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    $\begingroup$ Don’t we have to treat it as Karush-Kuhn-Tucker rather than Lagrange, since the constraint is an inequality? $\endgroup$ Commented Sep 24, 2020 at 11:44
  • $\begingroup$ @Dave, Would this make any real difference? (It has been quite a while since I was in a calculus class and don't remember the details of how they compare). From a practical standpoint (understanding rather than rigorous proof), any meaningful constraints considered will have the solution equal to the constraint, so I don't think it will really matter. $\endgroup$ Commented Sep 24, 2020 at 13:38
  • $\begingroup$ @Dave even if you take the KKT formulation, your lambda hyperparameter is positive, which means all the KKT inequality constraints are activ, i.e. they are on the boundary g(x)=0. This comes down to the Lagrangian formulation because we only have equality constraints left. $\endgroup$ Commented Aug 21, 2021 at 12:23