Explanation Given: The bank has set aside $25 million for commercial and home loans. Every million of commercial loan requires two forms, whereas home loans require three forms and the bank cannot process more than 72 forms.It is the bank’s policy to loan at least four times as much as home loans as for commercial loans.At least $10 million will be used for these two loans. The bank earns 10% on commercial loans and 12% home loans. Explanation: Let there be $ x 1 millions of commercial loans and $ x 2 millions of bank loans. Then, form a system of linear inequalities using the provided statement. The system of inequalities is: x 1 + x 2 ≤ 25 2 x 1 + 3 x 2 ≤ 72 x 1 − 4 x 2 ≤ 0 x 1 + x 2 ≥ 10 \ The objective function to be maximized is z = 0.1 x 1 + 0.12 x 2 . Introduce slack variables s 1 , s 2 , and s 3 . Introduce surplus variable s 4 . The above system of linear inequalities becomes the following system of linear equations: x 1 + x 2 + s 1 = 25 2 x 1 + 3 x 2 + s 2 = 72 x 1 − 4 x 2 + s 3 = 0 x 1 + x 2 − s 4 = 10 z − 0.1 x 1 − 0.12 x 2 = 0 Also, x 1 , x 2 ≥ 0 This produces the following tableau: [ x 1 x 2 s 1 s 2 s 3 s 4 z 1 1 1 0 0 0 0 25 2 3 0 1 0 0 0 72 1 − 4 0 0 1 0 0 0 1 1 0 0 0 − 1 0 10 − 0.1 − 0.12 0 0 0 0 1 0 ] _ The solution of this tableau gives s 4 = − 10 , which is an unwanted solution; hence, we would select a nonnegative entry farthest on the left in the fourth row. Use ratio test to find the pivot element. 25 1 = 25 , 72 2 = 36 , 0 1 = 0 , 10 1 = 10 Hence, use 1 as the pivot element as it gives the least ratio. Use the first element of the fourth row as the pivot element. Use row operation R 1 − R 3 → R 1 , R 2 − 2 R 3 → R 2 , R 4 − R 3 → R 4 to change the remaining elements in the pivot column to 0. [ x 1 x 2 s 1 s 2 s 3 s 4 z 0 5 1 0 − 1 0 0 25 0 11 0 1 − 2 0 0 72 1 − 4 0 0 1 0 0 0 0 5 0 0 − 1 − 1 0 10 0 − 0.52 0 0 0.1 0 1 0 ] _ Since the basic variable s 4 is negative, choose a positive entry farthest to the left in the fourth row. Use ratio test to find the pivot element in this column. 25 5 = 5 , 72 11 = 6.5 , 0 − 4 = 0 , 10 5 = 2 Since -4 is negative and choosing 5 as the pivot eliminates the negative basic variable, choose the second element in the fourth row as the pivot element. Use row operation 1 5 R 4 → R 4 to change the pivot element to1, use R 1 − 5 R 4 → R 1 , R 2 − 11 R 4 → R 2 , R 3 + 4 R 4 → R 3 , R 5 + 0.52 R 4 → R 5 to change the remaining elements in the pivot column to 0. [ x 1 x 2 s 1 s 2 s 3 s 4 z 0 5 1 0 − 1 0 0 25 0 11 0 1 − 2 0 0 72 1 − 4 0 0 1 0 0 0 0 1 0 0 − 1 5 − 1 5 0 2 0 − 0.52 0 0 0

11th Edition

ISBN: 9780321979438

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

Publisher: PEARSON

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Chapter 4.4, Problem 27E

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The amount of money to be allotted to each type of loan to maximize the interest income.

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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 amount of money to be allotted to each type of loan to maximize the interest income.

Solution Summary: The author explains how the bank makes a maximum return of 2.94 million when they provide 3 million for commercial and 22 million on home loans.

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The amount of money to be allotted to each type of loan to maximize the interest income.

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Chapter 4 Solutions