Confusion with Lagrange Multipliers I am numerically solving an optimization problem of the form: Maximize $z$ subject to $f(\alpha,z)=c$.
Using the method of Lagrange Multipliers I first write down the Lagranian
$$
\mathscr L(\alpha,z,\lambda)=z-\lambda(f(\alpha,z)-c),
$$
for which upon setting the gradient equal to zero yields the system of equations
$$
\begin{aligned}
\lambda\partial_\alpha f(\alpha,z)&=0\\
\lambda\partial_z f(\alpha,z)&=1\\
f(\alpha,z) &=c.
\end{aligned}
$$
Here is my confusion: I have already proven that $\partial_z f(\alpha,z)>0$ for all $\alpha$ and $z$; thus, according to the second equation $\lambda$ will always be some positive constant. If this is the case, then why do I need the Lagrange multiplier at all?  Wouldn't it suffice to simply solve the system
$$
\begin{aligned}
\partial_\alpha f(\alpha,z)&=0\\
f(\alpha,z) &=c.
\end{aligned}
$$
I proceeded to (numerically) solve this system of two equations and did indeed verify that the solution solves my maximization problem.  So do I need the original system of three equations?  What am I missing?
 A: Your observations are correct, though they apply quite specifically to your problem. It is not uncommon for the method of Lagrange multipliers to yield equations that either you already knew, or are useless (like $0 = 0$).
What is true in general is that you never have to use the method of Lagrange multipliers. It's always possible (perhaps not algebraically, but definitely numerically) to use the constraint to eliminate one of the variables, but this method may be disadvantageous for a couple reasons (it may complicate the calculations, for one). For your problem, in many cases we could use the constraint $f(\alpha, z) = c$ and solve for $z$ as a function of $c$ and $\alpha$ and then set the derivative of $z$ with respect to $\alpha$ to zero like in a normal one-variable extremization problem. These will lead to the exact same equations you have already deduced.
The moral of the story? There is no hands-down most efficient way for solving many extremization problems; it will depend on the nature of the problem.
