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 $F_0\cap G_0=\emptyset$ ?

Notice that $d_1+d_2<0$ and $-d_1+d_2<0$ implies that $d_2<-|d_1|$, or $-d_2>|d_1|$. Hence $-2d_1-3d_2>-2d_1+3|d_1|\geqslant0$ for all $d_1\in\mathbb{R}$. 
You can also draw a picture (good for these 2d geometric problems):

