Converting the given inequation into equations, we get
2x + 3y = 6, x – 2y = 2, 3x + 2y = 12, –3x + 2y = 3, x = 0 and y = 0
Region represented by –2x – 3y ≤ –6: The line –2x – 3y = –6 or 2x + 3y = 6 cuts OX and OY at A1 (3, 0) and B1 (0, 3) respectively. Join these points to obtain the line 2x + 3y = 6.
Since O (0, 0) does not satisfy the inequation –2x – 3y ≤ –6. So, the represented –2x – 3y ≤ –6 is that part of XOY-plane which does not contain the origin.
Region represented by x – 2y ≤ 2: The line x – 2y = 2 meets the coordinate axes at A2 (2, 0) and B2 (0, –1). Join these points to obtain x – 2y = 2. Since (0, 0) satisfies the inequation x – 2y ≤ 2, so the region containing the origin represents the solution set of this inequation.
Region represented by 3x + 2y ≤ 12. The line 3x + 2y ≤ 12 intersects OX andOY at A3 (4, 0) and B3 (0, 6). Join these points to obtain the line 3x + 2y = 12. Clearly, (0, 0) satisfies the inequation 3x + 2y ≤ 12. So, the region containing the origin is the solution set of the given inequations.
Region represented by –3x + 2y ≤ 3: The line –3x + 2y = 3 intersects OX andOY at A4 (–1, 0) and B4 (0, 3/2). Join these points to obtain the line –3x + 2y = 3. Clearly, (0, 0) satisfies this inequation. So, the region containing the origin represents the solution set of the given inequation.
Region represented by x ≥ 0, y ≥ 0: Clearly, XOY quadrant represents the solution set of these two inequations.
The shaded region shown in fig. represents the common solution set of the above inequations. This region is the feasible region of the given LPP.
The coordinates of the corner-points (vertices) of the shaded feasible regionP1P2P3P4 are
. These points have been obtained by solving the equations of the corresponding intersecting lines, simultaneously.
The value of the objective function at these points are given in the following table:
Point (x, y)
|
Value of the objective function
Z = 5x + 2y
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|
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Clearly, Z is min at and maximum at . The minimum and maximum values of Z are 63/13 and 19 respectively.