Consider z=ax+2y, because the optimal solution must be reached at the feasible vertex, so
(1) If a=0, then z=2y, obviously there is a minimum at point (1, 0), and z=0.
(2) If a≠0, the values at (1, 0), (0, 1) and (3, 4) are respectively:
z=a
z=2
z=3a+8
Therefore, when -4
When 0
To sum up, the point at which the minimum value is reached depends on the value of a.
When restricted
-4 & lt; A<2, the minimum value exists only at point (1, 0).
So the range of a is (-4,2).