Complementary slackness conditions (Kuhn-Tucker)
It is possible to have
$$g(x^*) = b\; {\rm and}\; \lambda^* = 0$$.
When the multiplier is zero and the constraint is equal to zero, then
a) The constraint does not really &quot;bind&quot;
b) That's why the multiplier is zero.
What does it mean &quot;the constraint does not really bind&quot;?
It means that the solution $x^*$, that makes $g(x^*) = b$, would be chosen even if the constraint was not imposed. In that sense, the constraint is not really binding, because it does not really forbid us to go where we wanted to be, because we are already there.
Consider the simple example
$$\max_x \{-ax^2 + bx\},\qquad s.t. \;x \geq \frac{b}{2a} $$
The Lagrangean is
$$\Lambda = -ax^2 + bx + \lambda \left(x -\frac{b}{2a}\right)$$
and the f.o.c is
$$x = \frac{b+\lambda}{2a}$$.
Try cases:
a) $\lambda^* = 0$ leads to $x^* = \frac{b}{2a}$, which is the unconstrained f.o.c. also.
b) $\lambda^* &gt; 0$ the f.o.c. initially indicates that $x^* &gt; b/2a$. But then the constraint is not binding and we should have $\lambda^* =0$ : contradiction.
So we see that in this case the solution is
$$x^* = \frac{b}{2a},\;\;\; \lambda^* = 0.$$
So the constraint appears to be binding, but it really is not.