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How can Lagrange multiplier method be used when inequality conditions are given instead of equality conditions?

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  • $\begingroup$ Your optimum may well not be at a constraint when using this method, as the inequality constraint will specify LESS THAN or equal to a constraint, or GREATER THAN or equal to a constraint. However, note that when I worked on a large industrial optimizer with 20 degrees of freedom, that beast was normally at 20 constraints when it converged, meaning that the normal case seems to be that the number of constraints equals the number of degrees of freedom. $\endgroup$ – David White May 20 '19 at 19:11

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