Explanation Given information: A confectioner sells standard-mixture package that contains 200 g of peanuts and 100 g of cashews and sold it for $ 1.95 . A confectioner sells deluxe mixture package that contains 50 g of peanuts and 150 g of cashews and sold it for $ 2.25 . The confectioner has 20 kg of peanuts and 15 kg of cashews. There should be at least as many standard packages as deluxe packages. Calculation: Consider the number of bags of standard package as x and number of bags of deluxe package as y . The confectioner sold the standard mixture package for $ 1.95 and the deluxe mixture package for $ 2.25 . P = 1.95 x + 2.25 y ....... (1) Here, P is the total profit and called as objective function. The number of bags of each mixture cannot be negative. x ≥ 0 and y ≥ 0 The confectioner has 15 kg of cashews and the standard package contains 100 g of cashews and the deluxe mixture package contains 150 g of cashews. 100 x + 150 y ≤ 15000 Divide both sides of above inequality by 50 to simplify the above inequality as, 2 x + 3 y ≤ 300 ....... (2) The confectioner has 20kg of peanuts and the standard package contains 200g of peanuts and the deluxe mixture package contains 50g of peanuts. 200 x + 50 y ≤ 20000 Divide both sides of above inequality by 50 to simplify the above inequality as, 4 x + y ≤ 400 ....... (3) The confectioner has at least as many standard mixture packages as deluxe mixture packages. x ≥ y ....... (4) Therefore, x and y must satisfy the constraints as, { x ≥ 0 , y ≥ 0 x ≥ y 2 x + 3 y ≤ 300 4 x + y ≤ 400 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 ≤ 300 and 4 x + y ≤ 400 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 = 300 ....... (5) The corresponding equation of inequality (3) can be expressed as, 4 x + y = 400 ....... (6) The corresponding equation of inequality (4) can be expressed as, x = y ....... (7) Multiply equation (5) by 2 as, 4 x + 6 y = 600 ....... (8) Subtract equation (6) from equation (8) to obtain the value of y as, 6 y − y = 600 − 400 5 y = 200 y = 200 5 = 40 Substitute 40 for y in equation (5) to obtain the value of x as, 2 x + 3 ⋅ 40 = 300 2 x + 120 = 300 2 x = 180 x = 90 The intersection of the lines 2 x + 3 y = 300 and 4 x + y = 400 is the point ( 90 , 40 ) . Substitute 0 for y in equation (6) to obtain the corresponding value of x as, 4 x + 0 = 400 4 x = 400 x = 400 4 = 100 The intersection of the lines y = 0 and 4 x + y = 400 is the point ( 100 , 0 )

7th Edition

ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 5, Problem 11P

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The number of bags of standard package and the number of bags of deluxe package to maximize the revenue.

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Chapter 5, Problem 10P

Chapter 5, Problem 12P

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The number of bags of standard package and the number of bags of deluxe package to maximize the revenue.

Solution Summary: The author calculates the number of bags of standard and deluxe packages to maximize the revenue.

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To calculate: The number of bags of standard package and the number of bags of deluxe package to maximize the revenue.

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Chapter 5 Solutions

To calculate:

The number of bags of standard package and the number of bags of deluxe package to maximize the revenue.