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The production manager for the Coory soft drink company is considering the production of two kinds of soft drinks: regular and diet. Two of her limited resources are production time (8 hours = 480 minutes per day) and syrup (1 of the ingredients), limited to 675 gallons per day. To produce a regular case requires 2 minutes and 5 gallons of syrup, while a Solve this model by using graphical analysis. A) R = 75, D = 90, Z = $405 B) R = 135, D = 0, Z = $405 C) R = 90, D = 75, Z = $420 D) R = 40, D= 100, Z = $320 Regular Diet R Profit $4 3.00 $ 2.00 Requirements/Constraints Regular Diet Limited Resources Production Time minutes <= 480 Syrup 5 3 gallons 675 <= 2R+4D = 480 5R+3D=675 Solution_2 D=D0 Solution_2 R=0 D = 120 Solution_1 2R+4D = 480 R= 240 R=240-2D 240*5 > 675 5(240-2D)+3D=675 Solution_3 R=0 is not feasible D = 675/3=225 Solution_3 D=0 135 7D=525 R = 675/5=135 D=75 R=240-2*75=90 225*4 = 900 is not feasible Optimal Solutions: 1 90 135 R D Profit 75 120 $420.00 $240.00 $405.00 Highest Profit

The production manager for the Coory soft drink company is considering the production of two kinds of soft drinks: regular and diet. Two of her limited resources are production time (8 hours = 480 minutes per day) and syrup (1 of the ingredients), limited to 675 gallons per day. To produce a regular case requires 2 minutes and 5 gallons of syrup, while a Solve this model by using graphical analysis. A) R = 75, D = 90, Z = $405 B) R = 135, D = 0, Z = $405 C) R = 90, D = 75, Z = $420 D) R = 40, D= 100, Z = $320 Regular Diet R Profit $4 3.00 $ 2.00 Requirements/Constraints Regular Diet Limited Resources Production Time minutes <= 480 Syrup 5 3 gallons 675 <= 2R+4D = 480 5R+3D=675 Solution_2 D=D0 Solution_2 R=0 D = 120 Solution_1 2R+4D = 480 R= 240 R=240-2D 240*5 > 675 5(240-2D)+3D=675 Solution_3 R=0 is not feasible D = 675/3=225 Solution_3 D=0 135 7D=525 R = 675/5=135 D=75 R=240-2*75=90 225*4 = 900 is not feasible Optimal Solutions: 1 90 135 R D Profit 75 120 $420.00 $240.00 $405.00 Highest Profit

Practical Management Science
Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter2: Introduction To Spreadsheet Modeling
Section: Chapter Questions
Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...
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The production manager for the Coory soft drink company is considering the production of two kinds of soft drinks: regular and diet. Two of her limited resources are production time (8 hours = 480 minutes per day) and syrup (1 of the ingredients), limited to 675 gallons per day. To produce a regular case requires 2 minutes and 5 gallons of syrup, while a Solve this model by using graphical analysis. A) R = 75, D = 90, Z = $405 B) R = 135, D = 0, Z = $405 C) R = 90, D = 75, Z = $420 D) R = 40, D= 100, Z = $320 Regular Diet D Profit 3.00 $ 2.00 Requirements/Constraints Regular Diet Limited Resources Production Time 4 minutes 480 <= Syrup 3 gallons 675 2R+4D = 480 5R+3D=675 Solution_1 2R+4D = 480 Solution_2 R=0 Solution_2 D=0 R=240-2D D = 120 R = 240 240*5 > 675 5(240-2D)+3D=675 Solution_3 R=0 is not feasible D = 675/3=225 Solution_3 D=0 135 7D=525 R = 675/5=135 D=75 R=240-2*75=90 225*4 = 900 is not feasible Optimal Solutions: 2 3 1 90 135 120 $240.00 75 $405.00 $420.00 Highest Profit Profit to download and review the examples of problems and solutions for Exam 1.
Transcribed Image Text:The production manager for the Coory soft drink company is considering the production of two kinds of soft drinks: regular and diet. Two of her limited resources are production time (8 hours = 480 minutes per day) and syrup (1 of the ingredients), limited to 675 gallons per day. To produce a regular case requires 2 minutes and 5 gallons of syrup, while a Solve this model by using graphical analysis. A) R = 75, D = 90, Z = $405 B) R = 135, D = 0, Z = $405 C) R = 90, D = 75, Z = $420 D) R = 40, D= 100, Z = $320 Regular Diet D Profit 3.00 $ 2.00 Requirements/Constraints Regular Diet Limited Resources Production Time 4 minutes 480 <= Syrup 3 gallons 675 2R+4D = 480 5R+3D=675 Solution_1 2R+4D = 480 Solution_2 R=0 Solution_2 D=0 R=240-2D D = 120 R = 240 240*5 > 675 5(240-2D)+3D=675 Solution_3 R=0 is not feasible D = 675/3=225 Solution_3 D=0 135 7D=525 R = 675/5=135 D=75 R=240-2*75=90 225*4 = 900 is not feasible Optimal Solutions: 2 3 1 90 135 120 $240.00 75 $405.00 $420.00 Highest Profit Profit to download and review the examples of problems and solutions for Exam 1.
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