Here is an example given in the book "Forst and Hoffmann - Optimization-Theory and Practice - 2010 - Springer", p.37
Let's define: \begin{align*} f(x) &:= x_1x_2^2\\ h(x) &:= h_1(x) = x_1^2 + x_2^2 - 2 \text{ for } x = (x_1, x_2)^\top \in D :=\mathbb{R}^2 \end{align*} we consider the problem $$ f(x)\rightarrow \text{min} \quad \text{subject to the constraint} \quad h(x) = 0 $$ Points $x$ meeting the constraint are different from 0 and thus also meet the rank condition. With $\mu := \mu_1$ the equation $\nabla f(x) + \mu \nabla h(x) = 0$ translates into $$ x_2^2 + \mu x_1 = 0\quad \text{and}\quad 2x_1x_2 + \mu 2 x_2 = 0 $$ Multiplication of the first equation by $x_2$ and the second by $x_1$ gives $$ x_2^3 + 2\mu x_1x_2 = 0 \quad \text{and} \quad 2x_1^2 x_2 + 2\mu x_1x_2 = 0 $$ and thus $$ x_2^3 = 2x_1^2x_2 $$ For $x_2 = 0$ the constraint yields $x_1 = \pm \sqrt{2}$. Of these to evidently $\star$ on ly $x_1 = \sqrt{2}$ remains as a potential minimizer. If $x_2 \neq 0$, we have $x_2^2 = 2x_1^2$ and hence with the constraint $3x_1^2 = 2 \star$, thus $x_1 = \pm \sqrt{2/3}$ and then $x_2 = \pm 2/\sqrt{3}$. In this case the distribution of the zeros and signs of th $f$ gives that only $x = (-\sqrt{2/3}, \pm 2/\sqrt{3})^\top$ remain as potential minimizers. Since $f$ is continuous on the compact set $\{x\in \mathbb{R}|h(x) = 0\}$, we know that there exists a global minimizer. Altogether, we get: $f$ attains its global minimum at $(-\sqrt{2/3}, \pm 2/\sqrt{3})^\top$, the point $(\sqrt{2}, 0)^\top$ yields a local minimum.
Here are the two points that I don't understand (marked by an asterisk $\star$ in the text):
- Of these to evidently on ly $x_1 = \sqrt{2}$ remains as a potential minimizer. Why $-\sqrt{2}$ is eliminated as potential minimizer?
- we have $x_2^2 = 2x_1^2$ and hence with the constraint $3x_1^2 = 2$. From where do we get the equation $3x_1^2 = 2$