A refinery has two smelters that extract metallic iron from iron ore:

- Smelter A processes 1000 tons of iron ore per hour and uses 7 megawatts of energy per hour.
- Smelter B processes 2000 tons of iron ore per hour and uses 13 megawatts of energy per hour.

Each smelter must operate at least 8 hours per day and no more than 24 hours.

If the refinery must process at least 30,000 tons of iron ore per day, how many hours should each smelter operate to minimize energy expenditure?

To minimize energy expenditure, only Smelter Operation Optimization B should be operated for 15 hours, as it alone can fulfill the 30,000 tons iron ore processing requirement within the 24-hour operating limit.

Explanation:

The problem involves finding the most energy-efficient way for the smelters to operate to process at least 30,000 tons of iron ore. Since Smelter B is more efficient (it processes 2000 tons per hour while consuming only 13 megawatts compared to Smelter A's 7 megawatts for only 1000 tons per hour), we would prioritize running Smelter B where possible.

To meet the minimum requirement of processing 30,000 tons of iron ore, Smelter B would run for 15 hours (since 2000 tons/hour * 15 hours = 30,000 tons) leaving us with 9 hours for Smelter A to operate if needed without exceeding the 24-hour limit. Since Smelter B can handle the entire requirement by itself, there is no need to run Smelter A in this case.

However, if there was a situation where Smelter B couldn't handle the entire workload, the remaining load could be accommodated by Smelter A by calculating the remaining tons divided by Smelter A's efficiency (in tons/hour).

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