Explanation Calculation: The given inequality set is { 2 x + y ≥ 12 x + y ≤ 8 2 x + y ≤ 14 x ≥ 0 , y ≥ 0 (1) The boundary equations of the above inequalities are { 2 x + y = 12 x + y = 8 2 x + y = 14 x = 0 , y = 0 (2) Sketch the graph of the straight lines 2 x + y = 12 , x + y = 8 , 2 x + y = 14 , x = 0 and y = 0 are as shown below in Figure 1. From Figure 1, it is observed that the equation (2) divides the plane into 6 regions. The solution region of the given inequality set is obtained as follows. Take any arbitrary test point in the region I, to test the solution region of the inequalities. Let ( 6 , 10 ) be the test point of the region I. Substitute the values x = 6 and y = 10 in equation (1), { 2 x + y ≥ 12 x + y ≤ 8 2 x + y ≤ 14 x ≥ 0 , y ≥ 0 = { 2 ( 6 ) + 10 ≥ 12 6 + 10 ≤ 8 2 ( 6 ) + 10 ≤ 14 6 ≥ 0 , 10 ≥ 0 On further simplification, the inequality set becomes { 32 ≥ 12 16 ≤ 8 22 ≤ 14 6 ≥ 0 , 10 ≥ 0 . Since the expression 16 ≤ 8 and 22 ≤ 14 are not true, the point ( 6 , 10 ) does not satisfy the equation (1). Thus, the region I is not a solution region of the inequalities. Take any arbitrary test point in the region II, to test the solution region of the inequalities. Let ( 4 , 6 ) be the test point of the region II. Substitute the values x = 4 and y = 6 in the equation (1), { 2 x + y ≥ 12 x + y ≤ 8 2 x + y ≤ 14 x ≥ 0 , y ≥ 0 = { 2 ( 4 ) + 6 ≥ 12 4 + 6 ≤ 8 2 ( 4 ) + 6 ≤ 14 4 ≥ 0 , 6 ≥ 0 The simplified form of the inequality set is { 14 ≥ 12 10 ≤ 8 14 ≤ 14 4 ≥ 0 , 6 ≥ 0 . Since the expression 10 ≤ 8 is not true, the point ( 4 , 6 ) does not satisfy the equation (1). Thus, the region II is not a solution region of the inequalities. Take any arbitrary test point in the region III, to test the solution region of the inequalities. Let ( 12 , 1 ) be the test point of the region III. Substitute the values x = 12 and y = 1 in the equation (1), { 2 x + y ≥ 12 x + y ≤ 8 2 x + y ≤ 14 x ≥ 0 , y ≥ 0 = { 2 ( 12 ) + 1 ≥ 12 12 + 1 ≤ 8 2 ( 12 ) + 1 ≤ 14 12 ≥ 0 , 1 ≥ 0 Thus, the inequalities become { 25 ≥ 12 13 ≤ 8 25 ≤ 14 12 ≥ 0 , 1 ≥ 0 . Since the expressions 13 ≤ 8 and 25 ≤ 14 are not true, the point ( 12 , 1 ) does not satisfy the equation (1)

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Chapter 8.1, Problem 16E

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To locate: The solution region of the inequality set {2x+y≥12x+y≤82x+y≤14x≥0,y≥0 from the given graph and identify it by finding the corner points.

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Chapter 8.1, Problem 15E

Chapter 8.1, Problem 17E

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Chapter 8 Solutions

EBK COLLEGE ALGEBRA IN CONTEXT

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Ch. 8.1 - Prob. 11E Ch. 8.1 - In Exercises 1116, the graph of the boundary...Ch. 8.1 - Prob. 13E Ch. 8.1 - Prob. 14E Ch. 8.1 - Prob. 15E Ch. 8.1 - Prob. 16E Ch. 8.1 - Prob. 17E Ch. 8.1 - Prob. 18E Ch. 8.1 - For each system of inequalities in Exercises 1722,...Ch. 8.1 - Prob. 20E Ch. 8.1 - Prob. 21E Ch. 8.1 - Prob. 22E Ch. 8.1 - Prob. 23E Ch. 8.1 - In Exercises 2326, graph the solution of the...Ch. 8.1 - Prob. 25E Ch. 8.1 - In Exercises 2326, graph the solution of the...Ch. 8.1 - Prob. 27E Ch. 8.1 - Prob. 28E Ch. 8.1 - Prob. 29E Ch. 8.1 - Sales Trix Auto Sales sells used cars. 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To locate: The solution region of the inequality set { 2 x + y ≥ 12 x + y ≤ 8 2 x + y ≤ 14 x ≥ 0 , y ≥ 0 from the given graph and identify it by finding the corner points.

Solution Summary: The author analyzes the solution region of the inequality set l2x+y's graph and determines it by finding the corner points.

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To locate: The solution region of the inequality set {2x+y≥12x+y≤82x+y≤14x≥0,y≥0 from the given graph and identify it by finding the corner points.

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Chapter 8 Solutions