1. Answer to set 1 Graphical Linear Programming

2. (1) 4X1 + 3X2 >96 X1X2 096/3=32 96/4=240

3. X1 X2 0,32 24,0 (1)

4. (2) X1 + X2 <50 X1X2 050 0

5. X1 X2 0,32 24,0 0,50 50,0 (1) (2)

6. (3) X1 +3X2 >60 X1X2 060/3=20 600

7. X1 X2 0,32 24,0 0,50 50,0 (1) (2) 0,20 60,0 (3)

8. Corner Point: (1),(3) (1) 4X1 + 3X2 = 96 (3) X1 + 3X2 = 60 (1)-(3) 3X1 = 36 X1 = 12 (3) 12 + 3X2 = 60, so X2 = 16

9. Corner Point: (2),(3) (2) X1 + X2 = 50 (3) X1 + 3X2 = 60 (3)-(2) 2X2= 10 X2 = 5 (2) X1 + 5 = 50 X2 = 45

10. X1 X2 0,32 24,0 0,50 50,0 (1) (2) 0,20 60,0 (3) Corner points Not feasible Not feasible 12,16 45,5

11. MINIMUM COST X1X22X1+7X2 1216136 455125=MIN 032224 050350

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