Explanation Given: Mario has three nursery business Japanese Maples, tricolor beech. Mario sells Japanese maple of a certain size for $ 350 in profit and certain size of tri-color beech trees for $ 500 in profit. The travel time to acquire the Japanese maple tree is 5 hours and travel time to acquire for beech tree is 7 hours. The digging process takes 1 hour for Japanese maple tree and 2 hours for the beech trees. Both kinds of tree require 4 hours of time to deliver to the client. Mario has 3600 work hours for travel time and 900 work-hour for digging and 2600 work-hours for delivery to client. Calculation: Suppose x 1 is the number of Japanese Maples, x 2 is the number of tri-color beech tree. z ismaximum profit. Consider the summarize the information as follows, The constraints can be expressed as: 5 x 1 + 7 x 2 ≤ 3600 x 1 + 2 x 2 ≤ 900 4 x 1 + 4 x 2 ≤ 2600 Travel time Digging time Delivry time Profit Japanese Maples ( x 1 ) 5 1 4 $ 350 Tri-Color beech tree ( x 2 ) 7 2 4 $ 500 Total working hours available 3600 900 2600 After simplification of third constraint, x 1 + x 2 ≤ 650 where x 1 and x 3 are nonnegative. The first of these constrains says that 5 x 1 + 7 x 2 ≤ 3600 is less than or perhaps equal to 3600 . Suppose s 1 is the slack variable, Therefore, the first equation is, 5 x 1 + 7 x 2 + s 1 = 3600 The second constrains x 1 + 2 x 2 ≤ 900 can be expressed by adding slack variable s 2 . Therefore, the second equation is, x 1 + 2 x 2 + s 2 = 900 The third constrains x 1 + x 2 ≤ 650 can be expressed by adding slack variable s 3 . Therefore, the third equation is, x 1 + x 2 + s 3 = 650 The profit is $ 350 on each Japanese Maples and $ 500 on tricolor beech tree. The constrains of maximum profit are: z = 350 x 1 + 500 x 2 After simplification, 350 x 1 + 500 x 2 − z = 0 Now linear programming problem can be written as follows, Maximize z = 350 x 1 + 500 x 2 Subject to 5 x 1 + 7 x 2 ≤ 3600 x 1 + 2 x 2 ≤ 900 x 1 + x 2 ≤ 650 With x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 , s 3 ≥ 0 Thus, the initial simplex tableau is x 1 x 2 s 1 s 2 s 3 [ 5 7 1 0 0 3600 1 2 0 1 0 900 1 1 0 0 1 650 − 350 − 500 0 0 0 0 ] The most negative element in the last row lies in the second column therefore it is the pivot column. Dividing the rightmost column by the pivot column to get the least quotient.The pivot is 2 , in row two, column two (b) To determine To calculation: The value of slack variables.

Mathematics with Applications, Books a la Carte, and MyLab Math with Pearson eText - Title-Specific Access Card Package (12th Edition)

12th Edition

ISBN: 9780134862651

Author: Lial

Publisher: PEARSON

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Chapter 7.5, Problem 9E

To determine

To calculate: The number of tree of each kind to be acquired for maximum profit and also the amount of maximum profit.

(b)

To determine

To calculation: The value of slack variables.

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Chapter 7.5, Problem 8E

Chapter 7.5, Problem 10E

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Mathematics with Applications, Books a la Carte, and MyLab Math with Pearson eText - Title-Specific Access Card Package (12th Edition)

Ch. 7.1 - Checkpoint 1 Graph the given...Ch. 7.1 - Prob. 2CP Ch. 7.1 - Prob. 3CP Ch. 7.1 - Prob. 4CP Ch. 7.1 - Prob. 5CP Ch. 7.1 - Checkpoint 6 Graph the feasible region of the...Ch. 7.1 - Prob. 7CP Ch. 7.1 - Match the inequality with its graph, which is one...Ch. 7.1 - Match the inequality with its graph, which is one...Ch. 7.1 - Match the inequality with its graph, which is one...

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