Using a Lagrange multiplier to handle an inequality constraint

Question

so I was trying to do a very basic convex optimization example using the method of Lagrange multipliers. So I wanted to: $$ \min f_{0}(x)=x^2 $$ $$ ensuring\space f_{1}(x) = x - 2 \leq 0 $$ So I wrote the equation... $$ L(x,\lambda) = f_{0}(x)-\lambda f_{1}(x)=x^2-\lambda(x-2)=x^2-\lambda x +\lambda2 $$ And then took the gradient... $$ \bigtriangledown_{x,\lambda}L(x,\lambda)=\Big(\frac{\partial L}{\partial x},\frac{\partial L}{\partial \lambda}\Big) = (0,0) $$ ...to get the two equations... $$ 2x-\lambda=0 \space and \space -x+2=0 $$ Which only seems to indicated that the solution is $$ x=2 $$

But this is obviously wrong. So did I make an egregious mistake or does the method only work for more than one dimensional space or ....?

Answer 1

We have $$2x-\lambda = 0$$ but we do not have $-x+2=0$ in this case because the constraint need not be active.

The right constraint is $\lambda \ge 0$ and the complementary slackness condition.

$$\lambda(-x+2)=0.$$

• If $\lambda=0$, then $x=0$. (The constraint is not active).

• If $\lambda \ne 0$, then $x=2$. (The constraint is active).

Answer 2

Introducing a slack variable $s \in \mathbb R$, we obtain the following equality-constrained quadratically constrained quadratic program (QCQP) in $x$ and $s$

$$\begin{array}{ll} \underset{x, s}{\text{minimize}} & x^2\\ \text{subject to} & x + s^2 = 2\end{array}$$

We define the Lagrangian

$$\mathcal L (x,s,\mu) := x^2 + \mu \left( x + s^2 - 2 \right)$$

Taking the partial derivatives and finding where they vanish, we obtain

$$\begin{array}{rl} 2 x + \mu &= 0\\ \mu s &= 0\\ x + s^2 &= 2\end{array}$$

From the 2nd equation, we have that $\mu = 0$ or $s = 0$. Let us consider each case:

• If $\mu = 0$, then $x = 0$ and $s^2 = 2$. This is the global minimum.
• If $s = 0$, then $x = 2$ and $\mu = -4$. This is at the boundary of the feasible region $(-\infty,2]$.