linear programming

14
GRADE XII SCIENCE IISY IISY LINEAR PROGRAMMING LINEAR PROGRAMMING Sutarman Sutarman 2007 2007

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IISY. IISY. LINEAR PROGRAMMING. GRADE XII SCIENCE. Sutarman. 2007. Linear Function. These are the examples of linear functions with two variables x and y :. 1. 1. 1. 1. Linear Inequalities in the Plane. Form of linear inequalities:. or. Solution of an Inequality (1). - PowerPoint PPT Presentation

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Page 1: LINEAR PROGRAMMING

GRADE XII SCIENCE

IISYIISY

LINEAR PROGRAMMINGLINEAR PROGRAMMING

SutarmanSutarman20072007

Page 2: LINEAR PROGRAMMING

These are the examples of linear functions These are the examples of linear functions with two variables with two variables xx and and yy : :

Linear FunctionLinear Function

yxyxf 32),(

yxyxf 4),(

1 1

1 1

Page 3: LINEAR PROGRAMMING

Linear Inequalities in the Plane

Form of linear inequalities:

cbyax

cbyax or

Page 4: LINEAR PROGRAMMING

Solution of an Inequality (1)

The solution of an inequality consists of all points in the plane R2 that satisfy the given inequality.

Page 5: LINEAR PROGRAMMING

Solution of an Inequality (2)

Steps to find the solution of inequality:Graph the line L : ax + by = cChoose a test point P(xo,yo) not on L and

substitute the point to the inequality.Suppose the test point P is a solution of the

inequality, that is, the derived statement is true. Then shade the side of L doesn’t contain P.

Suppose the test point P is not a solution of the inequality, that is, the derived statement is not true. Then shade the side of L that contains P.

Page 6: LINEAR PROGRAMMING

No

Yes

Test Point

SUBSTITUTION

START

TRUE ?

SHADE AREA DOESN’T CONSIST THE TEST POINT

SHADE AREA CONSISTS THE TEST POINT

END

Substitute the test point to the inequality.

Not shaded is the area of solutions.

Page 7: LINEAR PROGRAMMING

Example 1:Solve (graph) the inequality 4x

y

x0 4

Test point!

0 ≤ 4 is true x = 0 substitute to x ≤ 4

Shade half plane doesn’t consist the test point P(0,0) i.e. right side of the line L : x = 4. Left side is the area of solutions.

Answer :Draw line L : x = 4.Apply test point P (0,0).

x = 4

x = 0 is satisfy to x ≤ 4

The area of solutions

Page 8: LINEAR PROGRAMMING

Example 3:Solve (graph) the inequality 4y

y

x0

4

Test point!

0 ≤ 4 is true y = 0 substitute to y ≤ 4

Shade half plane doesn’t consist the test point P(0,0) i.e. upper side of line L : y = 4. Lower side is the area of solutions.

Answer :Draw line L : y = 4.Apply test point P (0,0).

y = 4

y = 0 is satisfy to y ≤ 4

The area of solutions

Page 9: LINEAR PROGRAMMING

3

Example 5:Solve (graph) the inequality 2x + 3y ≤ 6.

y

x0

2

Test point!

2(0)+3(0) ≤ 6 0 ≤ 6 is true

x=0, y = 0 substitute to 2x+3y ≤ 6

Shade half plane doesn’t consist the test point P(0,0) i.e. upper side of line L : 2x+3y = 6. Lower side is the area of solutions.

Answer :Draw line L : 2x+3y = 6.Apply test point P (0,0).

2x+3y=6

x=0, y = 0 are satisfies to 2x+3y ≤ 6

The area of solutions

xy

0

2

3

0

Page 10: LINEAR PROGRAMMING

3x+8y=7224

Example 8:Solve (graph) the inequality 3x + 8y ≥ 72.

y

x0

9

Test point!

3(0) + 8(0) ≥ 72 0 ≥ 72 is not true

x=0, y = 0 substitute to 3x+8y ≥ 72

Shade half plane consists the test point P(0,0) i.e. lower side of line L : 3x+8y = 72. Upper side is the area of solutions.

Answer :Draw line L : 3x+8y = 72.Apply test point P (0,0).

x=0, y = 0 are not satisfies to 3x+8y ≥ 72

The area of solutions

xy

(0,9) (24,0)

0

9

24

0

Page 11: LINEAR PROGRAMMING

Draw line L : x = 0, i.e. y-axis.

Example 9:Solve (graph) the inequality 0x

y

x0

(1,1)

Test point!

1 ≥ 0 is true x = 1 substitute to x ≥ 0

Shade half plane doesn’t consist the test point P(1,1) i.e. left side of line L : x = 0. Right side is the area of solutions.

Answer :

Apply test point P (1,1).

x = 0

x = 1 is satisfy to x ≥ 0

The area of solutions

Page 12: LINEAR PROGRAMMING

Draw line L : y = 0, i.e. x-axis.

Example 10:Solve (graph) the inequality 0y

y

x0

(1,1)

Test point!

1 ≥ 0 is true y = 1 substitute to y ≥ 0

Shade half plane doesn’t consist the test point P(1,1) i.e. lower side of line L : y = 0. Upper side is the area of solutions.

Answer :

Apply test point P (1,1).

y = 0y = 1 is satisfy to y ≥ 0

The area of solutions

Page 13: LINEAR PROGRAMMING

2(0)+3(0) ≤ 12 0 ≤ 12 is true

Shade half plane doesn’t consist the test point P(0,0) i.e. upper side of line L : 2x+y = 8. Lower side is the area of solutions.

2(0)+0 ≤ 8 0 ≤ 8 is true

x=0, y = 0 substitute to 2x+y ≤ 8

y = 1 is satisfy to y ≥ 0

y = 1 substitute to y ≥ 0

Shade half plane doesn’t consist the test point P(1,1) i.e. left side of line L : x = 0. Right side is the area of solutions.

x=0, y = 0 are satisfies to 2x+y ≤ 8

Shade half plane doesn’t consist the test point P(1,1) i.e. lower side of line L : y = 0. Upper side is the area of solutions.Shade half plane doesn’t consist the test point P(0,0) i.e. upper side of line L : 2x+3y = 12. Lower side is the area of solutions.

6

Example 12:Solve (graph) the inequalities 2x + 3y ≤ 12, 2x + y ≤ 8, x ≥0, y ≥ 0,

y

x0

4

Test point!

x=0, y = 0 substitute to 2x+3y ≤ 12

Answer :Draw line L : 2x+3y = 12.Apply test point P (0,0).

2x+3y=12

x=0, y = 0 are satisfies to 2x+3y ≤ 12

The area of solutions

xy

(0,4) (6,0)

0

4

6

0

4

8Test point!

Draw line L : 2x+y = 8.Apply test point P (0,0).

2x+y=8

The area of solutions

xy

(0,8) (4,0)

0

8

4

0Draw line L : x = 0, i.e. y-axis.

(1,1)

Test point!

1 ≥ 0 is true x = 1 substitute to x ≥ 0

Apply test point P (1,1).x = 0

x = 1 is satisfy to x ≥ 0The area of

solutions

Draw line L : y = 0, i.e. x-axis.

(1,1)

Test point!

1 ≥ 0 is true

Apply test point P (1,1).

y = 0The area of

solutions

Page 14: LINEAR PROGRAMMING

Bye bye …………

n Thank You