38. Refining A refinery has two smelters that extract metallic iron from iron or

Question

38. Refining 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 pro- cesses 2000 tons of iron ore per hour and uses 13 megawatts of energy per hour. Each refinery must be operated at least 8 hours per day and, of course, no more than 24 hou the refinery must process at least 30,000 tons of iron ore per day, how many hours should each smelter operate in order to expend as little energy as possible? rs. If

Explanation / Answer

Let the smelters A and B be operated for x and y hours respectively per day.Then 8 x 24 and 8 y24. Also, the production of metallic iron by the smelters A and B are 1000x MT and 2000y MT, per day so that 1000x+2000y = 30000 or, x+2y = 30…(1). Further, the energy consumptions by the smelters A and B are 7x MW and 13y MW per day respectively, so that the daily energy consumption by the refinery is 7x+13y. We have to minimize 7x+13y subject to 8 x 24 and 8 y24 and x+2y = 30. Since x+2y = 30, hence x = 30-2y so that 7x+13y = 7(30-2y)+13y = 210-14y+13y = 210-y. Further, since x = 30-2y 8, we have 30-8 2y or, y 11. Thus, subject to the given conditions, 7x+13y =210-y will be minimum if x = 8 and y = 11. Then the metallic iron prpduced will be 8000+22000 = 3000MT per day with the minimum energy consumption of 7*8 +13*11 = 56+143 = 199 MW per day. Hence, under the given conditions, the smelters A and B should be operated for 8 hours and 11 hours per day respectively for minimum energy consumption.

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