582 Words3 Pages

A. The linear program model in this case is solving: Profit= 1.05h+0.75p+1.35s subject to 16h+24.5p+25s ≤ 55296 which is the space constraint and 0.45h+0.75p+0.90s ≤ 1500 which is the cost constraint. Solving for h-p-s ≥ 0, ℎ − 2 ≥ 0, , ℎ, ≥ 0. Doing so produced an optimum solution of h=1250, p=1250, s=0 generating sales of $2250. Linear Program Model- Julia's Food Booth | | h | p | s | | | | Profit (per item) | $ 1.05 | $ 0.75 | $ 1.35 | | | | | | | | | | | | | | | Usage | | Space Available | Space available in warmers | 16 | 24.5 | 25 | 50625.00 | <= | 55296 | Cost | 0.45 | 0.75 | 0.9 | 1500.00 | <= | 1500 | Demand Constraint 1 | -1 | 1 | -1 | 0.00 | >= | 0 | Demand Constraint 2 | 1 | 0 | -2 | 1250.00 | >= | 0 | Units | h | p | s | | | Total | | 1250 | 1250 | 0 | | | $ 2,250.00 |
Therefore, Julia would make more money not selling any barbeque sandwiches and selling only hotdogs and pizza. Considering the cost per game, Julia will have to pay $1000 for booth rent and $100 for the warmers resulting in Julia’s profit per game to be $1150. I recommend for Julia to rent the booth for each game.
B. Sensitivity Report | Adjustable Cells | Final | Reduced | Objective | Allowable | Allowable | Name | Value | Cost | Coefficient | Increase | Decrease | Pizza | 1250 | 0 | 0.75 | 1 | 1.0000 | Hotdogs | 1250 | 0 | 1.05 | 1E+30 | 0.2727 | Barbeque Sandwiches | 0 | -0.3750 | 1.3500 | 0.3750 | 1E+30 | | | | | | | Constraints | Final | Shadow | Constraint | Allowable | Allowable | Name | Value | Price | R.H. Side | Increase | Decrease | Usage | 50625.00 | 0.00 | 55296 | 1E+30 | 4671 | Cost | 1500.00 | 1.50 | 1500 | 138.4 | 1500 | Demand Constraint 1 | 0.00 | -0.38 | 0 | 2000 | 3333.33 | Demand Constraint 2 | 1250.00 |

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