Explanation Given: The constraints are as follows: x ≥ 0 y ≥ 0 x + y ≤ 50 2 x + y ≥ 20 x + 2 y ≥ 30 The objective function is Z = 4 x + 2 y . Calculation: The given constraints are as follows: x ≥ 0 y ≥ 0 x + y ≤ 50 2 x + y ≥ 20 x + 2 y ≥ 30 For the constraint x ≥ 0 only positive values of x are allowed including 0 that is the region to the right side of the y axis. For the constraint y ≥ 0 only positive values of y are allowed including 0 that is the region to the upper side of the x axis. The common region or the shared region for the constraint x ≥ 0 and y ≥ 0 is the first quadrant of the cartesian plane including the positive values of x and y and the origin. Figure 1 shows the region shared under the constraint x ≥ 0 and y ≥ 0 . Draw the graph of the equation x + y = 50 . x y 0 50 50 0 Table 1 Table 1 shows the values of the coordinate as ( x , y ) which lies on the line x + y = 50 . The coordinates required to draw the graph of the equation x + y = 50 are as follows: ( 0 , 50 ) and ( 50 , 0 ) . To determine the region of the inequality x + y < 50 check whether the coordinate of upper or lower region of the line x + y = 50 satisfies the inequality x + y < 50 . Check for the coordinate ( − 50 , 50 ) . Substitute − 50 for x and 50 for y in x + y < 50 . − 50 + 50 < ? 50 0 < 50 The coordinate ( − 50 , 50 ) satisfies the inequality x + y < 50 . Therefore, the region in the graph which corresponds to the inequality x + y < 50 is the region below the line x + y = 50 . Figure 2 shows the graph of the equation x + y = 50 and the shaded region corresponds to the inequality x + y < 50 . Now draw the graph of the equation 2 x + y = 20 . x y 10 0 0 20 Table 2 Table 2 shows the values of the coordinate as ( x , y ) which lies on the line 2 x + y = 20 . The coordinates required to draw the graph of the equation 2 x + y = 20 are as follows: ( 10 , 0 ) and ( 0 , 20 ) . To determine the region of the inequality 2 x + y > 20 check whether the coordinate of upper or lower region of the line 2 x + y = 20 satisfies the inequality 2 x + y > 20 . Check for the coordinate ( 10 , 20 ) . Substitute 1 for x and 2 for y in 2 x + y > 20 . ( 2 ⋅ 10 ) + 20 > ? 20 20 + 20 > ? 20 40 > 20 The coordinate ( 10 , 20 ) satisfies the inequality 2 x + y > 20 . Therefore, the region in the graph which corresponds to the inequality 2 x + y > 20 is the region above the line 2 x + y = 20 . Figure 3 shows the graph of the equation 2 x + y = 20 and the shaded region corresponds to the inequality 2 x + y > 20 . Draw the graph of the equation x + 2 y = 30 . x y 30 0 0 15 Table 3 Table 3 shows the values of the coordinate as ( x , y ) which lies on the line x + 2 y = 30

A Graphical Approach to College Algebra (6th Edition)

6th Edition

ISBN: 9780321920300

Author: John Hornsby, Margaret L. Lial, Gary K. Rockswold

Publisher: PEARSON

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Chapter 6, Problem 82RE

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To find: The values of the coordinate (x,y) such that x≥0,y≥0,x+y≤50,2x+y≥20 and x+2y≥30 to minimize the objective function Z=4x+2y.

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Chapter 6, Problem 81RE

Chapter 6, Problem 83RE

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Chapter 6 Solutions

A Graphical Approach to College Algebra (6th Edition)

Ch. 6.1 - CHECKING ANALYTIC SKILLS Use each graph to...Ch. 6.1 - Prob. 8E Ch. 6.1 - CHECKING ANALYTIC SKILLS Use each graph to...Ch. 6.1 - Prob. 10E Ch. 6.1 - Prob. 11E Ch. 6.1 - Prob. 12E Ch. 6.1 - Prob. 13E Ch. 6.1 - Prob. 14E Ch. 6.1 - Prob. 15E Ch. 6.1 - Prob. 16E

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The values of the coordinate ( x , y ) such that x ≥ 0 , y ≥ 0 , x + y ≤ 50 , 2 x + y ≥ 20 and x + 2 y ≥ 30 to minimize the objective function Z = 4 x + 2 y .

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To find: The values of the coordinate (x,y) such that x≥0,y≥0,x+y≤50,2x+y≥20 and x+2y≥30 to minimize the objective function Z=4x+2y.

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