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My main question is that when I use Lagrange Multipliers/KKT conditions to perform optimization with threshold constraints, I seem to get contradictory FOC.

Here is a characteristic example: take an optimization problem like the following:

$$ \begin{align*} &\max_{x,y} 2x+y-(x-y)^2\\ &s.t.\\ &x,y \geq 0\\ &x+y \leq I \end{align*} $$ For some value $I>0$. If I translate this directly to a lagrangian I believe the KKT conditions should give that I need to optimize: $$ \mathcal{L}(x,y,\mu_1, \mu_2, \mu_3)=2x+y-(x-y)^2 + \mu_1(-x) + \mu_2(-y)+\mu_3(x+y-I) $$ However partial derivatives w.r.t $\mu_1, \mu_2,\mu_3$ gives: $$\begin{align*} &-x=0\\ &-y=0\\ &x+y-I=0\\ \end{align*}$$ which gives a contradiction for $I>0$.

Could someone help me understand why I seem to get this contradiction?

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1 Answer 1

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Instead of differentiating w.r.t. the multipliers you should properly write the complementary slackness condition (defined in your link). In your case this is

$$-\mu_1 x=0$$

$$-\mu_2 y=0$$

$$\mu_3 (x+y-I)=0$$

This says that either the constraint binds with equality or the multiplier is zero. If you just differentiate with respect to $\mu_1$ for instance you would incorrectly conclude that always $x=0$, while this can be strictly positive.

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  • $\begingroup$ Ahh got it. So at an optimum we check for FOC for the term we wish to optimize but complementary slackness for inequality terms? $\endgroup$
    – naveace
    Feb 21, 2022 at 0:19
  • $\begingroup$ @naveace Yep correct. $\endgroup$
    – fes
    Feb 21, 2022 at 9:03

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