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if I have a problem min(-f) s.t. g<0, I can rewrite it as -max(f) s.t. g<0. In this case, if I take Lagrangians, would my lagrangian be L=f- lambda(g-0) or would I have to have a negative in front of the f?

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Solving

$Min(-F[x])$ s.t. $G[x]\leq 0$

is same as solving

$Max(F[x])$ s.t. $G[x]\leq0 $

So, the Lagrangian for the minimizing problem will be:

$L = -F[x] - \lambda (0-G[x])$

For the maximization problem the Lagrangian will be:

$L = F[x] + \lambda (0-G[x])$

In both cases, $\lambda \geq 0$

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    $\begingroup$ I would have thought that $\lambda$ for the minimisation problem is non-positive, $\lambda \le 0$. The first Lagrangian is the negative of the second one, so the first mulitplier should be the negative of the second one. Also the inequalities in the maximisation and minmisation problem should be weak ($\le$) instead of strict. $\endgroup$
    – tdm
    Jan 27, 2022 at 6:29
  • $\begingroup$ Correct and correct! Thanks for pointing it out. $\endgroup$
    – erik
    Jan 27, 2022 at 19:01

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