# The Possibility Restaurant

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“The Possibility” Restaurant Problem Angela Fox and Zooey Caulfield were food and nutrition majors at State University, as well as close friends and roommates. Upon graduation Angela and Zooey decided to open a French restaurant in Draperton, the small town where the university was located. There were no other French restaurants in Draperton, and the possibility of doing something new and somewhat risky intrigued the two friends. They purchased an old Victorian home just off Main Street for their new restaurant, which they named "The Possibility". Angela and Zooey knew in advance that at least initially they could not offer a full varied menu of dishes. They had no idea what their local customers' taste in French cuisine would be, so…show more content…
A) Formulate a Linear programming model for Angela and Zooey that will help then estimate the number of meals they should prepare each night and solve this model graphically. B) If Angela and Zooey increased the menu price on the fish dinners so that the profit for both dinners was the same, what effect would that have on their solution? Suppose Angela and Zooey reconsidered the demand for beef dinners and decided that at least 20% of their customers would purchase beef dinners. What effect would this have on their meal preparation plan? Part A: Variables: X1 = Fish Dinners X2 = Beef Dinners The Objective Function: Max Z = 12X1 + 16X2 Constraints: X1 + X2 &lt;= 60 (Maximum meal each night) 0.25X1 + 0.50X2 &lt;= 20 (Maximum labor hours allowed each day) 2X1 – 3X2 &gt;= 0 (Ratio 3 Fish Dinners to 2 Beef Dinners) 0.9X2 – 0.1X1 &gt;= 0 (At least 10% of the customers will have the Beef Dinners) X1, X2 &gt;= 0 (non-negativity constraint) After inputting the variables and constraints in Excel and Solvers, the optimal point is point C, X1 = 40 and X2 = 20 and the maximum profit is \$800 a day. Point A: Point B: Point C: Point D: Point E: X1 = 0 X1 = 34 X1 = 40 X1 = 54 X1 = 60 X2 = 40 X2 = 22 X2 = 20 X2 = 6 X2 = 0 Z = \$640 Z =\$760 Z = \$800 Z = \$744 Z = \$720 Part B1 (20% more beef dinners): The Objective Function: Max Z = 12X1 + 16X2 Constraints: X1 + X2 &lt;= 60 (Maximum meal each