Lagrange Mutiplier for inequality constraint

I'm a bit confused about Lagrange multipliers. I know it works wonders if I only have equality constraints. Whenever I have inequality constraints, or both, I use Kuhn-Tucker conditions and it does the job.

But my question is, can I solve a inequality constraint problem using only Lagrange multiplier?

For example:

min $$f(x)=x^4$$ with $$x \leq -1$$.

$$L(x,\lambda)=x^4+ \lambda(x+1)$$, gives:

$$4x^3+ \lambda =0$$ and $$x+1=0 \Leftrightarrow x=-1 , \lambda =4$$

Is this correct? I just assumed equality.. What happens if the constraint is not active?

Introducing the slack variable $$\epsilon$$ we can transform an inequality into an equality. So with this idea we can write the equivalent Lagrangian

$$L(x,\lambda,\epsilon) = x^4+\lambda(x+1+\epsilon^2)$$

The stationary points are solved by

$$L_x = 4x^3+\lambda = 0\\ L_{\lambda} = x+1+\epsilon^2 = 0\\ L_{\epsilon} = 2\lambda\epsilon = 0$$

Now, depending of the values for $$\epsilon,\lambda$$ we can qualify those stationary points.

NOTE

After solving we have the feasible solution $$x = -1,\ \,\lambda = 4,\ \ \epsilon = 0$$ which indicates that the restriction actuates.

• I've seen this somewhere also. Is this the same as assuming that if constraint is active then $\lambda \geq 0$ and if its inactive we have $\lambda=0$? What I just said comes from what you did right? – Amateur Mathematician Dec 5 '18 at 23:03
• forgot to tag you – Amateur Mathematician Dec 5 '18 at 23:40
• @AmateurMathematician $\epsilon = 0$ indicates that the restriction is active. $\lambda > 0$ indicates that the objective function gradient $4 x^3$ points in the same direction as the restriction, at the solution point. – Cesareo Dec 5 '18 at 23:51
• The stationary points are just the KKT conditions. – LinAlg Dec 6 '18 at 1:22
• @Cesareo can tou recommend me a book on this 'slack variable'? Also why do the slack variable squared? – Amateur Mathematician Dec 6 '18 at 8:06