Explanation Given Information: The provided constraints are 4 x 1 + x 2 + x 3 ≤ 45 − x 1 + x 2 + 2 x 3 ≤ 0 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 provided objective function, Z = x 1 + 4 x 2 + 2 x 3 − 10 This function is not linear due to the term − 10 . Let p = x 1 + 4 x 2 + 2 x 3 − 10 the maximum value of p can also given maximum value of Z by subtracting 10 from it. Therefore we need to maximise p. Consider the constraints 4 x 1 + x 2 + x 3 ≤ 45 − x 1 + x 2 + 2 x 3 ≤ 0 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 . 4 x 1 + x 2 + x 3 + s = 45 − x 1 + x 2 + 2 x 3 + t = 0 Step 2: Write the equations in an augmented matrix. The equations are 4 x 1 + x 2 + x 3 + s = 45 − x 1 + x 2 + 2 x 3 + t = 0 − x 1 − 4 x 2 − 2 x 3 + p = 0 Write the equations in augmented matrix. x 1 x 2 x 3 s t p s 4 1 1 1 0 0 45 t − 1 1 2 0 1 0 0 p − 1 − 4 − 2 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 − 4 . Thus, pivot the x 2 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 45 1 = 45 and 0 1 = 0 . Thus, the pivot is in 2 nd row in column x 3 as, x 1 x 2 x 3 s t p s 4 1 1 1 0 0 45 t − 1 1 2 0 1 0 0 p − 1 − 4 − 2 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

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ISBN: 9781337280426

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 5, Problem 29RE

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To calculate: The maximum value of Z=x1+4x2+2x3−10 subject to constraints,

4x1+x2+x3≤45−x1+x2+2x3≤0x1≥0,x2≥0,x3≥0

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Chapter 5, Problem 28RE

Chapter 5, Problem 30RE

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The maximum value of Z = x 1 + 4 x 2 + 2 x 3 − 10 subject to constraints, 4 x 1 + x 2 + x 3 ≤ 45 − x 1 + x 2 + 2 x 3 ≤ 0 x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0

Solution Summary: The author explains how to calculate the maximum value of Z = x_1+4.

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To calculate: The maximum value of Z=x1+4x2+2x3−10 subject to constraints,

4x1+x2+x3≤45−x1+x2+2x3≤0x1≥0,x2≥0,x3≥0

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Chapter 5 Solutions

The maximum value of Z = x 1 + 4 x 2 + 2 x 3 − 10 subject to constraints, 4 x 1 + x 2 + x 3 ≤ 45 − x 1 + x 2 + 2 x 3 ≤ 0 x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0

Solution Summary: The author explains how to calculate the maximum value of Z = x_1+4.

Videos

To calculate: The maximum value of Z=x1+4x2+2x3−10 subject to constraints, 4x1+x2+x3≤45−x1+x2+2x3≤0x1≥0,x2≥0,x3≥0

Want to see the full answer?

Chapter 5 Solutions

To calculate: The maximum value of Z=x1+4x2+2x3−10 subject to constraints,

4x1+x2+x3≤45−x1+x2+2x3≤0x1≥0,x2≥0,x3≥0