linear programming
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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 PresentationTRANSCRIPT
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GRADE XII SCIENCE
IISYIISY
LINEAR PROGRAMMINGLINEAR PROGRAMMING
SutarmanSutarman20072007
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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
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Linear Inequalities in the Plane
Form of linear inequalities:
cbyax
cbyax or
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Solution of an Inequality (1)
The solution of an inequality consists of all points in the plane R2 that satisfy the given inequality.
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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.
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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.
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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
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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
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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
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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
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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
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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
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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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