linear programming
DESCRIPTION
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 PresentationTRANSCRIPT
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
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