Explanation Given Information: The given function is, f = 5 x + 6 y And the given inequalities are, x + 3 y ≤ 24 4 x + 3 y ≤ 42 2 x + y ≤ 20 Restrict x ≥ 0 , y ≥ 0 . Formula used: The following method is used, Step 1. Write the function from the problem. Step 2. Check if the region is closed or not. Step 3. Find the corners of the feasible region. Step 4. Evaluate the function at each corner of the feasible region. Step 5. Find the maximum value of the function. Calculation: Consider the given function and inequalities, f = 5 x + 6 y Subject to: x + 3 y ≤ 24 4 x + 3 y ≤ 42 2 x + y ≤ 20 Restrict x ≥ 0 , y ≥ 0 . The two inequalities x ≥ 0 , y ≥ 0 restrict the solution to first quadrant. To graph theses inequalities, first draw the graph for the following three lines as a solid line because points lying on the line satisfy the given inequalities, x + 3 y = 24 ......(1) 4 x + 3 y = 42 ......(2) 2 x + y = 20 ......(3) So, solve the first equation for y , x + 3 y = 24 3 y = 24 − x y = 24 − x 3 Now, find three solutions for this equation as, x y = 24 − x 3 y 0 24 − 0 3 8 3 24 − 3 3 = 21 3 7 6 24 − 6 3 = 18 3 6 The solution for the first equation are ( 0 , 8 ) , ( 3 , 7 ) , ( 6 , 6 ) . Solve the second equation for y, 4 x + 3 y = 42 3 y = 42 − 4 x y = 42 − 4 x 3 Now, find three solutions for this equation as, x y = 42 − 4 x 3 y 0 42 − 4 ( 0 ) 3 = 42 3 14 3 42 − 4 ( 3 ) 3 = 30 3 10 6 42 − 4 ( 6 ) 3 = 18 3 6 The solution for the second equation are ( 0 , 14 ) , ( 3 , 10 ) , ( 6 , 6 ) . Solve the third equation for y , 2 x + y = 20 y = 20 − 2 x Now, find three solutions for this equation as, x y = 20 − 2 x y 0 20 − 2 ( 0 ) 20 5 20 − 2 ( 5 ) 10 10 20 − 2 ( 10 ) 0 The solution for the third equation are ( 0 , 20 ) , ( 5 , 10 ) , ( 10 , 0 ) . Now plot these points for these three equations on the graph and connect them with a line as, Now pick ( 0 , 0 ) as a test point, because that is not in any of the three lines, and test this for the inequalities. Clearly, co-ordinates satisfy all the three inequalities. Hence, the solution region lies in first quadrant and towards the origin from the lines x + 3 y = 24 , 4 x + 3 y = 42 and 2 x + y = 20

Mathematical Applications for the Management, Life, and Social Sciences

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ISBN: 9781337625340

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 4, Problem 9RE

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To calculate: The maximize value of function f=5x+6y that subjected to constrains x+3y≤24, 4x+3y≤42, 2x+y≤20, x≥0 and y≥0

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Chapter 4, Problem 8RE

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The maximize value of function f = 5 x + 6 y that subjected to constrains x + 3 y ≤ 24 , 4 x + 3 y ≤ 42 , 2 x + y ≤ 20 , x ≥ 0 and y ≥ 0

Solution Summary: The author explains the formula used to calculate the maximum value of the function f=5x+6y.

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To calculate: The maximize value of function f=5x+6y that subjected to constrains x+3y≤24, 4x+3y≤42, 2x+y≤20, x≥0 and y≥0

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