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

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

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

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)

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

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

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

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

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

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

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)

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

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

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)

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

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

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

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

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)

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

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

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

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

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

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)

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

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

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