Julia’s Food Booth

A.Formulate and solve an L.P. model for this case. Decision Variables | X1 | # of Pizzas | X2 | # of Hotdogs | X3 | # of BBQ Sandwiches | Objective Function | | X1 | X2 | X3 | Selling Price | $ 1.50 | $ 1.50 | $ 2.25 | Cost | $ 0.75 | $ 0.45 | $ 0.90 | Profit | $ 0.75 | $ 1.05 | $ 1.35 |

Maximize total profit Z=$0.75X1+$1.05X2+$1.35X3
Budget Constraints: $0.75X1+$0.45X2+$0.90X3<=$1,500
Space Constraints: 3x4x16= 196 Sq. Ft., 192x12x12=27,648 In Sq.
Oven will be filled at the beginning and halftime of the game: 2x27, 648=55,296 In. Sq.
Space for Pizza: 14x14=196 In. Sq., 196/8(Slices) =24.5 In. Sq., Round down to 24
Space in oven constraint: 24X1+16X2+25X3<=55,296
She can sell at least as many pizzas as hotdogs and BBQ sandwiches: X1>=X2+X3
She can sell twice as many hotdogs as BBQ sandwiches: X2/X3>=2, X2-2X3>=0
X1, X2, X3 >=0

Linear Program: | | | | | Maximize total profit Z=$0.75X1+$1.05X2+$1.35X3 | Subject to the following constraints: | | | 1 | $0.75X1+$0.45X2+$0.90X3<=$1,500 | | 2 | 24X1+16X2+25X3<=55,296 inch squared | | 3 | X1-X2-X3>=0 | | | | 4 | X2-2X3>=0 | | | | | 5 | X1, X2, X3 >=0 | | | |

Linear Program Solved:
Maximize Total Profit Z=$0.75(1,250) +$1.05(1,250) +$1.35(0)
Maximize Total Profit Z=$2,250

Constraints: 1. $0.75(1,250)+$0.45(1,250)+$0.90(0)<=$1,500
$1,500<=$1,500
2. 24(1,250)+16(1,250)+25(0)<=55,296
50,000<=55,296
3. 1,250-1,250-0>=0
0>=0
4. 1,250-2(0)>=0
1,250>=0
5. 1,250, 1,250, 0>=0

The linear program that has been created above demonstrates what Julia needs to in order to turn a profit in her food business. The decision variables that need to be solved in this case is the number of pizza slices,