Question

Problem 01: Solve the LP problem using the graphic method Max Z = 30x1 + 50x2 St. 10x1 + 15x2 < 150 3x1 + 5x2 < 40 X1 2 3 X2 2 2 X1, X2 0

Answer 1

1. To draw constraint 10x4 + 15x2 < 150 - (1) Treat it as 10x4 + 15x2 = 150 When xı = 0 then x2 = ? = 10(0) + 15x2 = 150 = 15x2 = 150 150 = 10 2 x2 = 15 When xy = 0 then xy = ? = 10x4 +15(0) = 150 = 10x1 = 150 = x2 = 150 x1 = 10 = 15 x2 100

2. To draw constraint 3x4 + 5x2 < 40 → (2) Treat it as 3x1 + 5x2 = 40 When x1 = 0 then x2 = ? = 300) + 5x2 = 40 = 5x2 = 40 40 = x2= 5 = 8 When x2 = 0 then xy = ? = 3x1 +5(0) = 40 = 3x1 = 40 = x2 = 40 = 13.33 x1 0 13.33

3. To draw constraint x, 23 (3) Treat it as x1 = 3 Here line is parallel to Y-axis x 33 x2 0 1 4. To draw constraint x2 > 2 → (4) Treat it as x2 = 2 Here line is parallel to X-axis x 0 1

10x1 + 15x2 <= 150 X2 >= 2 3x1+5x2 <=40 X1 >= 3 10 15 20 25

The value of the objective function at each of these extreme points is as follows:

Extreme Point Coordinates (x1,2) Lines through Extreme Point Objective function value z = 30x 1 + 50x2 43.2) 30(3) + 50(2) = 190 B(10,2) 3 →X123 4 → X222 2 + 3x1 + 5x2 40 4 → X,> 2 2 – 3x1 + 5x2 5 40 - 3 → xq 23 30(10) + 50(2) = 400 C(3.6.2) 30(3) + 50(6.2) = 400

The maximum value of the objective function z=400 occurs at 2 extreme points.

Hence, problem has multiple optimal solutions and max z = 400.