For $c < 0$, we have no feasible solutions and hence no optimal solutions.
For $c=0$, our only feasible solution is $z=0$ obtained by $(0,0)$.
For $c > 0$, well...
I graphed the constraints besides $x_1 + x_2 = c$ and came up with a feasible region with corner points $(0,0), (0,3), (3,3), (6,2), (7,0)$.
For $c \in (0,3]$, it looks like we have a triangle for a feasible region and so need only inspect $(0,0), (c,0), (0,c)$
For $c=4$, we inspect $(0,0), (4,0), (0,3), (1,3)$
For $c=5$, we inspect $(0,0), (5,0), (0,3), (2,3)$
For $c=6$, we inspect $(0,0), (6,0), (0,3), (3,3)$
Generalising: For $c \in \color{red}{[}3,6]$, we inspect $(0,0), (c,0), (0,3), (c-3,3)$ where for $\color{red}{c=3}$, $(0,3) = (c-3,3)$
For $c=7$, we inspect $(0,0), (7,0), (0,3), (3,3)$ and the intersection between $x_1+x_2=c=7$ and $x_1+3x_2=12$ which is $(9/2,5/2)$
Generalising: For $c \in (6,7]$, we inspect $(0,0), (c,0), (0,3), (3,3)$ and the intersection between $x_1+x_2=c$ and $x_1+3x_2=12$ which is $(-6+3c/2,6-c/2)$
For $c = 7.5$, we inspect $(0,0), (7,0), (0,3), (3,3)$, the intersection between $x_1+x_2=c=7.5$ and $x_1+3x_2=12$ which is $(5.25,2.25)$ and the intersection between $x_1+x_2=c=7.5$ and $2x_1+x_2=14$ which is $(6.5,1)$
Generalising: For $c \in (7,8]$, we inspect, we inspect $(0,0), (7,0), (0,3), (3,3)$, the intersection between $x_1+x_2=c$ and $x_1+3x_2=12$ which is $(-6+3c/2,6-c/2)$ and the intersection between $x_1+x_2=c$ and $2x_1+x_2=14$ which is $(14-c,2c-14)$
For $c=8$, the constraint $x_1+x_2 \le c = 8$ is redundant as can be seen here:
- Generalising: For $c\ge8$, the constraint $x_1+x_2 \le c$ is redundant.
Is that right?
From Chapter 2 here.