Question (graphical solution technique): Consider the following linear programming problem:
P: Max z = 2x1 + x2
s.t. 4x1 + 5x2  20
2x1 + 5x2  10
x2  0
(1)
(2)
(3)
The variable x1 is unrestricted in this problem.
(a) Graph the feasible set and the gradient of the objective function.
(b) What happens if we replace the above objective function by Min z = 1x1 + 2x2?
(c) What happens if we use the objective function Min z = x1 + 2x2 instead?
Solution: (a) The figure below shows the feasible set as the shaded area. The gradient of the objective
function is shown as (a) The optimal solution is the point labeled “A.” Its coordinates are x = (1⅔, 2⅔)
with a value of the objective function of z = 6.
(b) The new objective function is shown as (b). It happens to be perpendicular to constraint (2). As a
result, all points on the line segment between A and B are optimal. This is a case of alternative optimal
solutions.
(c) The new objective function is shown as (c). It is apparent that this is a case of unbounded “optimal”
solutions.