(...) that $q_t$ is in fact an extremum of $A$, and not only a point at which its variation vanishes?
As discussed in the answer by contributor peek-a-boo: one should not pay too much attention to the customary name of the concept.
As you found out in the course of your exploration: the actual criterion is: find the point in variation space such that the derivative of Hamilton's action is zero.
So: a more descriptive name of the concept would be:
Hamilton's vanishing derivative
That raises the question: what is it about taking the derivative that makes it essential?
In order to access the necessary information the operation that is performed is taking the derivative with respect to the spatial coordinate(s).
The fact that taking a derivative (with respect to position) is necessary is correlated with the fact that potential energy does not have an intrinsic zero point. (Unlike, say, temperature, which does have an intrinsic zero point.) The choice of zero point of potential energy is arbitrary; whenever a potential energy is evaluated it is in terms of difference of potential.
To access the necessary information you take the derivative of the potential energy with respect to the spatial coordinate.
When you are using generalized coordinates: the derivative of the potential energy with respect to that generalized coordinate $q$ is referred to as a 'generalized force'.
Quoting the discussion of generalized forces by Richard FitzPatrick:
(...) a generalized force does not necessarily have the dimensions of force. However, the product $Q_i\,q_i$ must have the dimensions of work. Thus, if a particular $q_i$ is a Cartesian coordinate then the associated $Q_i$ is a force. Conversely, if a particular $q_i$ is an angle then the associated $Q_i$ is a torque.
The (generalized) force is a quantity that does have an intrinsic zero point.
Generalizing the relation between force and second time derivative of position
In cartesian coordinates:
$F=ma$
In polar coordinates:
Torque = moment of inertia $\times$ second time derivative of angle
For current through an inductor (LC-circuit):
Electromotive force = inductance $\times$ change of current
The general pattern is:
- There is an entity that tends to cause change of state, in mechanics referred to as 'force'.
- There is opposition to change of state, in mechanics referred to as 'inertia', in an inductor referred to as 'Inductance', etc.
- The relation between the above two is in terms of a second time derivative of the state, in mechanics referred to as 'acceleration'.
Integration of force/torque/etc. with respect to position coordinate gives work done, and potential energy is defined as the negative of work done.
Integral of acceleration with respect to position:
(Using the relations $ds=v \ dt$ and $a \ dt =dv$ to change the differential, with corresponding change of limits.)
$$ \int_{s_0}^s a \ ds = \int_{t_0}^t a \ v \ dt = \int_{t_0}^t v \ a \ dt = \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 $$
The pattern is:
Take a second derivative wrt position (here: acceleration), and integrate with respect to position. The result is an expression that is quadratic in the first time derivative of position: ($\tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2$)
Conversely, taking the derivative of $\tfrac{1}{2}v^2$ with respect to position recovers the acceleration.
As we know: the true trajectory has the property that throughout the trajectory the derivative with respect to time of the sum of potential and kinetic energy is zero.
$$ \frac{d(E_k + E_p)}{dt} = 0 \tag{1} $$
There is another derivative-is-zero property of the true trajectory: derivative with respect to the position coordinate $s$:
$$ \frac{d(E_k - E_p)}{ds} = 0 \tag{2} $$
Of course: the striking thing is:
In (1) the potential energy has a plus-sign, and in (2) the potential energy has a minus-sign.
The plus sign being replaced with a minus sign is because of the following:
With respect to time the two energies (potential and kinetic), are counter-changing, such that the sum remains the same value.
When sweeping out variation of the trial trajectory: with respect to the position coordinate the two energies (potential and kinetic), are co-changing. Hence (1) features a plus-sign and (2) features a minus sign.
The thing is: (1) and (2) are not independent equations. (1) and (2) are two ways of expressing the same physical property.
The effect of applying the Euler-Lagrange equation
In every situation where you apply the Euler-Lagrange equation you are applying the following operation: differentiation with respect to the position coordinate.
By differentiating with respect to the position coordinate you are accessing the information that you need.
The fact that Hamilton's stationary action holds good goes back to the property that is expressed by (1) and (2).