MTH-5101 Practice Test (1) x ≥ 40 (2) y ≥ 0 (3) y ≤ 140

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Presentation transcript:

MTH-5101 Practice Test (1) x ≥ 40 (2) y ≥ 0 (3) y ≤ 140 Name: ________________________ Construct the polygon of constraints for the following linear inequalities. Determine the exact ordered pairs for the critical points of the polygon of constraints. (Use the grids on the next page to construct graphs.) (1) x ≥ 40 (2) y ≥ 0 (4) x - y - 30 ≤ 0 (3) y ≤ 140 (5) x + y ≤ 200 (6) 5x + 4y ≥ 600

Construct the polygon of constraints for the following linear inequalities. (1) x ≥ 40 50 150 100 200 (2) y ≥ 0 (3) y ≤ 140 (4) x - y - 30 ≤ 0 (5) x + y ≤ 200 (6) 5x + 4y ≥ 600

x - y - 30 ≤ 0 -y = -x + 30 y = x - 30 x - y - 30 ≤ 0 Test (0,0) 50 150 100 200 x - y - 30 ≤ 0 -y = -x + 30 y = x - 30 x y 30 100 70 x - y - 30 ≤ 0 Test (0,0) 0 - 0 - 30 ≤ 0 -30 ≤ 0 True x + y ≤ 200 y = -x + 200 x y 200 5x + 4y ≥ 600 4y = -5x + 600 5x + 4y ≥ 600 Test (0,0) 5(0) + 4(0) ≥ 600 0 ≥ 600 False x + y ≤ 200 Test (0,0) 0 + 0 ≤ 200 true x y 150 120

Construct the polygon of constraints for the following linear inequalities. (1) x ≥ 40 (2) y ≥ 0 (3) y ≤ 140 (4) x - y - 30 ≤ 0 (5) x + y ≤ 200 (6) 5x + 4y ≥ 600

Construct the polygon of constraints for the following linear inequalities. (1) x = 40 (6) 5x + 4y = 600 Point A is defined by the intersection of the line having the equation x = 40 and the line having the equation 5x + 4y = 600. A (40,100) One of these equations (1) has only one variable. This means that one of the variables of the point of intersection is defined: x = 40. (40, ____ ) The other variable y can be determined by substituting the given x-value into the equation containing 2 variables. 5x + 4y = 600 5(40) + 4y = 600 4y = 600 – 200 4y = 400 y = 100 (40,100)

Construct the polygon of constraints for the following linear inequalities. (1) x = 40 (3) y = 140 Point B is defined by the intersection of the line having the equation x = 40 and the line having the equation y = 140. B (40,140) Both of these equations have only one variable. This means that both of the variables of the point of intersection is defined: x = 40; y = 140. (40, 140)

Construct the polygon of constraints for the following linear inequalities. (5) x + y = 200 Point C is defined by the intersection of the line having the equation y = 140 and the line having the equation x + y = 200. C (60,140) One of these equations (3) has only one variable. This means that one of the variables of the point of intersection is defined: y = 140. ( ____, 140) The other variable x can be determined by substituting the given y-value into the equation containing 2 variables. x + y = 200 x + 140 = 200 x = 200 – 140 x = 60 (60,140)

Construct the polygon of constraints for the following linear inequalities. (4) x - y - 30 = 0 (5) x + y = 200 Point D is defined by the intersection of the line having the equation x - y – 30 = 0 and the line having the equation x + y = 200. Both of these equations have 2 variables. To solve this either comparison, substitution or elimination can be used. D (115,85) (4) x - y = 30 (5) x + y = 200 2x = 230 x = 115 x + y = 200 115 + y = 200 y = 200 – 115 y = 85 (115,85)

Construct the polygon of constraints for the following linear inequalities. (4) x - y - 30 = 0 (6) 5x + 4y = 600 Point E is defined by the intersection of the line having the equation x - y – 30 = 0 and the line having the equation 5x + 4y = 600. Both of these equations have 2 variables. To solve this either comparison, substitution or elimination can be used. E (80,50) (4) (x - y = 30) х 4 (6) 5x + 4y = 600 4x - 4y = 120 9x = 720 x = 80 5x + 4y = 600 5(80) + 4y = 600 4y = 600 – 400 4y = 200 y = 50 (80,50)