GRADE XII SCIENCE

IISYIISY

LINEAR PROGRAMMINGLINEAR PROGRAMMING

SutarmanSutarman20072007

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

Linear Inequalities in the Plane

Form of linear inequalities:

cbyax

cbyax or

Solution of an Inequality (1)

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

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.

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.

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

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

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

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

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

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

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

Bye bye …………

n Thank You