The details for this method are as follows.

1. Formulate the mathematical model of the given linear programming problem (LPP).

2. Treat inequalities as equalities and then draw the lines corresponding to each equation and non-negativity restrictions.

3. Locate the end points (corner points) on the feasible region.

4. Determine the value of the objective function corresponding to the end points determined in step 3.

5. Find out the optimal value of the objective function.

The following examples illustrate the method.

Example 1

Maximize z = 18x_{1} + 16x_{2}

subject to

15x_{1} + 25x_{2} ≤
375

24x_{1} + 11x_{2} ≤
264

x_{1}, x_{2} ≥ 0

Solution.

If only x_{1} and no x_{2} is produced, the maximum
value of x_{1} is 375/15 = 25. If only x_{2} and no
x_{1} is produced, the maximum value of x_{2} is 375/25
= 15. A line drawn between these two points (25, 0) & (0, 15), represents
the constraint factor 15x_{1} + 25x_{2} ≤
375. Any point which lies on or below this line will satisfy this inequality
and the solution will be somewhere in the region bounded by it.

Similarly, the line for the second constraint 24x_{1} + 11x_{2} ≤ 264 can be drawn. The polygon *oabc* represents the region of values for x_{1} & x_{2} that satisfy all the constraints. This polygon is called the solution
set.

The solution to this simple problem is exhibited graphically below.

The end points (corner points) of the shaded area are (0,0), (11,0), (5.7, 11.58) and (0,15). The values of the objective function at these points are 0, 198, 288 (approx.) and 240. Out of these four values, 288 is maximum.

The optimal solution is at the extreme point b, where x_{1} = 5.7 & x_{2} = 11.58, and z = 288.

Maximize z = 6x_{1} - 2x_{2}

subject to

2x_{1} - x_{2} ≤ 2

x_{1} ≤ 3

x_{1}, x_{2} ≥ 0

Solution.

First, we draw the line 2x_{1} - x_{2} ≤
2, which passes through the points (1, 0) & (0, -2). Any point which
lies on or below this line will satisfy this inequality and the solution
will be somewhere in the region bounded by it.

Similarly, the line for the second constraint x_{1} ≤
3 is drawn. Thus, the optimal solution lies at one of the corner points
of the dark shaded portion bounded by these straight lines.

Optimal solution is x_{1} = 3, x_{2} = 4.2, and the
maximum value of z is 9.6.