Karush–Kuhn–Tucker conditions

Write inequalities as $f_i(x)\leq0$ and equalities as $h_j(x)=0$. Primal feasibility says the candidate obeys these rules. Dual feasibility says inequality multipliers are nonnegative. Stationarity balances the objective gradient with a weighted combination of constraint gradients. Complementary slackness requires each inequality multiplier times its constraint value to vanish. Together these conditions connect a feasible decision to an admissible lower-bound construction.

An inequality is active when its value is exactly zero. Complementary slackness forces the multiplier of every strictly inactive inequality to be zero. The converse is false: an active inequality may have zero multiplier if it is locally irrelevant to the optimum's value or redundant with other constraints. Thus counting nonzero prices is not the same as listing every geometric boundary touching the answer. Equality multipliers remain unrestricted in sign.

Suppose the objective and inequality functions are differentiable and convex, the equalities are affine, and the candidate lies in their common domain. Nonnegative multipliers make the Lagrangian convex in the primal variable. Stationarity then gives its global minimum. Feasibility and complementary slackness make that minimum equal the primal objective at the candidate, matching a dual lower bound. This proves global optimality of any KKT point in this setting.

A constraint qualification is not needed for that sufficiency argument once a valid KKT certificate has been supplied. It matters when asking whether an optimum must possess such multipliers. For convex problems, Slater's condition is a standard sufficient qualification. For general smooth constrained problems, conditions such as linear independence of active constraint gradients can provide necessity at a local optimum. Without qualification, an optimum can exist while no KKT multiplier vector exists.

Consider minimizing x subject to $x^2\leq0$. The only feasible point is zero, so it is automatically optimal. Yet stationarity would require $1+\lambda(2x)=0$ at zero, an impossibility. The constraint has zero gradient there and admits no strictly feasible point. This example shows why writing down KKT equations and finding no solution does not, by itself, prove that the original problem has no optimum.

Numerical solvers report residuals rather than exact symbolic equalities. Feasibility errors, stationarity errors, and complementarity errors need scaling-aware tolerances; a tiny product can hide a large multiplier paired with a slightly infeasible constraint. The lab has no constraints or multipliers. Use its gradient readout as the unconstrained special case, then compare that zero-gradient stopping rule with the nonzero constrained gradient in the worked example. It cannot test KKT conditions for a user-supplied problem.