our teacher gave us a problem in the exam that I failed to answer it even after passing it, and I ask for an explanation from people here please...
this is the problem :
let K be a subset of $\Bbb R^2$ such that : $$K=\{(x,y) \in \Bbb R^2, \mathrm{e}^{-x} \le y, \quad y \le 2-x,\quad y \le x\}$$ and $$ f : \Bbb R^2 \to \Bbb R $$ the function given by $$ f(x,y) = y\cdot\mathrm{e}^{-x} $$
1/ Show the existence of a solution (x*,y*) for the problem $$\mathbf{min} \underset{(x,y) \in K} \quad f(x,y) $$
I know there are many ways for doing that, like showing that f(x,y) is continuous and coercive and showing that K is closed, or showing that K is bounded and closed, means compact (I don't know if it is an enough condition for the existence), or showing that f(x,y) is strictly convex on the convex set K, please tell me which way it is the most efficient.
2/ justify and write the optimality conditions of KKT and solve the problem
and here the calamity, I find up to 9 cases that I couldn't at anyone find x or y --"
If $(x,y) \in K$ then $0<e^{-x} \leq y \leq 2-x$ which gives $x \leq 2$. Also $0<y \leq x\leq 2$. Hence $K$ is bounded. It is also closed, so is $K$ is compact. Since $f$ is continuous it attains its minimum at some point. To minimize $f$ we can minimize over $y$ first and then minimize over $x$. Hence we have to minimize $e^{-2x}$ subject to the conditions $e^{-x} \leq \min \{x,2-x\}$. It is easy to see that there is a unique $t$ such that $e^{t}(2-t)=1$ and $f$ is minimized when $x=t, y=e^{-t}$.