1. Standard of Competency
Solving Linear Programming Problem
Basic Competency
2.1 Solving System of Linear Inequalities in Two Variables
2.2 Mathematics modelling from linear programming problem

2. Solving System of Linear Inequalities in Two Variables
Symbol : ≤ (less than or equal to),
≥ (more than or equal to),
< (less than),
> (more than)
x – 3y < 6
3x – 5y ≤ -1

3. PROGRAM LINEAR (Linear Programming) Kusuma Bangsa Senior High School
CLASS XII NATURAL SCIENCE
Kusuma Bangsa Senior High School

4. x y (x,y) Solving of Linear Inequalities in Two Variables Example :
Determine the area of solution set from linear inequalities in
two variable following :
2x + 3y < 6
4x – y ≥ 12
a. 2x + 3y < 6 x y
(x,y) 3 2
(0,2)
(3,0)

5. P(0,0)
2(0) + 3(0) < 6
0 < 6
Solution Area

6. b. 4x – y ≥ 12 x y
(x,y) 3
-12
(3,0)
(0,-12)

7. P(0,0)
4(0) - (0) ≥12
0 ≥ 12
Solution Area

8. Y = 2x- 8
Y – 2x = -8
Solution Area
P(0,0)
(0) - 2(0) ….-8
0 ≥ -8
Y – 2x ≤-8

9. 5x + 6y = 30
Solution Area
P(0,0)
5(0) +6(0) ….30
0 < 30
5x + 6y < 30

10. Solving of System Linear Inequalities in Two Variables
1. 2x -2y ≥4 and x – y ≥ 6
2. x + 3y ≤ 4 and 2x + y ≤ 8, and 5x – 8y ≤ 0

11. x y (x,y) x y (x,y) Sketch the system of linear inequalities below
1. 2x + 3y ≤ 6, 4x + y ≤ 4, x ≥0, and y ≥ 0
2x + 3y ≤ 6
4x + y ≤ 4 x y
(x,y) x y
(x,y) 3 1 2 4
(0,2)
(3,0)
(0,4)
(1,0)

12. x = 0
4x+y=4
Solution Area
2x+3y=6
y = 0

13. x = 0
x = 0
4x+y=4
Solution Area
2x+3y=6
y = 0

14. Sketch the system of linear inequalities below
2. x + y ≥ 4, x + y ≤ 9, 2x-3y ≤6, x ≥0, and y ≥ 0
x + y ≥ 4
x + y ≤ 9 x y
(x,y) x y
(x,y) 9 4 9 4
(4,0)
(0,9)
(9,0)
(0,4) x y
(x,y)
2x-3y ≤6 3 -2
(3,0)
(0,-2)

15. x = 0
x + y = 9
2x-3y = 6
Solution Area
y = 0
x + y = 4

16. x = 0
x + y = 9
2x-3y = 6
Solution Area
y = 0
x + y = 4

17. Solution Area 5x+4y ≤20 3x+6y≤ 18 x ≥0, and y ≥ 0 (0,5) (0,3) (6,0)
(4,0)

18. Optimum solution from objective function of
system of linear inequalities in two variables
ax + by ≤ c and rx + sy ≤ p as constraints,
Z = cx + dy as the objective function

19. Determine the maximum value Z = 2x + 3y at the system of linear inequalities
2x + 3y ≤ 6, 4x + y ≤ 4, x ≥ 0, y ≥ 0
2x + 3y ≤ 6
4x + y ≤ 4 x y
(x,y) x y
(x,y) 3 1 2 4
(0,2)
(3,0)
(0,4)
(1,0)

20. Determine the maximum value Z = 2x + 7y at the system of linear inequalities
4x + 6y ≤ 3, x + y ≤ 2, x ≥ 0, y ≥ 0
Determine the minimum value Z = 5x + 7y at the system of linear inequalities
x + 3y ≥ 9, 2x - y ≤ 12, x ≥ 0, y ≥ 0

21. x = 0
x = 0
4x+y=4
Solution Area
(0,2) ?
2x+3y=6
(0,0)
(1,0)
y = 0

22. 4x+y=4 2x+3y=6 x 1 4x + y = 4 4x + 6y = 12 - x 2 -5y = -8 Y= 8/5

23. Solution Area (3/5,8/5) x = 0 x = 0 4x+y=4 2x+3y=6 y = 0 (0,2) (0,0)
(1,0)
y = 0

24. Titik pojok Nilai Z = 2x + 3y (0,0) 2(0) + 3(0) = 0 2(1) + 3(0) = 2
(1,0)
2(0) + 3(2) = 6
(0,2)
(3/5,8/5)
2(3/5) + 3(8/5) = 6
the maximum value is 6

25. Determine the minimum value Z = 4x + 3y at the system of linear inequalities
2x + y ≥ 6, x + 2y ≥ 6, 6x + y ≥12, x ≥ 0, y ≥ 0
2x + y ≥ 6
x + 2y ≥ 6 x y
(x,y) x y
(x,y) 3 6 6 3
(0,6)
(3,0)
(0,3)
(6,0) x y
(x,y) 2
6x + y ≥12 12
(2,0)
(0,12)

26. x = 0
6x + y = 12
(0,12)
2x + y = 6
Solution Area B A
(6,0)
x + 2y = 6

27. 2x + y = 6 x + 2y = 6 2x + y = 6 6x + y = 12 x 1 2x + y = 6-
-3y = -6
Y= 2
A(2,2)
X = 2
2x + y = 6
6x + y = 12 -
-4x = -6
x= 3/2
B(3/2,3)
y = 3

28. Titik pojok Nilai Z = 4x + 3y (6,0) 4(6) + 3(0) = 24 4(0) + 3(12) = 36
(0,12)
4(2) + 3(2) = 14
(2,2)
(3/2,3)
4(3/2) + 3(3) = 15
the minimum value is 14