If $F_0\cap G_0=\emptyset$ then $x$ is a local minimum of function Consider the theorem:


Consider the following linear optimization problem $$\max 2x_1+3x_2$$ $$\text{s.t.} x_1+x_2\le8\\ -x_1+2x_2\le4\\ x_1,x_2\ge0$$
a) For each extreme point verify if necessary condition of theorem is satisfied.
b)Find the optimal solution and justify the optimality of the solution.

First we switch the problem to $\min$. Thus we have $$\min -2x_1-3x_2$$ $$s.t. x_1+x_2\le8\\ -x_1+2x_2\le4\\ x_1,x_2\ge0$$
Drawing the feasible region we found that there are only 3 extreme points: $A=(0,2),B=(4,4),C=(0,8)$.
Notice that $A$ is the only point that satisfies the constraint conditions.
Now we try to see that $F_0\cap G_0=\emptyset$. We first calculate the gradients
$\nabla f(A)=(-2,-3)^t,\nabla g_1(A)=(1,1)^t,\nabla g_2(A)=(-1,2)^t$.
And $\nabla f(A)^td=-2d_1-3d_2$
$\nabla g_1(A)^td=d_1+d_2$
$\nabla g_2(A)^td=-d_1+d_2$
We ask the 3 of them to be less than zero.
My question is how can I check that $F_0\cap G_0=\emptyset$ ?

More systematic way to do that is to use Gordan's theorem (e.g. you can find it in Bazaraa et al, Theorem 2.4.9). You need to check that the system
$$
Ad=\begin{bmatrix}1 & 1\\-1 & 1\\-2 & -3\end{bmatrix}d<0
$$
is inconsistent. It is equivalent to existence of a non-zero solution to
$$
A^Tp=0,\quad p\ge 0.
$$
It is easier in general to deal with equalities because one may use the standard elimination technique:
$$
A^T=\begin{bmatrix}
1 & -1 & -2\\ 1 & 1 & -3
\end{bmatrix}\sim
\begin{bmatrix}
1 & -1 & -2\\ 0 & 2 & -1
\end{bmatrix}.
$$
Thus all the solutions to $A^Tp=0$ are
$$
\begin{cases}
p_1&=p_2+2p_3,\\
2p_2&=p_3.
\end{cases}
$$
It is easy to find a positive solution e.g. if we take $p_3=2$, we get $p=(5,1,2)$, hence, the original system of inequalities is inconsistent and $F_0\cap G_0=\emptyset$.
