From what I read, the Karush-Kuhn-Tucker conditions are a generalization of the Lagrange Multiplier Method.

For the Lagrange Multiplier Method I have been able to find a serie of steps I must do to find the result, but I don't see quite clearly what I am supposed to do with the KKT conditions.

Lagrange Multiplier method:

  1. Set the constraint function equal to 0
  2. Create another function called the Lagrangian function by combining the objective function and the constraint
  3. Take the partial derivatives of the Lagrangian and set them to 0
  4. Solve the system of equations to get the values that minimize the objective function

KKT conditions method:

  1. Define the Lagrangian function
  2. Write the additionnal conditions
  3. ??

What should I do? Do I need to solve a system of equations as before ? Do I need to check that the condition hold ? What if they don't hold ?

  • 2
    $\begingroup$ So, suppose you have constraints $h_j (x) = 0$ and $g_i (x) \geqslant 0$ and you maximize the function. In KKT method it is essential to enumerate all situations when inequality constraints become active (i.e., $g_i (x) = 0$). In each of these cases you set up Lagrangian function as previously, but you add active constraints to it. When you find stationary points of this Lagrangian, you must check 1) whether they are feasible (i.e., all other constraints $g_i (x) \geqslant 0 $ hold); 2) whether gradient of target function has negative dot product with gradients of active constraints. $\endgroup$
    – Evgeny
    Dec 26, 2015 at 23:16
  • $\begingroup$ If 1) or 2) is not true, then the stationary points is not a local optimum? $\endgroup$
    – Octoplus
    Dec 26, 2015 at 23:49
  • 2
    $\begingroup$ If (1) is not true, then this point just doesn't belong to the domain of interest. Note that you need to check only inequalities, because equality constraints are satisfied automatically by Lagrangian method. If (2) is not true then you might find a feasible point with greater value of target function and thus point is not a local maximum. $\endgroup$
    – Evgeny
    Dec 27, 2015 at 0:15


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