# Lagrange Multipliers Formulation

Suppose we have the following problem: $$\text{minimize } \ f(x) \\ \text{subject to } \ Ax = b$$

How do we know whether to write the Lagrangian Dual as $$\text{minimize } f(x) + \lambda(Ax-b)$$ versus $$\text{minimize } f(x) + \lambda(b-Ax)?$$

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We don't, and it does not matter. The $\lambda$ you are going to find will change sign. See also here.
Then why is that in minimization problems I see equality constraints $Ax = b$ being written as $Ax-b$ but in maximization problems it is written as $b-Ax$? In a minimization problem I could write it as $b-Ax$? –  robbie Jun 15 '12 at 6:07
When you minimize, $(x, \lambda)$ is variable. In one case you'll get $(x_0,\lambda_0)$ as a solution, in the other case $(x_0,-\lambda_0)$. You are interested in $x_0$. The solution is the same, yes. (And both problems are equally difficult). –  user20266 Jun 15 '12 at 6:10