1 / 14

LINEAR PROGRAMMING

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).

teryl
Download Presentation

LINEAR PROGRAMMING

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. IISY IISY LINEAR PROGRAMMING GRADE XII SCIENCE Sutarman 2007

  2. Linear Function • These are the examples of linear functions with two variables x and y : 1 1 1 1

  3. Linear Inequalities in the Plane • Form of linear inequalities: or

  4. Solution of an Inequality (1) • The solution of an inequality consists of all points in the plane R2 that satisfy the given inequality.

  5. Solution of an Inequality (2) • Steps to find the solution of inequality: • Graph the line L : ax+ by = c • Choose 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 Lthat contains P.

  6. Test Point START Substitute the test point to the inequality. SUBSTITUTION TRUE ? SHADE AREA CONSISTS THE TEST POINT No Yes SHADE AREA DOESN’T CONSIST THE TEST POINT END Not shaded is the area of solutions.

  7. Example 1: Solve (graph) the inequality Answer : Draw line L : x = 4. y x = 4 Apply test point P (0,0). Test point! The area of solutions x = 0  substitute to x≤ 4 0 ≤ 4 is true x = 0 is satisfy tox≤ 4 x 0 4 Shade half plane doesn’t consist the test point P(0,0) i.e. right side of the lineL : x= 4. Left side is the area of solutions.

  8. Example 3: Solve (graph) the inequality Answer : Draw line L : y = 4. y Apply test point P (0,0). Test point! y = 4 4 The area of solutions y = 0  substitute to y≤ 4 0 ≤ 4 is true y = 0 is satisfy toy≤ 4 x 0 Shade half plane doesn’t consist the test point P(0,0) i.e. upper side of lineL : y= 4. Lower side is the area of solutions.

  9. Example 5: Solve (graph) the inequality 2x + 3y≤ 6. Answer : 0 3 Draw line L : 2x+3y = 6. y 0 2 Apply test point P (0,0). Test point! x=0, y = 0  substitute to 2x+3y ≤ 6 2 2(0)+3(0) ≤ 6 0 ≤ 6 is true x The area of solutions x=0, y = 0 are satisfies to2x+3y≤6 3 0 2x+3y=6 Shade half plane doesn’t consist the test point P(0,0) i.e. upper side of lineL : 2x+3y = 6. Lower side is the area of solutions.

  10. Example 8: Solve (graph) the inequality 3x + 8y≥ 72. Answer : 0 24 Draw line L : 3x+8y = 72. y 0 9 Apply test point P (0,0). (0,9) (24,0) Test point! x=0, y = 0  substitute to 3x+8y ≥ 72 The area of solutions 9 3(0) + 8(0) ≥ 72 0 ≥ 72 is not true x x=0, y = 0 are not satisfies to3x+8y≥ 72 0 24 3x+8y=72 Shade half plane consists the test point P(0,0) i.e. lower side of lineL : 3x+8y = 72. Upper side is the area of solutions.

  11. Example 9: Solve (graph) the inequality Answer : Draw line L : x = 0, i.e. y-axis. x = 0 y Apply test pointP (1,1). Test point! The area of solutions x = 1  substitute to x≥ 0 1 ≥ 0 is true (1,1) x = 1 is satisfy tox≥ 0 x 0 Shade half plane doesn’t consist the test point P(1,1) i.e. left side of lineL : x= 0. Right side is the area of solutions.

  12. Example 10: Solve (graph) the inequality Answer : Draw line L : y = 0, i.e. x-axis. y Apply test point P (1,1). The area of solutions Test point! y = 1  substitute to y≥ 0 (1,1) y = 0 1 ≥ 0 is true x y = 1 is satisfy toy≥0 0 Shade half plane doesn’t consist the test point P(1,1) i.e. lower side of lineL : y= 0. Upper side is the area of solutions.

  13. Example 12: Solve (graph) the inequalities 2x+ 3y≤ 12, 2x + y ≤ 8, x≥0, y≥ 0, Answer : 0 0 6 4 Draw line L : y = 0, i.e. x-axis. x = 0 Draw line L : x = 0, i.e. y-axis. Draw line L : 2x+3y = 12. Draw line L : 2x+y = 8. y 0 0 4 8 Apply test point P (1,1). Apply test pointP (1,1). Apply test point P (0,0). Apply test point P (0,0). (6,0) (4,0) (0,4) (0,8) Test point! Test point! Test point! 8 Test point! x=0, y = 0  substitute to 2x+3y ≤ 12 x=0, y = 0  substitute to 2x+y ≤ 8 x = 1  substitute to x≥ 0 y = 1  substitute to y≥ 0 1 ≥ 0 is true 1 ≥ 0 is true 2(0)+3(0) ≤ 12 0 ≤ 12 is true 2(0)+0 ≤ 8 0 ≤ 8 is true 4 y = 1 is satisfy toy≥ 0 x=0, y = 0 are satisfies to2x+3y≤12 x = 1 is satisfy tox≥ 0 x=0, y = 0 are satisfies to2x+y≤ 8 The area of solutions The area of solutions The area of solutions The area of solutions (1,1) y = 0 (1,1) x 0 6 4 2x+y=8 2x+3y=12 Shade half plane doesn’t consist the test point P(1,1) i.e. lower side of lineL : 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 lineL : 2x+3y = 12. Lower side is the area of solutions. Shade half plane doesn’t consist the test point P(0,0) i.e. upper side of lineL : 2x+y = 8. Lower side is the area of solutions. Shade half plane doesn’t consist the test point P(1,1) i.e. left side of lineL : x= 0. Right side is the area of solutions.

  14. Bye bye ………… n Thank You

More Related