Explanation If one shades the regions of points that do satisfy the inequalities, then the feasible region for more than one inequality will correspond to the overlap of the shaded regions. This becomes difficult when there are several inequalities. Example: Let the inequalities are − 3 x + 2 y ≤ 5 , 3 x − 2 y ≤ 6 , x ≤ 2 y and x ≥ 0 , y ≥ 0 . Consider the inequality − 3 x + 2 y ≤ 5 . Assume the equation as, − 3 x + 2 y = 5 Now, substitute the values of x in order to get the corresponding values of y and sketch the graph of the line. For x = 0 , − 3 ( 0 ) + 2 y = 5 2 y = 5 y = 5 2 Therefore, the point is ( 0 , 5 2 ) . For x = − 5 3 , − 3 ( − 5 3 ) + 2 y = 5 2 y = 0 y = 0 Therefore, the point is ( − 5 3 , 0 ) . For x = 1 , − 3 ( 1 ) + 2 y = 5 2 y = 8 y = 4 Therefore, the point is ( 1 , 4 ) . The points for the equation are as follows: x y ( x, y ) 0 5 2 ( 0 , 5 2 ) − 5 3 0 ( − 5 3 , 0 ) 1 4 ( 1 , 4 ) The graph of the inequality is as follows: Hence, the boundary points for the equation are ( − 1.66 , 0 ) and ( 0 , 2.5 ) . Consider the inequality 3 x − 2 y ≥ 6 . Assume the equation as, 3 x − 2 y = 6 Now, substitute the values of x in order to get the corresponding values of y and sketch the graph of the line. For x = 0 , 3 ( 0 ) − 2 y = 6 − 2 y = 6 y = − 3 Therefore, the point is ( 0 , − 3 ) . For x = 2 , 3 ( 2 ) − 2 y = 6 − 2 y = 0 y = 0 Therefore, the point is ( 2 , 0 ) . For x = 1 , 3 ( 1 ) − 2 y = 6 − 2 y = 3 y = − 3 2 Therefore, the point is ( 1 , − 3 2 ) . The points for the equation are as follows: x y ( x, y ) 0 − 3 ( 0 , − 3 ) 2 0 ( 2 , 0 ) 1 − 3 2 ( 1 , − 3 2 ) The graph of the equation is as follows:

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ISBN: 9781337280426

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 5.1, Problem 48E

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Chapter 5.1, Problem 47E

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The advantage of shading the region of points that do not satisfy the given inequalities with an example.

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