# Julia’s Food Booth Case Problem

Topics: Food, Linear programming, Money Pages: 3 (649 words) Published: March 14, 2013
Julia’s Food Booth Case Problem
MAT 540- Quantitative Methods
February 23, 2013

(A) Formulate and solve an L.P. model for this case.
The following variables were be used:
X1 = Slices of Pizza
X2 = Hot Dogs
X3 = BBQ Sandwiches
The objective is to maximize profit.
maximize Z= 0 .75X1+1.05X2+1.35X3
Subject to:
0.75X1+1.05X2+1.35X3≤1,500 (Budget)
24X1+16X2+25X3≤55,296in2 (Oven Space)
X1≥X2+X3
X2X3≥2.0
X1, X2, X3≥0
(B) Evaluate the prospect of borrowing money before the first game. Following table shows the sensitivity report from excel solver | | | | | | |
| | | | | | | | |
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 | | | | | | | |
| | | | | | | | |

| | | | | | | |
Adjustable Cells| | | | | |
|  |  | Final| Reduced| Objective| Allowable| Allowable| | Cell| Name| Value| Cost| Coefficient| Increase| Decrease| | \$B\$12| Pizza=| 1250| 0| 0.75| 1| 1.00|
| \$B\$13| Hot Dogs=| 1250| 0| 1.05| 1E+30| 0.27|
| \$B\$14| Barbecue=| 0| 0| 1.35| 0.375000011| 1E+30| | | | | | | | |
Constraints| | | | | |
|  |  | Final| Shadow| Constraint| Allowable| Allowable| | Cell| Name| Value| Price| R.H. Side| Increase| Decrease| | \$G\$9| Demand Usage| 1,250.00 | - | 0| 1250| 1E+30| | \$G\$7| Oven...

References: Taylor, B. M. (2010). Introduction to management science (11th ed.). Upper Saddle River, NJ: Pearson/Prentice Hall.