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HOTaco1837
22.09.2020 •
Mathematics
A metal refining factory has a waste disposal problem. For 1 kg of metal produced, 3 kg of waster are created. The waste is contained in wastewater at a concentration of 2 kg/m3. The regulatory authority has imposed an effluent maximum standard of 100,000 kg/week that the company may discharge into a nearby river. The factory has a production capacity of 55,000 kg of metal per week. Metal is sold at a price of $1.30/kg and production costs are $0.90/kg.
The factory's wastewater treatment facility has a maximum capacity of 70,000 m^3/week, however the facility s efficiency (fraction of waste removed) varies with the waste loading. If W is the wastewater inflow in 104 m^3/week, then for W between 0-70,000 m%3/week, the treatment efficiency is 1-0.06*W. Thus more wastewater in the treatment plant means less efficiency at removing the waste. Wastewater treatment costs are $0.20/m^3.
The metal factory needs to meet two objectives:
Obj. 1: discharge no more than 100,000 kg/week of waste into the river
Obj. 2: maximize profits (S/week)
Consider the following alternatives:
a. For a production of 50,000 kg/week of metal and treatment of 100,000 kg/week of waste in the treatment plant, does this meet both objectives? What are the expected total effluent (kg/week) to the river and the total profits ($/week)?
b. Suppose the metal factory decides to prioritize minimizing waste to the river, but Still needs to be profitable to cover labor, etc. (profit $1000/week.) How much metal can they produce and how much effluent to the river is this?
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Ответ:
36 months or 3 years.
Step-by-step explanation:
Her Monthly expenses are $3,000.
Her 6 month expenses will be = 3,000 * 6 = $18,000
She saves 10% of her $5,000 income each month = 10% * 5,000 = $500
Saving $500 to get to $18,000;
= 18,000/500
= 36 months
= 3 years