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Does anyone know why the KKT Conditions are considered such a fundamental result in optimization?

As far as I understand, the KKT Conditions appear to be a set of conditions that if satisfied - suggest that the "constraints" within the optimization problem are not necessary in determining the solution?

Is this why the KKT Conditions are so important?

Thanks!

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  • $\begingroup$ In calculus, we learn that if $x^*$ is a minimizer for an unconstrained optimization problem then $\nabla f(x^*) = 0$. (Here $f:\mathbb R^n \to \mathbb R$ is the objective function that we're minimizing. I'm assuming $f$ is differentiable.) So we can solve unconstrained optimization problems by setting the derivative or gradient of $f$ equal to $0$ and solving for $x$. The analogous fact for constrained optimization problems is that a minimizer must satisfy the KKT conditions. The analogous strategy for solving constrained optimization problems is to solve the KKT conditions. $\endgroup$
    – littleO
    Commented Jan 13, 2022 at 4:27

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No, that's not true at all. The KKT conditions are necessary conditions for a point to be locally optimal. The point must satisfy all the constraints: that's part of the KKT conditions. But some of the inequality constraints may be equalities for this point. The KKT conditions tell you how the gradient of the objective must be related to the gradients of the constraints that are equalities for the point in question.

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  • $\begingroup$ Might be worth noting that the equation $\nabla f(x^*) + \lambda^* \nabla g(x^*) = 0$ can be interpreted as saying that $x^*$ is a critical point of the Lagrangian $L(x,\lambda^*) = f(x) + \lambda^* g(x)$. If our optimization problem is convex, then $x^*$ is a minimizer of $L(x,\lambda^*)$. So if we magically know the value of $\lambda^*$, and if the function $x \mapsto L(x,\lambda^*)$ has a unique minimizer, then we can find $x^*$ by solving an unconstrained optimization problem. $\endgroup$
    – littleO
    Commented Jan 13, 2022 at 4:47

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