Consider the following: find the absolute max and absolute min of $f(x, y, z) = yz + xz$ over $V = \{(x, y, z) \in\mathbb{R}^3; 2x + y + z \leq 1; x \geq 0, y \geq 0, z \geq 0\}$
Attempts
So first of all, $V$ is bounded and closed, hence it's compact. This means there do exist max and min (global). Even without starting any kind of calculation, I can smell the global min is at $(0, 0, 0)$, since we have a restriction over the three variables but they all can start from zero. The function cannot be negative, due to the conditions over $(x, y, z)$ so this "proves" the origin is the global minimum.
Yet I am having troubles in finding the global max. Also I wonder if there is a suitable way to find the global min too, without intuition.
The Lagrangian reads
$$L = yz + xz - \lambda(2x + y + z - 1)$$
And the associated system of equations (gradient plus constraint) reads
$$\begin{cases} z - 2\lambda = 0 \\\\ z - \lambda = 0 \\\\ y + x - \lambda = 0 \\\\ \lambda(2x + y + z - 1) = 0 \\\\ 2x + y + z - 1 \leq 0 \end{cases} $$
I avoided to write also the positive conditions on $(x, y, z)$.
Now from the first two equations it's obvious that $z = 0$. But then from the third and from the constraint I get $x = 1$, which turns into $y = -1$.
The point $(1, -1, 0)$ is not valid though.
I don't really know how to continue here, because I cannot say "suppose $\lambda \neq 0$ and then reason on the constraint for the first two equations implies $\lambda = 0$ and $z = 0$.
I am not capable to use KKT here, have no clue on how I would proceed with this. Any help? Also $(0, 0, 0)$ does not emerge from here.
Considering the Lagrangian
$$ L(x,y,z,\lambda,s) = x z+y z+b x+\lambda _1 \left(s_1^2+2 x+y+z-1\right)+\lambda _2 \left(x-s_2^2\right)+\lambda _3 \left(y-s_3^2\right)+\lambda _4 \left(z-s_4^2\right) $$
The stationary conditions gives
$$ \nabla L = 0 = \cases{ b+2 \lambda _1+\lambda _2+z\\ \lambda _1+\lambda _3+z\\ \lambda _1+\lambda _4+x+y\\ s_1^2+2 x+y+z-1\\ x-s_2^2\\ y-s_3^2\\ z-s_4^2\\ \lambda _1 s_1\\ \lambda _2s_2\\ \lambda _3 s_3\\ \lambda _4 s_4} $$
after substitution of $b = 0$ into the formal solutions, we have
$$ \left[ \begin{array}{cccccccccccc} f & x & z & y & \lambda_1 & \lambda_2 & \lambda_3 & \lambda_4 & s_1^2 & s_2^2 & s_3^2 & s_4^2\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & -1 & -1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 1 & 0 \\ 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ \frac{1}{8} & \frac{1}{4} & 0 & \frac{1}{2} & -\frac{1}{4} & 0 & -\frac{1}{4} & 0 & 0 & \frac{1}{4} & 0 & \frac{1}{2} \\ \frac{1}{4} & 0 & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ \end{array} \right] $$
Here $s_i = 0$ means that the $i$-th constraint is active and $s_i > 0$ means that the solution point is internal to the corresponding constraint. The problem here is to administrate the $2^4$ possibilities of $\lambda_is_i = 0$