Solution

Given objective function is

$z=-2x_1-3x_2$

To maximize $z=-2x_1-3x_2$, we need to minimize the negative loss.

The main lower constraint is

$x_1+x_2\geq 2$

Now check the feasible corner points on $x_1+x_2=2$:

If $x_2=0$, then

$x_1=2$

So, point is $(2,0)$.

At $(2,0)$,

$z=-2(2)-3(0)$

$z=-4$

If $x_1=0$, then

$x_2=2$

So, point is $(0,2)$.

At $(0,2)$,

$z=-2(0)-3(2)$

$z=-6$

Since $-4>-6$, maximum value is $-4$.

Also, $(2,0)$ satisfies the other constraints:

$2x_1+x_2=2(2)+0=4\leq 10$

$x_1+x_2=2+0=2\leq 8$

Therefore, optimal solution is

$x_1=2,\ x_2=0,\ z=-4$