Explanation Given Information: The provided equation of line is y = 3 x , y = − 1 4 x + 2 and y = 1 2 x and the objective function is 4 x + 3 y . The provided diagram of feasible set is shown below, Calculation: Consider the provided objective function is, 4 x + 3 y Now, for the intersection of the line y = 3 x and y = − 1 4 x + 2 . Equate the both line and solve for x . 3 x = − 1 4 x + 2 3 x + 1 4 x = 2 13 x = 8 x = 8 13 Substitute x = 8 13 in the equation of line y = 3 x . y = 3 ⋅ 8 13 = 24 13 Then, the intersection point of the line y = 3 x and y = − 1 4 x + 2 is ( 8 13 , 24 13 ) . Now, for the intersection of the line y = 3 x and y = 1 2 x . Equate the both line and solve for x . 3 x = 1 2 x 3 x − 1 2 x = 0 5 2 x = 0 x = 0 Substitute x = 0 in the equation of line y = 3 x . y = 3 ⋅ 0 = 0 Then, the intersection point of the line y = 3 x and y = 1 2 x is ( 0 , 0 ) . Now, for the intersection of the line y = − 1 4 x + 2 and y = 1 2 x . Equate the both line and solve for x . − 1 4 x + 2 = 1 2 x − 1 4 x − 1 2 x = − 2 − 1 − 2 4 x = − 2 − 3 4 x = − 2 Further solve the above equation. − 3 x = − 2 ⋅ 4 x = − 8 − 3 x = 8 3 Substitute x = 8 3 in the equation of line y = 1 2 x

Finite Mathematics & Its Applications (12th Edition)

12th Edition

ISBN: 9780134437767

Author: Larry J. Goldstein, David I. Schneider, Martha J. Siegel, Steven Hair

Publisher: PEARSON

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Chapter 3.3, Problem 4E

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To calculate: The value of x and y for which the objective function 4x+3y is maximized.

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Chapter 3.3, Problem 3E

Chapter 3.3, Problem 5E

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Chapter 3 Solutions

Finite Mathematics & Its Applications (12th Edition)

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The value of x and y for which the objective function 4 x + 3 y is maximized.

Solution Summary: The author calculates the objective function 4x+3y for x, y, and the intersection point of the line.

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To calculate: The value of x and y for which the objective function 4x+3y is maximized.

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Chapter 3 Solutions