Explanation Given Information: The provided constraints are; 3 x + 8 y ≤ 24 6 x + 4 y ≤ 30 x ≥ 0 , y ≥ 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, 3 x + 8 y ≤ 24 6 x + 4 y ≤ 30 x ≥ 0 , y ≥ 0 Step 1: Use slack variables and write the inequalities in equation form. Add the slack variables s and t as; 3 x + 8 y + s = 24 6 x + 4 y + t = 30 Step 2: Write the equations in an augmented matrix. The equations are; 3 x + 8 y + s = 24 6 x + 4 y + t = 30 − 2 x − 3 y + p = 0 Write the equations in augmented matrix as; x y s t p s 3 8 1 0 0 24 t 6 4 0 1 0 30 p − 2 − 3 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 − 3 . Thus, pivot the y 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 30 4 = 7.5 and 24 8 = 3 . Thus, the pivot is 8 in column y as; x y s t p s 3 8 1 0 0 24 t 6 4 0 1 0 30 p − 2 − 3 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 y and replace s with y . Thus, x is the active variable. Use row operations to clear the pivot column of the augmented matrix. Use the row operations 2 R 2 − R 1 and 8 R 3 + 3 R 1 as; x y s t p y 3 8 1 0 0 24 t 9 0 − 1 2 0 36 p − 7 0 3 0 8 72 Step 6: Repeat the above steps till all the entries in the bottom row are non-negative

Finite Mathematics and Application Calculus (Looseleaf) - Text Only

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ISBN: 9781337291484

Author: Waner

Publisher: Cengage

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Chapter 6.3, Problem 4E

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To calculate: The maximum value of $p=2x+3y$ subject to constraints $3x+8y≤24,6x+4y≤30,x≥0,y≥0$ .

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Chapter 6.3, Problem 3E

Chapter 6.3, Problem 5E

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Finite Mathematics and Application Calculus (Looseleaf) - Text Only

Ch. 6.1 - In Exercises 1-26, sketch the region that...Ch. 6.1 - In Exercises 1-26, sketch the region that...Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E Ch. 6.1 - Prob. 7E Ch. 6.1 - In Exercises 1-26, sketch the region that...Ch. 6.1 - Prob. 9E Ch. 6.1 - Prob. 10E

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The maximum value of p = 2 x + 3 y subject to constraints 3 x + 8 y ≤ 24 , 6 x + 4 y ≤ 30 , x ≥ 0 , y ≥ 0 .

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To calculate: The maximum value of $p=2x+3y$ subject to constraints $3x+8y≤24,6x+4y≤30,x≥0,y≥0$ .

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