Explanation Given Information: The provided constraints are; 5 x 1 − x 2 + x 3 ≤ 1500 2 x 1 + 2 x 2 + x 3 ≤ 2500 4 x 1 + 2 x 2 + x 3 ≤ 2000 x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 Formula used: Step 1: Use slack variables and write the inequalities in equation form. Step 2: Write the equations in an augmented matrix. Step 3: Choose the smallest negative number on the left side of the bottom row and pivot the column. Step 4: Select the pivot which is the smallest of the test ratios a b , where a is the entry in the right most column and b is the corresponding entry in the pivot column. Step 5: Clear the pivot column using the pivot and label the pivot row with the label of the pivot column. Step 6: Repeat the above steps till all the entries in the bottom row are non-negative. Use row operations to clear the column of the augmented matrix. Calculation: Consider the constraints, 5 x 1 − x 2 + x 3 ≤ 1500 2 x 1 + 2 x 2 + x 3 ≤ 2500 4 x 1 + 2 x 2 + x 3 ≤ 2000 x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 Step 1: Use slack variables and write the inequalities in equation form. Add the slack variables s , t and u as; 5 x 1 − x 2 + x 3 + s = 1500 2 x 1 + 2 x 2 + x 3 + t = 2500 4 x 1 + 2 x 2 + x 3 + u = 2000 Step 2: Write the equations in an augmented matrix. The equations are; 5 x 1 − x 2 + x 3 + s = 1500 2 x 1 + 2 x 2 + x 3 + t = 2500 4 x 1 + 2 x 2 + x 3 + u = 2000 − 3 x 1 − 7 x 2 − 8 x 3 + z = 0 Write the equations in augmented matrix as; x 1 x 2 x 3 s t u p s 5 − 1 1 1 0 0 0 1500 t 2 2 1 0 1 0 0 2500 u 4 2 1 0 0 1 0 2000 z − 3 − 7 − 8 0 0 0 1 0 Step 3: Choose the smallest negative number on the left side of the bottom row and pivot the column. Since, the smallest negative number on the left side of the bottom row is − 8 . Thus, pivot the x 3 column. Step 4: Select the pivot which is the smallest of the test ratios a b , where a is the entry in the right most column and b is the corresponding entry in the pivot column. Since, the test ratios are neither 0 nor negative. Thus, the test ratios are 1500 1 = 1500 , 2500 1 = 2500 and 2000 1 = 2000 . Thus, the pivot is 1 in column x 3 as; x 1 x 2 x 3 s t u p s 5 − 1 1 1 0 0 0 1500 t 2 2 1 0 1 0 0 2500 u 4 2 1 0 0 1 0 2000 z − 3 − 7 − 8 0 0 0 1 0 Step 5: Clear the pivot column using the pivot and label the pivot row with the label of the pivot column. Since, the pivot is s in column x 3 and replace s with x 3

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ISBN: 9781337280426

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 5.3, Problem 9E

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To calculate: The maximum value of z=3x1+7x2+8x3 subject to constraints

5x1−x2+x3≤1500,2x1+2x2+x3≤2500,4x1+2x2+x3≤2000 and x1≥0,x2≥0,x3≥0.

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Chapter 5.3, Problem 8E

Chapter 5.3, Problem 10E

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The maximum value of z = 3 x 1 + 7 x 2 + 8 x 3 subject to constraints 5 x 1 − x 2 + x 3 ≤ 1500 , 2 x 1 + 2 x 2 + x 3 ≤ 2500 , 4 x 1 + 2 x 2 + x 3 ≤ 2000 and x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 .

Solution Summary: The author explains how to calculate the maximum value of z=3x_1+7 x 2x +8.

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To calculate: The maximum value of z=3x1+7x2+8x3 subject to constraints

5x1−x2+x3≤1500,2x1+2x2+x3≤2500,4x1+2x2+x3≤2000 and x1≥0,x2≥0,x3≥0.

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Chapter 5 Solutions

The maximum value of z = 3 x 1 + 7 x 2 + 8 x 3 subject to constraints 5 x 1 − x 2 + x 3 ≤ 1500 , 2 x 1 + 2 x 2 + x 3 ≤ 2500 , 4 x 1 + 2 x 2 + x 3 ≤ 2000 and x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 .

Solution Summary: The author explains how to calculate the maximum value of z=3x_1+7 x 2x +8.

Videos

To calculate: The maximum value of z=3x1+7x2+8x3 subject to constraints 5x1−x2+x3≤1500,2x1+2x2+x3≤2500,4x1+2x2+x3≤2000 and x1≥0,x2≥0,x3≥0.

Want to see the full answer?

Chapter 5 Solutions

To calculate: The maximum value of z=3x1+7x2+8x3 subject to constraints

5x1−x2+x3≤1500,2x1+2x2+x3≤2500,4x1+2x2+x3≤2000 and x1≥0,x2≥0,x3≥0.