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Consider the following constrained optimization problem:

$$\min_{x} f = x_1 \\ \text{such that} \quad g_1=x_1^2+x_2^2-9 \leq0 \\ \qquad \qquad \ \ \quad \ g_2=-x_1^2-x_2^2+4 \leq0 \\ \qquad \ \ \ \ \ g_3 = x_1-x_2 \leq0$$

How many stationary points and minimizers are there?

The sketch below shows the feasible domain in green. The monotonicity analysis shows that $g_2$ (the inner circle) is an active constraint. enter image description here

For the stationary points we determine the gradient of $f$ and set it equal to $0$. Which is $\nabla f = \begin{bmatrix} 1 & 0 \end{bmatrix} = 0$ Which does not yield an $x_1$ or $x_2$ value.

So I dont know how many stationary points and minimizers there are.

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    $\begingroup$ No, for stationary points we DO NOT determine the gradient of $f$ but the gradient of the Lagrangian function. You must take into account the constraints. $\endgroup$
    – Alex Silva
    Commented Apr 2, 2019 at 13:18
  • $\begingroup$ so i would have to use this formula: $\nabla f + \mu^{\top} \nabla g $? $\endgroup$
    – user463102
    Commented Apr 3, 2019 at 12:51
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    $\begingroup$ Yes! Exactly :) $\endgroup$
    – Alex Silva
    Commented Apr 3, 2019 at 13:01
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    $\begingroup$ Let $L = f+\mu^{T}g$. Then $$\partial_{x1}(L) = 1+2x_1\mu_1-2x_1\mu_2+\mu_3=0$$ and $$\partial_{x2}(L) = 2x_2\mu_1-2x_2\mu_2-\mu_3=0$$ $\endgroup$
    – Alex Silva
    Commented Apr 5, 2019 at 6:49
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    $\begingroup$ Actually, by looking at your figure, it is clear that the minimum occurs when $(x_1,x_2) = (-3,0)$. $\endgroup$
    – Alex Silva
    Commented Apr 5, 2019 at 7:24

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