I am trying to understand how for a constrained system the introduction of Lagrange multipliers facilitates the incorporation of the holonomic constraints. I am using Classical Mechanics by John Taylor (2005). On page 276 it says that
$$\delta S = \int \left( \frac{\partial L}{\partial x} - \frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot x}\right)\,\delta x\,\mathrm dt + \int \left( \frac{\partial L}{\partial y} - \frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot y}\right)\,\delta y \,\mathrm dt = 0 \tag 1$$
for any displacements $\delta x$ and $\delta y$ consistent with the constraints. This I understand.
By writing the constraint equation $f(x,y)=C$ as
$$\delta f = \frac{\partial f}{\partial x}\delta x + \frac{\partial f}{\partial y}\delta y = 0, \tag 2$$
this constraint is incorporated in the Hamilton principle, giving
$$\delta S = \int \left( \frac{\partial L}{\partial x} +\lambda (t)\frac{\partial f}{\partial x} - \frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot x}\right)\,\delta x\,\mathrm dt + \int \left( \frac{\partial L}{\partial y} +\lambda (t)\frac{\partial f}{\partial y} - \frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot y}\right)\,\delta y\,\mathrm dt = 0. \tag 3$$
At this point I would expect this equation hold for all $\delta x$ and $\delta y$, though the author notes that still this integral is zero for any displacement consistent with the constraints. What is the point of putting it in the integrand?
Apparently, by some choice of $\lambda (t)$, this equation can be reasoned to lead to two (new) Lagrange equations in this 2-coordinate case, i.e.
$$\frac{\partial L}{\partial x} + \lambda \frac{\partial f}{\partial x} = \frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot x} \tag 4$$
$$\frac{\partial L}{\partial y} + \lambda \frac{\partial f}{\partial y} = \frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot y} \tag 5$$
which is said to still have to be supplemented with the original constraint equation $f(x,y)=C$ in order to have enough equations to be able to solve the system.
I understand the basic ideas that one needs enough equations for the number of unknowns that a problem has, and that - conveniently - Lagrange multipliers used in this way can be related to the magnitude of constraint forces. But why make the original formulation more complicated when you still need to supply the original constraint equation?
Also, to include the constraints (by means of $\delta f$) in $\delta S$, would you really need to multiply with an unknown $\lambda$? Why not set $\lambda=1$?
I suspect that my difficulty in understanding this is related to me not understanding either why, in the more purely mathematical problem of maximizing $h(\vec x)$ given $g(\vec x)=0$ by means of $\Lambda(\vec x,\lambda)=f(\vec x)-\lambda g(\vec x)$, one needs more than just the equations
$$\frac{\partial \Lambda}{\partial \vec x}=\vec 0 \tag{*}$$
(viz. $\frac{\partial \Lambda}{\partial \lambda}=0$, too)
Isn't the problem already contained in Eqs. $(*)$?