(a) Explanation Given: The Fancy Fashions, the local boutique, has $ 8000 available each month for advertisement. Newspaper ads cost $ 400 each, and no more than 30 can run per month. Internet banner ads cost $ 20 each, and no more than 60 can run per month. TV ads cost $ 2000 each, with a maximum of 10 available per month. Approximately, 4000 women see each newspaper ad, 3000 see each internet banner, and 10,000 see each TV advertisement. Explanation: Consider The number of newspaper advertisement is equal to x 1 . The number of internet banner advertisement is equal to x 2 . The number of TV advertisement is equal to x 3 . The constraints from problemare as follows: 400 x 1 + 20 x 2 + 2000 x 3 ≤ 8000 x 1 ≤ 30 x 2 ≤ 60 x 3 ≤ 10 The objective function is z = 4000 x 1 + 3000 x 2 + 10000 x 3 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 Now, apply the simplex method for maximization. Step 1: Add slack variables or basic variable to each constraint to convert it into linear equation. Basic variables are the variables that are nonzero. As a result, 400 x 1 + 20 x 2 + 2000 x 3 + s 1 = 8000 x 1 + s 2 = 30 x 2 + s 3 = 60 x 3 + s 4 = 10 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 Here s 1 , s 2 , s 3 , a n d s 4 are basic variables greater than or equal 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 s 1 s 2 s 3 s 4 z s 1 400 20 2000 1 0 0 0 0 8000 s 2 1 0 0 0 1 0 0 0 30 s 3 0 1 0 0 0 1 0 0 60 s 4 0 0 1 0 0 0 1 0 10 z − 4000 − 3000 − 10000 0 0 0 0 1 0 ] _ This tableau leads to the solution x 1 , x 2 , x 3 ≥ 0 and s 1 = 8000 , s 2 = 30 , s 3 = 60 , s 4 = 10 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 − 10 , 000 .Hence, the column corresponding to − 10 , 000 is the pivot column, which follows that x 3 is enter variable. Step 4: Select the pivot. Since the pivot must be a positive number, apply test ratio for the pivot. 8000 2000 = 4 , 10 1 = 10 Therefore, 2000 is pivot because the smallest positive ratio is obtained by the element of R 1 , which follows that s 1 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 1 → R 1 2000 : [ x 1 x 2 x 3 s 1 s 2 s 3 s 4 z x 3 1 5 1 100 1 1 2000 0 0 0 0 4 s 2 1 0 0 0 1 0 0 0 30 s 3 0 1 0 0 0 1 0 0 60 s 4 0 0 1 0 0 0 1 0 10 z − 4000 − 3000 − 10 , 000 0 0 0 0 1 0 ] _ Now make other entries in the pivot column zero. R 4 → R 4 − R 1 R 5 → R 5 + 10 , 000 R 1 This leads to [ x 1 x 2 x 3 s 1 s 2 s 3 s 4 z x 3 1 5 1 100 1 1 2000 0 0 0 0 4 s 2 1 0 0 0 1 0 0 0 30 s 3 0 1 0 0 0 1 0 0 60 s 4 − 1 5 − 1 100 0 − 1 2000 0 0 1 0 6 z − 2000 − 2900 0 5 0 0 0 1 40 , 000 ] _ Again, choose pivot column. Column 2 is the pivot. Pivot number is 4 1 100 = 400 , 60 1 = 60 Thus, 1 is pivot; that is, x 2 is enter variable and s 3 a depart variable (b) To determine Each type of advertisement required to obtain the maximum profit.

11th Edition

ISBN: 9780321979438

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

Publisher: PEARSON

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Chapter 4.2, Problem 29E

(a)

To determine

Each type of advertisement required to obtain the maximum profit.

(b)

To determine

Each type of advertisement required to obtain the maximum profit.

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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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Each type of advertisement required to obtain the maximum profit.

Solution Summary: The author explains how each type of advertisement required to obtain maximum profit. They add slack variables or basic variables to each constraint to convert it into linear equation.

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Each type of advertisement required to obtain the maximum profit.

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