( ) ( 200, 0 ) and 0, 200. Linear programming 6C. ( 40, 0) and ( 0, 80) ( 100.8,16.8) F = 168. ( 37.5, 22.5) m=97.5. Objective line passes through
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1 Linear programming 6C 1 a Need intersection of 4x+y= 1400 and 3x+2y= 1200 ( 320,120) m=760 b ( 0, 400) N = 1600 c Need intersection of x+ 3y= 1200 and 3x+2y= 1200 ( 171 3, 342 6) 7 7 P= d ( 350, 0) Q= 2100 ( 200, 0) and ( 0, 400) ( 400, 0) and ( 100, 0) ( ) ( 200, 0 ) and 0, 200 ( 100, 0) and ( 0, 600) 2 a ( 0, 90) E= 90 b Need intersection of 6y=x and 3x+ 7y=420 ( 100.8,16.8) F = 168 c Need intersection of 9x+10y= 900 3x+ 7y=420 ( 63 7, 32 8 ) G= d Same intersection as in b ( 100.8,16.8) H = ( 80, 0) and ( 0, 60) ( 120, 0) and ( 0, 20) 3 a Need intersection of 3x+y=60 and 5y= 3x ( 16 2,10 ) 3 J = b Need intersection of y=4x and 9x+5y=450 ( 15 15, 62 2 ) K = c Need intersection of 3x+y=60 and y=4x ( 8 4, 34 2) 7 7 L= d Need intersection of 9x+5y=450 and 5y= 3x ( 37.5, 22.5) m=97.5 ( 10, 0) and ( 0, 60) ( 40, 0) and ( 0, 80) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 1
2 4 a C b A c B d D e C f A g B h D i C j D 5 Let x be the mass of indoor feed and y be the mass of outdoor feed, in kilograms. Recall that we want to maximise P= 7x+ 6y, subject to x+ 2y 500, 2x+ y 500, x+ y 300, y 3 x, y 0 x 50, Draw the diagram including all constraints and mark the feasible region as R. (0, 350) and (300, 0). Maximum point is (200, 100). P max = 2000 Using the ruler method, we identify that optimal point is (200,100). At this pointp = 2000 Hence, we conclude that in order to maximise its profit, the company should produce 200kg of indoor feed and 100kg of outdoor feed. The profit will be Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 2
3 6 Decision variables: x= hours of work for factory R, y= hours of work for factory S Recall that we wish to minimise C= 300x+ 400y subject to: 5x+ 4y100 2x+ 3y60 2xy 2yx x, y0 Draw the diagram including all constraints and mark the feasible region as R. We can use the ruler method; in the picture we have drawn an objective line passing through points (0,15) and(20,0) This way, we identify the optimal point as the intersection of lines 5x+ 4y= 100 and 2x+ 3y= By solving simultaneous equations, we find the optimal point 8,14 and optimal value C= We conclude that in order to minimise operating cost, factory R should work for 8 h and factory S 7 for h Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 3
4 7 a b Vertices C= 3x+2y (0, 160) 320 (40, 80) 280 (90, 30) 330 (180, 0) 540 so minimum is (40, 80) value of C= 280 c (90, 30) C 1 = 270 d C 2 is parallel to x+y= 120 so all points from A to B are optimal points. Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 4
5 8 a 9 a b i ( 13 1, ) ii ( 34 2,17 1 ) P= Q= Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 5
6 9 b i We can identify 4 vertices of the feasible region as intersections of respective lines 5 12 y= 10 x,2y x= 100 A= 5, y= 10 x, 2x+ y= 400 B= 33, x= 120, 2x+ y= 400 C = (120,160) x= 120, 2y x= 100 D= (120,110) Now we apply the vertex testing method. Vertex Coordinates Value of z= 5x+ y A x= 5, y= B 1 1 x= 33, y= C x= 120, y= D x= 120, y= Maximal value of z in the feasible region is ii Minimal value of z in the feasible region is c We apply the vertex testing method to the points identified in part b. Vertex Coordinates Value of x+ 2y A x= 5, y= B 1 1 x= 33, y= C x= 120, y= D x= 120, y= Maximal value of x+ 2y is 700 Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 6
7 Challenge a Sketch the feasible region by noting the boundary line corresponding to 2 x + y 10 is a parabola. b Objective function is 3x+ y, so gradient of an objective line is 3 When we apply the ruler method we observe that at the optimal point parabola is tangent to the objective line. 2 For x + y= 10 we have d y 2x dx = 2x= 3 x= 1.5, y= 10 ( 1.5) 2 = 7.75 Maximal value of P= 3 (1.5) = Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 7
Linear programming 6D
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![Ch8: Straight Line Graph. x f(x) p 0.5 q 2 4 R 16. (a) Find the values of p, q and r. [3]](/thumbs/78/77995389.jpg "Ch8: Straight Line Graph. x f(x) p 0.5 q 2 4 R 16. (a) Find the values of p, q and r. [3]")
Ch8: Straight Line Graph. Answer the whole of this question on a sheet of graph paper. The table gives values of f() =, for 4. 0 3 4 f() p 0.5 q 4 R 6 (a) Find the values of p, q and r. [3] Using a scale
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