Lagrange multiplier for QCQP with $1$ equality constraint

Question

I want to find the maximum of $f(x,y) = x^2 - y^2$ under the constraint $\frac12 x^2 + y^2 - 1 = 0$.

I defined Lagrange function:

$$ L= x^2 - y^2 + \lambda \left( \frac12 x^2 + y^2 - 1 \right) $$

Then caculated $L'_x, L'_y, L'_{\lambda}$ and got the following equations system:

$$ \begin{aligned} x (2 + \lambda) &= 0 \\ 2 y (\lambda - 1) &= 0 \\ \frac12 x^2 + y^2 - 1 &= 0 \end{aligned} $$

How to solve this system so I can find the maximizer $(x,y)$?

From the first equation, you get $x =0 \lor \lambda = -2$. Eventually, you obtain a conjunction of disjunctions. Check each case. Note that equality-constrained QCQPs are used to encode combinatorial problems. Thus, do not be surprised if you encounter a small-scale "combinatorial explosion". — Rodrigo de Azevedo, Aug 13 '22 at 14:08

score 1 · Answer 1 · answered Aug 13 '22 at 14:58

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($\lambda = 1$, $x=0$ and $y=\pm 1$) Or ($\lambda = -2$, $y=0$ and $x=\pm \sqrt 2$)

answered Aug 13 '22 at 14:58

Kroki

Yes, but can you please explain shortly what are the cases i should consider @Youem – Aug 13 '22 at 15:31
The second equation gives you $\lambda =1$ or $y=0$. Do you see that? – Kroki Aug 13 '22 at 15:32
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@Youem Note that $(x=0, y=\pm 1)$ is a minimum of $f$. Only $(x=\pm\sqrt 2 , y=0)$ is a maximum. – Andreas Tsevas Aug 13 '22 at 15:36
Can someone explain how to look at each one of the possible cases in the system I wrote? – Aug 14 '22 at 08:14

1 Answers1