Explanation Approach: In order to make a bounded region, the linear equalities should be chosen in a way that the solution region is closed by the lines. For the formation of such region minimum of three equation are required. The first two equations can be the x axis and the y axis. The equation of line in slope intercept form is given by, y = m x + c Here, m is the slope and c is the y intercept. Calculation: The equation of x axis is. y = 0 The equation of y axis is, x = 0 For the third equation assume a line having slope − 4 and the y intercept as 8 Substitute − 4 for m and 8 for c in the above expression. y = − 4 x + 8 Rearrange the above equation. 4 x + y = 8 Plot the graph for the equation 4 x + y = 8 , x = 0 and y = 0

8th Edition

ISBN: 9781305104174

Author: David B. Johnson, Thomas A. Mowry

Publisher: Cengage Learning

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Chapter 12.0, Problem 22E

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The system of linear inequalities with a region of solution that is bounded.

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Chapter 12.0, Problem 21E

Chapter 12.0, Problem 23E

Chapter 12 Solutions

Mathematics: A Practical Odyssey

Ch. 12.0 - In Exercises 1-8, graph the region of the...Ch. 12.0 - Prob. 2E Ch. 12.0 - Prob. 3E Ch. 12.0 - In Exercises 1-8, graph the region of the...Ch. 12.0 - Prob. 5E Ch. 12.0 - Prob. 6E Ch. 12.0 - In Exercises 1-8, graph the region of the...Ch. 12.0 - Prob. 8E Ch. 12.0 - Prob. 9E Ch. 12.0 - Prob. 10E

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When you graph a system of inequalities, will there always be a feasible region? If so, explain why. If not, give an example of a graph of inequalities that does not have a feasible region. Why does it not have a feasible region?

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