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Example 2 Sketch the feasible region for the following set of constraints: 3y - 2x ≥ 0 y + 8x ≤ 52 y - 2x≤2 x ≥ 3. Then find the maximum and minimum values of the objective function z = 5x + 2y.

Example 2 Sketch the feasible region for the following set of constraints: 3y - 2x ≥ 0 y + 8x ≤ 52 y - 2x≤2 x ≥ 3. Then find the maximum and minimum values of the objective function z = 5x + 2y.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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m 18 d Example 2 A y - 2x = 2 x = 3 Then find the maximum and minimum values of the objective function z = 5x + 2y. Solution Graph the feasible region, as in Figure 7.19. To find the corner points, you must solve these four systems of equations: 10 5 y x=3 A ( The first two systems are easily solved by substitution, which shows that A = (3, 8) and B = (3,2). The other two systems can be solved either graphically (as in Figure 7.20) or algebraically (see Checkpoint 3). Hence, C = (6, 4) and D = (5, 12). B Sketch the feasible region for the following set of constraints: 3y - 2x ≥ 0 y + 8x ≤ 52 y - 2x ≤ 2 x ≥ 3. B 3y - 2x = 0 x = 3 y-2x = 2 Page 349 y+ 8x = 52 Figure 7.19 3y - 2x = 0 Corner Point (3,8) (3,2) 10 C 3y - 2x = 0 y + 8x = 52 -1 5(3) + 2(8) = 31 5(3) + 2(2) = 19 349 15 ធ D y-2x = 2 y + 8x = 52 Intersection X=6 -5 Value of z = 5x + 2y Use the corner points from the graph to find the maximum and minimum values of the objective function. Y=4 Figure 7.20 EG - 10 (minimum)
Transcribed Image Text:m 18 d Example 2 A y - 2x = 2 x = 3 Then find the maximum and minimum values of the objective function z = 5x + 2y. Solution Graph the feasible region, as in Figure 7.19. To find the corner points, you must solve these four systems of equations: 10 5 y x=3 A ( The first two systems are easily solved by substitution, which shows that A = (3, 8) and B = (3,2). The other two systems can be solved either graphically (as in Figure 7.20) or algebraically (see Checkpoint 3). Hence, C = (6, 4) and D = (5, 12). B Sketch the feasible region for the following set of constraints: 3y - 2x ≥ 0 y + 8x ≤ 52 y - 2x ≤ 2 x ≥ 3. B 3y - 2x = 0 x = 3 y-2x = 2 Page 349 y+ 8x = 52 Figure 7.19 3y - 2x = 0 Corner Point (3,8) (3,2) 10 C 3y - 2x = 0 y + 8x = 52 -1 5(3) + 2(8) = 31 5(3) + 2(2) = 19 349 15 ធ D y-2x = 2 y + 8x = 52 Intersection X=6 -5 Value of z = 5x + 2y Use the corner points from the graph to find the maximum and minimum values of the objective function. Y=4 Figure 7.20 EG - 10 (minimum)
nts in Page 348 This theorem simplifies the job of finding an optimum value. First, graph the feasible region and find all corner points. Then test each point in the objective function. Finally, identify the corner point producing the optimum solution. With the theorem, the problem in Example 1 could have been solved by identifying the five corner points of Figure 7.16: (0, 1), (0, 2), (.5, 2.25), (2, 0), and (1, 0). Then, sub- stituting each of these points into the objective function z = 2x + 5y would identify the corner points that produce the maximum and minimum values of z. Corner Point (0, 1) (0, 2) (.5, 2.25) (2,0) (1, 0) 2(0)+ 5(1) = 5 2(0)+ 5(2) = 10 2(.5) + 5(2.25) = 12.25 2(2) + 5(0) = 4 2(1) + 5(0) = 2 Value of z = 2x + 5y 348 ■ (maximum) From these results, the corner point (.5, 2.25) yields the maximum value of 12.25 and the cor- ner point (1, 0) gives the minimum value of 2. These are the same values found earlier. A summary of the steps for solving a linear programming problem by the graphical method is given here. (minimum)
Transcribed Image Text:nts in Page 348 This theorem simplifies the job of finding an optimum value. First, graph the feasible region and find all corner points. Then test each point in the objective function. Finally, identify the corner point producing the optimum solution. With the theorem, the problem in Example 1 could have been solved by identifying the five corner points of Figure 7.16: (0, 1), (0, 2), (.5, 2.25), (2, 0), and (1, 0). Then, sub- stituting each of these points into the objective function z = 2x + 5y would identify the corner points that produce the maximum and minimum values of z. Corner Point (0, 1) (0, 2) (.5, 2.25) (2,0) (1, 0) 2(0)+ 5(1) = 5 2(0)+ 5(2) = 10 2(.5) + 5(2.25) = 12.25 2(2) + 5(0) = 4 2(1) + 5(0) = 2 Value of z = 2x + 5y 348 ■ (maximum) From these results, the corner point (.5, 2.25) yields the maximum value of 12.25 and the cor- ner point (1, 0) gives the minimum value of 2. These are the same values found earlier. A summary of the steps for solving a linear programming problem by the graphical method is given here. (minimum)
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