Rayhoon Restaurant Essay

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Rayhoon Restaurant Quantitative Methods Math 540 September 02, 2014 Rayhoon Restaurant This is a Maximization Linear Programming problem because we need to work out the combination of meals using given constraints to maximize profits. There are four constraints in given problem. 1. Maximum number of meals which can sold each night 2. Kitchen staff labor hours is another constraint 3. Time required to prepare a Chicken Meal & Pork Meal. 4. Number of Chicken Meals which will be sold relative to Pork Meals Let X1 be Number of Chicken Meals & X2 be Number of Pork meals. Let Maximum Profit be Z. They make $12 profit per Chicken meal & $16 per Pork meal. So Linear programming model formulation is Maximize = Z = $12x1 + 16x2 x1 + x2 ≤ 60 (No of meals) Chicken requires 15/60 Min = 0.25 hrs. as compared to Pork which takes twice as long = 30/60 Min = 0.50 hrs. 0.25x1 + 0.50x2 ≤ 20 (Total Kitchen Hrs.) x1/x2 ≥ 3/2 or 2x1 – 3x2 ≥ 0 (Sell at least 3 Chicken for 2 pork meals) x2/(x1 + x2) ≤ 0.10 (At least 10% will order Pork) or x2 ≤ 0.10x1 + 0.10X2 or -0.10x1 + 0.90x2 ≥ 0 Non-Negativity constraints x1,x2 ≥ 0 Rayhoon Restaurant X1 X2 Number of Meals 40 20 USED RHS Profit 12 16 Meals 1 1 <= 60 60 Kitchen Hrs. 0.25 0.5 <= 20 20 Sell 2 -3 >= 20 0 Orders -0.1 0.9 >= 14 0 Maximize $ 800 They should prepare 40 Chicken & 20 Pork meals each day for maximum profit of $800. A. They are considering investing in some advertising to increase the maximum number of meals they serve. They estimate that if they spend $30 per day on a newspaper ad, it will increase the maximum number of meals they serve per day from 60 to 70. Should they make the investment? The linear programming model formulation with advertising cost is Maximize = Z = $12x1 + 16x2 - 30 x1 + x2 ≤ 70 (No of meals

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