Assignment 3: Julia’s Food Booth

Misti Brinson

Dr. Matt Johnson

Quantitative Methods - MAT

Assignment #3: Julia’s Food Booth

A. Formulate and solve an L.P. model for this case:

X1 = pizza slices

X2 = hot dogs

X3 = B B Q sandwiches

This model is set up for the first home game.

Next, you would maximize Z = $0.75 x1 + 1.05 x2 + 1.35 x3 which is subject to the following: $0.75x1 + 0.45x2 + 0.90x3 <= 1,500 thus leading to:

24x1 + 16x2 + 25x3 <= 55,296 in^2 of even space: thus you have x 1>=*x2 + x3 giving you X2

_____ >= 2.0 so x1, x2, x3 >=0

X3

All of which is displayed in the graph below. (Note: The oven space needed for a slice of pizza is determined by dividing the total space required by the slice of pizza, 14 x 14 = 196 in^2 by 8 or 24 in^2 per slice. The total space available would be the dimensions of the shelf, 36 in. x 48 in. = 1,728 in^2, multiply that by 2, the times before kickoff and halftime, thus the oven will be filled 55,296in^2.

Solution:X1 = 1,250 slices of pizza

X2 = 1,250 hot dogs

X3 = 0 bbq sandwiches

Z = $2,250

| | | | | | | | |

Food items:| | Pizza| Hot Dogs| Barbecue| | | | | Profit per item:| 0.75| 1.05| 1.35| | | | |

Constraints:| | | | Available| Usage| Left over| | Budget ($)| 0.75| 0.45| 0.90| 1,500 | 1,500.00 | 0| | Oven space (sq. in.)| 24| 16| 25| 55,296 | 50,000.00 | 5296| | Demand| 1| -1| -1| 0| - | 0| |

Demand| 0| 1| -2| 0| 1,250.00 | -1250| |

| | | | | | | | |

Stock| | | | | | | | |

Pizza=| 1250| slices| | | | | | |

Hot Dogs=| 1250| hot dogs| | | | | | |

Barbecue=| 0| sandwiches| | | | | | |

Profit=| 2,250.00 | | | | | | | |

Julia’s profit before expenses should be $2,250. The lease is $1,000 per game thus giving a profit of $1,250.00....

Misti Brinson

Dr. Matt Johnson

Quantitative Methods - MAT

Assignment #3: Julia’s Food Booth

A. Formulate and solve an L.P. model for this case:

X1 = pizza slices

X2 = hot dogs

X3 = B B Q sandwiches

This model is set up for the first home game.

Next, you would maximize Z = $0.75 x1 + 1.05 x2 + 1.35 x3 which is subject to the following: $0.75x1 + 0.45x2 + 0.90x3 <= 1,500 thus leading to:

24x1 + 16x2 + 25x3 <= 55,296 in^2 of even space: thus you have x 1>=*x2 + x3 giving you X2

_____ >= 2.0 so x1, x2, x3 >=0

X3

All of which is displayed in the graph below. (Note: The oven space needed for a slice of pizza is determined by dividing the total space required by the slice of pizza, 14 x 14 = 196 in^2 by 8 or 24 in^2 per slice. The total space available would be the dimensions of the shelf, 36 in. x 48 in. = 1,728 in^2, multiply that by 2, the times before kickoff and halftime, thus the oven will be filled 55,296in^2.

Solution:X1 = 1,250 slices of pizza

X2 = 1,250 hot dogs

X3 = 0 bbq sandwiches

Z = $2,250

| | | | | | | | |

Food items:| | Pizza| Hot Dogs| Barbecue| | | | | Profit per item:| 0.75| 1.05| 1.35| | | | |

Constraints:| | | | Available| Usage| Left over| | Budget ($)| 0.75| 0.45| 0.90| 1,500 | 1,500.00 | 0| | Oven space (sq. in.)| 24| 16| 25| 55,296 | 50,000.00 | 5296| | Demand| 1| -1| -1| 0| - | 0| |

Demand| 0| 1| -2| 0| 1,250.00 | -1250| |

| | | | | | | | |

Stock| | | | | | | | |

Pizza=| 1250| slices| | | | | | |

Hot Dogs=| 1250| hot dogs| | | | | | |

Barbecue=| 0| sandwiches| | | | | | |

Profit=| 2,250.00 | | | | | | | |

Julia’s profit before expenses should be $2,250. The lease is $1,000 per game thus giving a profit of $1,250.00....

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