Question

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

Answer 1

To minimize costs, the foundry should purchase two railroad cars from Birmingham and one car from Atlanta daily, totaling $40,000 per day. This satisfies the daily requirement of at least 4 tons of lead and 2.5 tons of copper.

Step-by-step explanation:

To minimize the long-range scrap metal cost for a foundry that needs at least 4 tons of lead and 2.5 tons of copper per day, we must determine the optimal number of railroad cars to be purchased daily from each source, Atlanta and Birmingham. A car from Atlanta yields 1 ton of copper and 1 ton of lead for $10,000, while a car from Birmingham yields 1 ton of copper and 2 tons of lead for $15,000.

The foundry requires at least 2.5 tons of copper. Purchasing two cars from Birmingham gives 2 tons of copper and 4 tons of lead, fulfilling the lead requirement but falling short of the copper requirement by 0.5 tons. Buying one more car from Atlanta will provide the additional 0.5 tons of copper and another ton of lead, which will meet both material requirements.

The total cost for this approach would be buying two cars from Birmingham for $30,000 and one car from Atlanta for $10,000, resulting in a total of $40,000 per day. This is the minimum cost to meet the daily requirement of at least 4 tons of lead and 2.5 tons of copper.

Answer 2

The answer is below

Step-by-step 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).