Graphical Method Of Linear Programming Problems

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Graphical Method Of Linear Programming Problems

We will now present the graphical method for solving linear programming problems that involve only two variables. This method is based on a theorem, called extreme point theorem, which states as follows:

Extreme point theorem. An optimum solution to a linear programming problem, if it exists, occurs at one at one of the comer (or extreme ) points of the area of feasible region.

The graphical method involves three basic steps.

Step. The first step in the graphical method is to determine the area of feasible region. This area will contain the set of all points that simultaneously satisfy all constraints (including non-negative restrictions).

Step. Note the coordinates of each corner point of the feasible region and substitute the coordinates of the corner points into the objective function.

Step. The optimum solution occurs in a maximization case at the corner point yielding the largest value of the objective function and in a minimization case at the corner point yielding smallest value of the objective function.

Example. Solve the following LP problem by the graphical method.

Maximize                   Z = 3x + 4y

subject to the constraints:

2x + 2y ≤ 80
2x + 4y ≤ 120
      x,y  ≥ 0.

Solution. First we find the feasible region which consists of the set of all points whose coordinates simultaneously satisfy all constraints in-eluding non-negative restrictions. Application the techniques of section to the above system of inequalities leads to the feasible region shown in.


The corner points of the feasible region are O (0,0), A (40,0), B (20, 20) and C (0,30). Note that the coordinates of the point B have been obtained by solving 2x + 2y =  80 and 2x + 4y = 120 simultaneously for x and y. The values of the objective function at these corner points are summarized in the following table:

   Corner Point____ Value of the objective function         (x,y)____________________ Z = 3x + 4y
O (0,0)                          3 (0) + 4 (0) = 0
A (40,0)                         3 (40) + 4 (0) = 120
B (20, 20)                      3 (20) + 4 (20) = 140
C (0,20)                         3 (0 ) + 4 (30) = 120

The maximum value of Z is thus given by the coordinates of B. Hence optimum solution occurs at X = 20 and y = 20, resulting a maximum value of Z as 140.

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