(a) Explanation Given: A biologist has 500 kg of nutrient A, 600 kg of nutrient B, and 300 kg of nutrient C. These nutrients will be used to make four types of food. The table provides growth value and content of food P Q R S A 0 0 37.5 62.5 B 0 75 50 37.5 C 100 25 12.5 0 Growth value 90 70 60 50 Explanation: Let The quantity of food Pbe x 1 The quantity of food Qbe x 2 The quantity of food Rbe x 3 The quantity of food Sbe x 4 The constraints from were followed as 0.375 x 3 + 0.625 x 4 ≥ 500 0.75 x 2 + 0.5 x 3 + 0.375 x 4 ≥ 600 x 1 + 0.25 x 2 + 0.125 x 3 ≥ 300 Simplify the above inequalities: 3 8 x 3 + 5 8 x 4 ≤ 500 3 4 x 2 + 1 2 x 3 + 3 8 x 4 ≤ 600 x 1 + 1 4 x 2 + 1 8 x 3 ≤ 300 Multiply 8 to both sides of each inequality: 3 x 3 + 5 x 4 ≤ 4000 6 x 2 + 4 x 3 + 3 x 4 ≤ 4800 8 x 1 + 2 x 2 + x 3 ≤ 2400 The objective function is z = 90 x 1 + 70 x 2 + 60 x 3 + 50 x 4 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 , and x 4 ≥ 0 Now, apply the simplex method for maximization. Step 1: Add slack variables or basic variable to each constraint to convert it into linear equations. Basic variables are the variables that are nonzero. As a result, 3 x 3 + 5 x 4 + s 1 = 4000 6 x 2 + 4 x 3 + 3 x 4 + s 2 = 4800 8 x 1 + 2 x 2 + x 3 + s 3 = 2400 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 , and x 4 ≥ 0 Here s 1 , s 2 , a n d s 3 are basic variables greater than orequal to zero and x 1 , x 2 , x 3 , and x 4 are nonbasic variables. Step 2: Represent the system of linear equations obtained above in simplex tableau. This is initial tableau. [ x 1 x 2 x 3 x 4 s 1 s 2 s 3 z s 1 0 0 3 5 1 0 0 0 4000 s 2 0 6 4 3 0 1 0 0 4800 s 3 8 2 1 0 0 0 1 0 2400 z − 90 − 70 − 60 − 50 0 0 0 1 0 ] _ This tableau leads to the solution x 1 , x 2 , x 3 , x 4 ≥ 0 and s 1 = 4000 , s 2 = 4800 , s 3 = 2400 with z = 0 . Step 3: Select a pivot column. Select the negative entry in the bottom row with the largest numerical value. In initial tableau, the negative entry in the bottom row with the largest numerical value is − 90 .Hence, the column corresponding to − 90 is pivot column, which follows that x 1 is enter variable. Step 4: Select the pivot. The pivot must be a positive number and only nonzero positive number in pivot column is 8. Therefore, 8 is pivot because the smallest positive ratio is obtained by the element of R 3 , which follows that s 3 is a depart variable. Step 5: Use the pivot to make other entries in the pivot column zero. For this, first, if pivot element is not 1 , make pivot element 1 by R 3 → 1 8 R 3 . [ x 1 x 2 x 3 x 4 s 1 s 2 s 3 z s 1 0 0 3 5 1 0 0 0 4000 s 2 0 6 4 3 0 1 0 0 4800 x 1 1 1 4 1 8 0 0 0 1 8 0 300 z − 90 − 70 − 60 − 50 0 0 0 1 0 ] _ Now make other entries in the pivot column zero: R 4 → R 4 + 90 R 3 This leads to [ x 1 x 2 x 3 x 4 s 1 s 2 s 3 z s 1 0 0 3 5 1 0 0 0 4000 s 2 0 6 4 3 0 1 0 0 4800 x 1 1 1 4 1 8 0 0 0 1 8 0 300 z 0 − 95 2 − 195 4 − 50 0 0 45 4 1 27 , 000 ] _ Again, choose pivot column. Column 4 is the pivot. Pivot number is 4000 5 = 800 , 4800 3 = 1600 Thus, 5 is pivot; that is, x 4 is enter and s 1 depart variable. Now, make pivot element 1 by R 1 → R 1 5 : [ x 1 x 2 x 3 x 4 s 1 s 2 s 3 z x 4 0 0 3 5 1 1 5 0 0 0 800 s 2 0 6 4 3 0 1 0 0 4800 x 1 1 1 4 1 8 0 0 0 1 8 0 300 z 0 − 95 2 − 195 4 − 50 0 0 45 4 1 27 , 000 ] _ Make other entry in pivot column equal to zero by R 2 → R 2 − 3 R 1 R 4 → R 4 + 50 R 1 [ x 1 x 2 x 3 x 4 s 1 s 2 s 3 z x 4 0 0 3 5 1 1 5 0 0 0 800 s 2 0 6 11 5 0 − 3 5 1 0 0 2400 x 1 1 1 4 1 8 0 0 0 1 8 0 300 z 0 − 95 2 − 75 4 0 10 0 45 4 1 67 , 000 ] _ Again, choose pivot column, which is column 2 (b) To determine The maximum growth value. (c) To determine The quantity of nutrient left over.

11th Edition

ISBN: 9780321979438

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 4.2, Problem 38E

(a)

To determine

The quantity of each food required to obtain maximum growth value.

(b)

To determine

The maximum growth value.

(c)

To determine

The quantity of nutrient left over.

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Chapter 4.2, Problem 37E

Chapter 4.2, Problem 39E

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Chapter 4 Solutions

Finite Mathematics (11th Edition)

Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - Prob. 2E Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - Prob. 4E Ch. 4.1 - For Exercises 5-8. (a) determine the number of...Ch. 4.1 - Prob. 6E Ch. 4.1 - For Exercises 5-8, (a) determine the number of...Ch. 4.1 - For Exercises 5-8, (a) determine the number of...Ch. 4.1 - Introduce slack variables as necessary, then...Ch. 4.1 - Introduce slack variables as necessary, then write...

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The quantity of each food required to obtain maximum growth value.

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The quantity of each food required to obtain maximum growth value.

The maximum growth value.

The quantity of nutrient left over.

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Chapter 4 Solutions