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Problem 392

# In order to produce 1000 tons of non-oxidizing steel for engine valves, at least the following units of manganese, chromium and molybdenum, will be needed weekly: 10 units of manganese, 12 units of chromium, and 14 units of molybdenum (1 unit is 10 pounds). These metals are obtainable from dealers in non-ferrous metals, who, to attract markets make them available in cases of three sizes, $$\mathrm{S}, \mathrm{M}$$ and $$L$$. One S case costs $$\ 9$$ and contains 2 units of manganese, 2 units of chromium and 1 unit of molybdenum. One $$\mathrm{M}$$ case costs $$\ 12$$ and contains 2 units of manganese, 3 units of chromium, and 1 unit of molybdenum. One $$L$$ case costs $$\ 15$$ and contains 1 unit of manganese, 1 unit of chromium and 5 units of molybdenum. How many cases of each kind should be purchased weekly so that the needed amounts of manganese, chromium and molybdenum are obtained at the smallest possible cost? What is the smallest possible cost?

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To produce the necessary amount of manganese, chromium, and molybdenum at the smallest possible cost, the company should purchase no S cases, 4 M cases, and 2 L cases per week. The lowest cost for these purchases will be $78 per week. See the step by step solution ### Step by step solution ## Unlock all solutions Get unlimited access to millions of textbook solutions with Vaia Premium Over 22 million students worldwide already upgrade their learning with Vaia! ## Step 1: Define the variables Let's define the variables: $$x_s$$ be the number of S cases purchased weekly, $$x_m$$ be the number of M cases purchased weekly, and $$x_l$$ be the number of L cases purchased weekly. ## Step 2: Set up the objective function We want to minimize the cost, which is the sum of the cost of the S, M, and L cases. Our objective function will be built as follows: Cost = $$9x_s + 12x_m + 15x_l$$ Our goal: Minimize Cost($$C) = 9x_s + 12x_m + 15x_l$$ ## Step 3: Set up the constraints Now, we will formulate the constraints based on the given requirements: 1. Manganese constraint: $$2x_s + 2x_m + 1x_l \ge 10$$ 2. Chromium constraint: $$2x_s + 3x_m + 1x_l \ge 12$$ 3. Molybdenum constraint: $$1x_s + 1x_m + 5x_l \ge 14$$ 4. Non-negativity constraint: $$x_s \ge 0, x_m \ge 0, x_l \ge 0$$ ## Step 4: Solve using linear programming techniques At this point, we need to use linear programming techniques to solve the problem. You can use a graphical method or software to find the solution. In this case, solving using linear programming techniques, we obtain the optimal solution: $$x_s = 0$$ $$x_m = 4$$ $$x_l = 2$$ ## Step 5: Calculate the minimum cost We found the number of cases to purchase for each type: S = 0, M = 4, and L = 2. Using the cost function, we can compute the minimum cost: Minimum Cost = $$9(0) + 12(4) + 15(2) = 0 + 48 + 30 =$$78 ## Step 6: Conclusion To obtain the needed amounts of manganese, chromium, and molybdenum at the smallest possible cost, there should be no S cases purchased, 4 M cases, and 2 L cases. The smallest possible cost is$78 per week.

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