Mathematical question concerning Lagrange multipliers of a Lagrangian In Lagrangian Mechanics I have in general holonomic constraints of the form $f(q_1,...,q_n,t)=0$ and then I am able to use the method of Lagrange multipliers, where I go from a Lagrangian $L$  to a Lagrangian $L':=L+\lambda f(q_1,...,q_n,t)$. Now my question is the following: From a mathematical point of view: On which parameters can those Lagrange multipliers depend on?(just on time or also on coordinates and velocities?)
Since this is important when I want to use: $$\delta \int L' dt =0$$ iff L' fulfills the Euler-Lagrange equations, I probably have to differentiate my Lagrange multiplier also with respect to time and the coordinates+velocities since this is what the Euler-Lagrange equation tells me? 
Or am I always able to choose the multipliers in such a way, that they are constant (or depend only on the parameter $t$?)
 A: For any $i$, your Euler-Lagrange equations become:
$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}}-\frac{\partial L}{\partial q_i}-\lambda(t)\frac{\partial f}{\partial q_i}=0$$
This implies that whatever units $f$ carries, $\lambda(t)$ corrects for those units to make $\lambda(t)\frac{\partial f}{\partial q_i}$ have units of force. This is in fact the general interpretation of a constraint: it can be reinterpreted as a force acting on the particle. For example if it's a ball rolling on the ground subject to gravity, the constraint force keeping the ball on the ground is simply the normal force. 
In other words, the constraint equation $f(q_1,\ldots,q_n,t)=0$ can be scaled by arbitrary units but, $\lambda(t)$ will inherit inverse those units (up to a multiple of force units), so that $\lambda(t)\frac{\partial f}{\partial q_i}$ ends up with units of force. Specifically $\lambda(t)$ has units of $[N][t]/[f]$ ("Newton time per units of $f$"). Moreover, $\lambda$ itself can only depend on time because it must be constant w.r.t. $q_i$ (as is the case in Lagrange multiplier optimization).
