I have this function $(x, y) \rightarrow e^{(x-1)²+(y-1)²} $ for $f: K \rightarrow\mathbb{R}$ with $K=\{(x,y)\in\mathbb{R²} | x²+y²\leq4,x\geq0\text{\}} $.
I know that $(1,1)$ is a global minimum and I suspect that $(0, \pm2)$ are the other extreme points, but how do I show that mathematically precise?
Calling
$$ \cases{ p = (x,y)\\ p_0=(1,1)\\ f(p) = e^{\|p-p_0\|^2} } $$
we can define the Lagrangian
$$ \mathcal{L}(p,\lambda, s) = f(p) + \lambda(\|p\|^2-4+s^2) $$
with $\lambda$ a lagrange multiplier and $s$ a slack variable to transform the restriction inequality into an equation. Now the stationary points are the solutions for
$$ \nabla \mathcal{L}=\cases{ 2(p-p_0)e^{\|p-p_0\|^2}+2\lambda p = 0\\ \|p\|^2-4+s^2 = 0\\ \lambda s = 0 } $$
or equivalently
$$ \nabla \mathcal{L}=\cases{ (p-p_0)+\Lambda p = 0\\ \|p\|^2-4+s^2 = 0\\ \Lambda s = 0 } $$
with $\Lambda = \lambda e^{-\|p-p_0\|^2}$. Solving for $\{p,\Lambda,s\}$ we have
$$ \left[ \begin{array}{ccccc} f & x & y & \Lambda & s^2\\ 1 & 1 & 1 & 0 & 2 \\ e^{2 \left(\sqrt{2}-1\right)^2} & \sqrt{2} & \sqrt{2} & \frac{1}{2} \left(\sqrt{2}-2\right) & 0 \\ \end{array} \right] $$
Here $s=0$ means that the restriction is actuating