Explanation Given information: The housing contractor divides a farm into 100 building lots and designed colonial style homes and ranch style homes. The colonial styled home requires $ 30 , 000 of capital and the ranch styled home requires $ 40 , 000 of capital. The colonial lot produces a profit of $ 4000 and the ranch lot produces a profit of $ 8000 if the lots are sold. The housing contractor has $ 3.6 million of capital on hand. Calculation: Consider the number of crates of oranges as x and number of crates of grapefruit as y . The profit on each crate of orange is $ 2.50 and the profit on each crate of grapefruit is $ 4 P = 2.5 x + 4 y ....... (1) Here, P is the total profit and called as objective function. The number of lots cannot be negative. x ≥ 0 and y ≥ 0 The volume of each crate of oranges is 4 ft 3 and the volume of each crate of grapefruit is 6 ft 3 . The maximum capacity of truck is 300 ft 3 . 4 x + 6 y ≤ 300 Divide both sides of above inequality by 2 to simplify the above inequality as, 2 x + 3 y ≤ 150 ....... (2) The weight of each crate of oranges is 80 lb and the weight of each crate of grapefruit is 100 lb . The truck cannot carry more than 5600 lb of weight. 80 x + 100 y ≤ 5600 Divide both sides of above inequality by 20 to simplify the above inequality as, 4 x + 5 y ≤ 280 ....... (3) A trucker cannot carry more crates of grapefruit than the crates of oranges. x ≥ y ....... (4) Therefore, x and y must satisfy the constraints as, { x ≥ 0 , y ≥ 0 x ≥ y 2 x + 3 y ≤ 150 4 x + 5 y ≤ 280 From inequality x ≥ 0 and y ≥ 0 it is observed that the solution set lies in I quadrant. The graph of the inequalities x ≥ 0 , y ≥ 0 , x ≥ y , 2 x + 3 y ≤ 150 and 4 x + 5 y ≤ 280 is shown in Figure 1. From Figure 1 it is observed that the solution set is represented by the darkest shade. The corresponding equation of inequality (2) can be expressed as, 2 x + 3 y = 150 ....... (5) The corresponding equation of inequality (3) can be expressed as, 4 x + 5 y = 280 ....... (6) The corresponding equation of inequality (3) can be expressed as, x = y ....... (7) Multiply the equation (5) by 2 as, 4 x + 6 y = 300 ....... (8) Subtract equation (6) from equation (8) to obtain the value of y as, y = 300 − 280 y = 20 Substitute 20 for y in equation (5) to obtain the value of x as, 2 x + 3 ⋅ 20 = 150 2 x + 60 = 150 2 x = 90 x = 45 The intersection of the lines 2 x + 3 y = 150 and 4 x + 5 y = 280 is the point ( 45 , 20 ) . Substitute 0 for y in equation (6) to obtain the corresponding value of x as, 4 x + 5 ⋅ 0 = 280 4 x = 280 x = 280 4 = 70 The intersection of the lines y = 0 and 4 x + 5 y = 280 is the point ( 70 , 0 )

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ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 5, Problem 7P

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The number of crates of oranges and number of crates of grapefruit to maximize the profit.

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Chapter 5, Problem 6P

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Ch. 5.1 - The system of equations {2x+3y=75xy=9} is a system...Ch. 5.1 - A system of equation in two variable can be solved...Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Elimination method Use the elimination method to...Ch. 5.1 - Elimination method Use the elimination method to...Ch. 5.1 - Elimination method Use the elimination method to...Ch. 5.1 - Elimination method Use the elimination method to...Ch. 5.1 - Elimination method Use the elimination method to...Ch. 5.1 - Elimination method Use the elimination method to...

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The number of crates of oranges and number of crates of grapefruit to maximize the profit.

Solution Summary: The author calculates the number of crates of oranges and grapefruit to maximize the profit. The housing contractor divides a farm into 100 building lots and designed colonial and ranch style homes.

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To calculate: The number of crates of oranges and number of crates of grapefruit to maximize the profit.

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Chapter 5 Solutions

To calculate:

The number of crates of oranges and number of crates of grapefruit to maximize the profit.