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A foundry is developing a long-range strategic plan for buying scrap metal for its operations. The foundrycan buy scrap metal in unlimited quantity from two sources: Atlanta and Birmingham, and it receives thescrap daily by railroad cars.The scrap is melted down, and lead and copper are extracted. Each railroad car from Atlanta yields 1 ton ofcopper and 1 ton of lead, and costs $10,000. Each railroad car from Birmingham yields 1 ton of copper and2 tons of lead, and costs $15,000. The foundry needs at least 4 tons of lead and at least 2.5 tons of copperper day for the foreseeable future.1. In order to minimize the long-range scrap metal cost, how many raiload cars of scrap should bepurchased per day from each source

Sagot :

raphealnwobi

Answer:

The answer is below

Explanation:

Let x represent the number of railroad cars of scrap purchased per day from Atlanta and let y represent the number of railroad cars of scrap purchased per day from Birmingham.

Since Atlanta yields 1 ton of copper and 1 ton of lead while Birmingham yields 1 ton of copper and 2 tons of lead.

The foundry needs at least 2.5 tons of copper per day. Hence:

x + y ≥ 2.5      (1)

The foundry needs at least 4 tons of lead per day. Hence:

x + 2y ≥ 4      (2)

Plotting equations 1 and 2 using geogebra online graphing tool, we get the points that is the solution to the problem as:

(0, 2.5), (4, 0), (1, 1.5)

Car from Atlanta cost $10000 while car from Birmingham costs $15000. Therefore the cost equation is:

Cost = 10000x + 15000y

We are to find the minimum cost:

At (0, 2.5): Cost = 10000(0) + 15000(2.5) = $37500

At (4, 0): Cost = 10000(4) + 15000(0) = $40000

At (1, 1.5): Cost = 10000(1) + 15000(1.5) = $32500

The minimum cost is at (1, 1.5).

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