SOLUTION: Find the maximum and minimum value of z = x+3y subject to the constraints -x + 3y <= 6, x - 3y >= 6, and x+ y >= 6 and also draw a graph

Let me, first, to rewrite your constraints in equivalent form


    y <=  2 +     (1)    (first  given constraint)

    y >= -2 +     (2)    (second given constraint) 

    y >=  6 - x    (3)     (third  given constraint) 


The plot is shown below.


    


    Plot  x-3y = -6  (red),  x-3y = 6  (green) and  x+y = 6 (blue).


So, your feasible area is the set of points between two parallel lines (red and green; constraint (1) and (2)), that are ABOVE blue line.


The given linear objective function  z = x + 3y,  OBVIOUSLY, HAS NO maximum

    (its value can be made arbitrary large).


It has a minimum value, instead.


This minimum value is its value in one of the two intersection (corner) points.


The intersection points are  (3,3)  and (6,0).


The values of the objective function  z = x + 3y  at these points are


    at (3,3)  z = 3 + 3*3 = 12

    at (0,6)  z = 6 + 3*0 = 6.


ANSWER.  The given objective function HAS NO maximum value.

         It has minimum value at the point (x,y) = (6,0).