direction, which means eliminating $y$ between the objective function and the constraint and extremizing this new single variable objective function.
Expanding on this last.  From the constraint, either $y = \sqrt{5 - x^2}$ or $y = - \sqrt{5 - x^2}$.  In the first case, $h_1(x) = x^2 + 3\sqrt{5 - x^2}$, 

$h_1'(x) = 2x - \frac{3x}{\sqrt{5-x^2}}$, and $h_1'(x) = 0$ when $x \in \{ -\sqrt{11}/2, 0, \sqrt{11}/2 \}$ where $h_1'(\pm \sqrt{11}/2) < 0$ and $h'(0) > 0$.  This finds your points $R$, $P$, and $S$ and assigns the correct local extremization to each of them.  Switching to $h_2(x) = x^2 - 3\sqrt{5 - x^2}$, 

only $h_2'(0) = 0$ and $h_2''(0) > 0$, correctly identifying $Q$ as a local minimum.    We should also check $h_1(\pm \sqrt{5})$ and $h_2(\pm \sqrt{5})$ because calculus does not generally tell you about the endpoints (consider the extremization of a nonconstant line -- the derivative is never zero and the extrema occur at the endpoints of the search interval).  The first derivative is mute on these points because infinitesimal displacement from these points in the $x$ direction (along the red axis in your picture) leaves the constraint cylinder, so the first and second derivatives in $x$ tell you nothing about these points.
A: There is a second derivative test for constrained problems, based on the bordered Hessian matrix. You can find the statement in my book Multivariable Mathematics: Linear Algebra, Multivariable Calculus, and Manifolds in problem 34 on p. 225. This handout which I found with a search may do the trick for you.
