    Question

# Solve the following systems of inequations graphically: (i) 2x+y≥8,x+2y≥8,x+y≤6 (ii) 12x+12y≤840,3x+6y≤300,8x+4y≤480 x≥0,y≥0 (iii) x+2y≤40,3x+y≥30,4x+3y≥60,x≥0,y≥0 (iv) 5x+y≥10,2x+2y≥12,x+4y≥12,x≥0,y≥0

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## (i) We have, 2x+y≥8,x+2y≥8,x+y≤6 Converting the given inequation into equations,we obtain, 2x+y=8,x+2y=8,and x+y=6 Region represented by 2x+y≥8 Putting x=0 in2x+y=8,we get y=8 Putting y=0 in 2x+y=8,we get x=82=4 ∴ The line 2x+y=8 meets the coordinates axes at (0,8) and (4,0).Join these points by a thick line. Now ,putting x=0 and y=0 in 2x+y≥8,we get 0≥8 This is not possible. ∴ We find that (0,0) is not satisifes the inequation 2x+y≥8 So,the portion not containing the origin is represented by the given inequation Region represented by x+2y≥8 Putting x=0 in x+2y=8,we get y=82=4 Putting y=0 in x+2y=8,we get x=8 ∴ The line x+2y=8 meets the coordinates axes at (0,4) and (8,0).Joining these points by a thick line. Now ,putting x=0 and y=0 in x+2y≥8,we get 0≥8 This is not possible. ∴ We find that (0,0) is not satisifes the inequation x+2y≥8 So,the portion not containing the origin is represented by the given inequation. Region represented by x+y≤6 Putting x=0 in x+y=6,we get y=6 Putting y=0 in x+y=6,we get x=6 ∴ The line x+y=6 meets the coordinates axes at (0,6) and (6,0). Joining these points by a thick line. Now ,putting x=0 and y=0 in x+y≤6,we get 0≥6 Therefore,(0,0) satisfies x+y≤6,so the portion containing the origin is represented by the given inequation.The common region of the above three regions represents the solution set of the given inequations as shown below: (ii) We have, 12x+12y≤840,3x+6y≤300,8x+4y≤480 x≥0,y≥0 Converting the given inequation into equations,we obtain, 12x+12y= 840, 3x+6y= 300,8x+4y=480, x = 0 and y=0 Region represented by 12x+12y≤840 Putting x=0 in 12x+12y= 840,we get y=84012=70 Putting y=0 in by 12x+12y≤840,we get x=84012=70 ∴ The line 12x+12y= 840meets the coordinates axes at (0,70) and (70,0).Join these points by a thick line., Now ,putting x=0 and y=0 in 12x+12y≤840,we get 0≤840 Therefore,(0,0) satisfies the inequality 12x+12y≤840, So,the portion not containing the origin is represented by the given inequation 12x+12y≤840 Region represented by 3x+6y≤300 Putting x=0 in 3x+6y≤300,we get y=3006=50 Putting y=0 in x=3003=100 ∴ The line 3x+6y= 300meets the coordinates axes at (0,50) and (100,0).Joining these points by a thick line. Now ,putting x=0 and y=0 in 3x+6y≤300,we get 0≤300 Therefore ,(0,0) satisfies the inequality 8x+4y≤480 Putting x=0 in 8x+4y=480,we get, y=4804=120 Putting y=0 in 8x+4y=480,we get, x=4808=60 ∴ The line 8x+4y=480 meets the coordinates axes at (0,120) and (60,0).Join these points by a thick line. Now ,putting x=0 and y=0 in 8x+4y=480,we get 0≤480 Therefore,(0,0) satisfies the inequality 8x+4y=480 So, the portion containing the origin is represented the solution set of the inequation 8x+4y=480 Region represented by x≥0 and y≥0 Clearly x≥0 and y≥0 represents the first quadrant. The common region of the above five regions represents the solution set of the given inequations as shown below: (iii) We have, x+2y≤40,3x+y≥30,4x+3y≥60,x≥0,y≥0 Converting the given inequation into equations,we obtain, x+2y=40,3x+y=30,4x+3y=60,x=0 and y=0 Region represented by x+2y≤40 Putting x=0 in x+2y=40,we get y=402=20 Putting y=0 in by x+2y=40,we get x=40 ∴ The line x+2y=40,meets the coordinates axes at (0,20) and (40,0).Join these points by a thick line. Now ,putting x=0 and y=0 in x+2y≤40,we get 0≤40 therefore,(0,0) satisfies the inequality x+2y≤40, So,the portion not containing the origin is represented by the given inequation x+2y≤40 Region represented by 3x+y≥30 Putting x=0 in 3x+y=30,we get y=30 Putting y=0 in 3x+y=30,we get x=303=10 ∴ The line 3x+y=30,meets the coordinates axes at (0,30) and (10,0).Join these points by a thick line. Now,putting x=0 and y=0 in 3x+y≥30,we get 0≥30 This is not possible. Therefore ,(0,0) satisfies the inequality 3x+y≥30. So,the portion not containing the origin is represented by the given inequation 3x+y≥30. Region represented by 4x+3y≥60 Putting x=0 in 4x+3y= 60,we get, y=603=20 Putting y=0 in 4x+3y= 60,we get, x=604=15 ∴ The line 4x+3y= 60,meets the coordinates axes at (0,20) and (15,0).Join these points by a thick line. Now ,putting x=0 and y=0 in 4x+3y≥60 This is not possible. Therefore,(0,0) satisfies the inequality 4x+3y≥60 So, the portion containing the origin is represented the solution set of the inequation \4x+3y≥60. Region represented by x≥0 and y≥0 Clearly x≥0 and y≥0 represents the first quadrant. The common region of the above five regions represents the solution set of the given inequations as shown below: (iv) We have, 5x+y≥10,2x+2y≥12,x+4y≥12,x≥0,y≥0 Converting the given inequation into equations,we obtain, 5x+y=10,2x+2y=1,x+4y=12,x=0 and y=0 Region represented by 5x+y≥10 Putting x=0 in 5x+y=10,we get y=10 Putting y=0 in by 5x+y=10,we get x=105=2 ∴ The line 5x+y=10,meets the coordinates axes at (0,10) and (2,0).Join these points by a thick line. Now ,putting x=0 and y=0 in 5x+y≥10,we get 0≥10 therefore,(0,0) satisfies the inequality 5x+y≥10, So,the portion not containing the origin is represented by the given inequation 5x+y≥10 Region represented by 2x+2y≥12 Putting x=0 in 2x+2y=12,we get y=122=6 Putting y=0 in 2x+2y=12,we get x=122=6 ∴ The line 2x+2y=12,meets the coordinates axes at (0,6) and (6,0).Join these points by a thick line. Now,putting x=0 and y=0 in 2x+2y=12,we get 0≥12,which is not possible. Therefore ,(0,0) satisfies the inequality 2x+2y=12. So,the portion not containing the origin is represented by the given inequation 2x+2y=12. Region represented by x+4y≥12 Putting x=0 in x+4y= 12,we get, y=124=3 Putting y=0 in x+4y= 12,we get, x=12 ∴ The line x+4y= 12,meets the coordinates axes at (0,3) and (12,0).Join these points by a thick line. Now ,putting x=0 and y=0 in x+4y= 12,which is not possible. Therefore,(0,0) satisfies the inequality x+4y≥12 So, the portion containing the origin is represented the solution set of the inequation x+4y≥12 Region represented by x≥0 and y≥0 Clearly x≥0 and y≥0 represents the first quadrant. The common region of the above five regions represents the solution set of the given inequations as shown below:   Suggest Corrections  0      Related Videos   Graphical Method of Solving LPP
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