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To Draw: The regions in the first quadrant of x 1 x 2 -plane. The given inequalities are { x 1 − 3 x 2 ≥ − 6 x 1 + x 2 ≤ 6 . For the graph, the inequality sign is converted into equality as shown below: x 1 − 3 x 2 = − 6 x 1 + x 2 = 6 Substitute the value x 1 = 0 in the equation x 1 − 3 x 2 = − 6 . 0 − 3 x 2 = − 6 3 x 2 = 6 x 2 = 2 It means that equation x 1 − 3 x 2 = − 6 cut the x 2 axis at x 2 = 2 . Substitute the value x 2 = 0 in the equation x 1 − 3 x 2 = − 6 . x 1 − 3 ( 0 ) = − 6 x 1 − 0 = − 6 x 1 = − 6 It means that equation x 1 − 3 x 2 = − 6 cut the x 1 axis at x 1 = − 6 . Substitute ( 0 , 0 ) in the x 1 − 3 x 2 ≥ − 6 to get, ( 0 ) − 3 ( 0 ) ≥ − 6 0 ≥ − 6 which is true. The region defined by the inequality contains the origin. Use an online graphing calculator to sketch the region as shown in below in figure 1. From the Figure 1, it is observed that origin lies above the line x 1 − 3 x 2 = − 6 . Substitute the value x 1 = 0 in the equation x 1 + x 2 = 6 . 0 + x 2 = 6 x 2 = 6 Substitute the value x 2 = 0 in the equation x 1 + x 2 = 6 . x 1 + 0 = 6 x 1 = 6 It means that equation x 1 + x 2 = 6 cut the axis x 2 ,and x 1 axis at ( 0 , 6 ) , and ( 6 , 0 ) respectively. Substitute ( 0 , 0 ) in the x 1 + x 2 ≤ 6 to get, 0 + 0 ≤ 6 0 ≤ 6 which is true. The region defined by the inequality contains the origin. Use an online graphing calculator to sketch the region as shown in below in figure 1. From the Figure 2, it is observed that origin lies below the line x 1 + x 2 = 6 . Use an online graphing calculator to sketch the region as shown in the below figure. From the Figure 3, it is observed that the shaded region is bounded by points ( 0 , 0 ) , ( 0 , 2 ) , ( 3 , 3 ) and ( 6 , 0 ) .

To Draw: The regions in the first quadrant of x 1 x 2 -plane. The given inequalities are { x 1 − 3 x 2 ≥ − 6 x 1 + x 2 ≤ 6 . For the graph, the inequality sign is converted into equality as shown below: x 1 − 3 x 2 = − 6 x 1 + x 2 = 6 Substitute the value x 1 = 0 in the equation x 1 − 3 x 2 = − 6 . 0 − 3 x 2 = − 6 3 x 2 = 6 x 2 = 2 It means that equation x 1 − 3 x 2 = − 6 cut the x 2 axis at x 2 = 2 . Substitute the value x 2 = 0 in the equation x 1 − 3 x 2 = − 6 . x 1 − 3 ( 0 ) = − 6 x 1 − 0 = − 6 x 1 = − 6 It means that equation x 1 − 3 x 2 = − 6 cut the x 1 axis at x 1 = − 6 . Substitute ( 0 , 0 ) in the x 1 − 3 x 2 ≥ − 6 to get, ( 0 ) − 3 ( 0 ) ≥ − 6 0 ≥ − 6 which is true. The region defined by the inequality contains the origin. Use an online graphing calculator to sketch the region as shown in below in figure 1. From the Figure 1, it is observed that origin lies above the line x 1 − 3 x 2 = − 6 . Substitute the value x 1 = 0 in the equation x 1 + x 2 = 6 . 0 + x 2 = 6 x 2 = 6 Substitute the value x 2 = 0 in the equation x 1 + x 2 = 6 . x 1 + 0 = 6 x 1 = 6 It means that equation x 1 + x 2 = 6 cut the axis x 2 ,and x 1 axis at ( 0 , 6 ) , and ( 6 , 0 ) respectively. Substitute ( 0 , 0 ) in the x 1 + x 2 ≤ 6 to get, 0 + 0 ≤ 6 0 ≤ 6 which is true. The region defined by the inequality contains the origin. Use an online graphing calculator to sketch the region as shown in below in figure 1. From the Figure 2, it is observed that origin lies below the line x 1 + x 2 = 6 . Use an online graphing calculator to sketch the region as shown in the below figure. From the Figure 3, it is observed that the shaded region is bounded by points ( 0 , 0 ) , ( 0 , 2 ) , ( 3 , 3 ) and ( 6 , 0 ) .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 22.1, Problem 10P

Chapter 22.2, Problem 2P

Chapter 22 Solutions

Advanced Engineering Mathematics

Ch. 22.1 - Prob. 1P Ch. 22.1 - Prob. 2P Ch. 22.1 - Prob. 3P Ch. 22.1 - Prob. 4P Ch. 22.1 - Prob. 5P Ch. 22.1 - Prob. 6P Ch. 22.1 - Prob. 7P Ch. 22.1 - Prob. 8P Ch. 22.1 - Prob. 9P Ch. 22.1 - CAS EXPERIMENT. Steepest Descent. (a) Write a...

Ch. 22.2 - Prob. 1P Ch. 22.2 - Prob. 2P Ch. 22.2 - Prob. 3P Ch. 22.2 - Prob. 4P Ch. 22.2 - Prob. 5P Ch. 22.2 - Prob. 6P Ch. 22.2 - Prob. 8P Ch. 22.2 - Prob. 9P Ch. 22.2 - Prob. 10P Ch. 22.2 - Prob. 11P Ch. 22.2 - Prob. 12P Ch. 22.2 - Prob. 13P Ch. 22.2 - Prob. 14P Ch. 22.2 - Prob. 15P Ch. 22.2 - Prob. 16P Ch. 22.2 - Prob. 17P Ch. 22.2 - Prob. 18P Ch. 22.2 - Prob. 19P Ch. 22.2 - Prob. 20P Ch. 22.2 - Prob. 21P Ch. 22.2 - Prob. 22P Ch. 22.3 - Prob. 1P Ch. 22.3 - Prob. 2P Ch. 22.3 - Prob. 3P Ch. 22.3 - Prob. 4P Ch. 22.3 - Prob. 5P Ch. 22.3 - Prob. 6P Ch. 22.3 - Prob. 7P Ch. 22.3 - Prob. 8P Ch. 22.3 - Prob. 9P Ch. 22.3 - Prob. 10P Ch. 22.3 - Prob. 11P Ch. 22.3 - Prob. 12P Ch. 22.3 - Prob. 13P Ch. 22.3 - Prob. 14P Ch. 22.4 - Prob. 1P Ch. 22.4 - Prob. 2P Ch. 22.4 - Prob. 3P Ch. 22.4 - Prob. 4P Ch. 22.4 - Prob. 6P Ch. 22.4 - Prob. 7P Ch. 22.4 - Prob. 8P Ch. 22.4 - Prob. 9P Ch. 22 - Prob. 1RQ Ch. 22 - Prob. 2RQ Ch. 22 - Prob. 3RQ Ch. 22 - Prob. 4RQ Ch. 22 - Prob. 5RQ Ch. 22 - Prob. 7RQ Ch. 22 - Prob. 8RQ Ch. 22 - Prob. 9RQ Ch. 22 - Prob. 10RQ Ch. 22 - Prob. 11RQ Ch. 22 - Prob. 13RQ Ch. 22 - Prob. 14RQ Ch. 22 - Prob. 15RQ Ch. 22 - Prob. 16RQ Ch. 22 - Prob. 17RQ Ch. 22 - Prob. 18RQ Ch. 22 - Prob. 19RQ Ch. 22 - Prob. 20RQ

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