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?
Best Answer
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.