Question 1123218: A small petroleum company owns two refineries. Refinery 1 costs $20,000 per day to operate, and it can produce 400 barrels of high-grade oil, 300 barrels of medium-grade oil, and 200 barrels of low-grade oil each day. Refinery 2 is newer and more modern. It costs $25,000 per day to operate, and it can produce 300 barrels of high-grade oil, 400 barrels of medium-grade oil, and 500 barrels of low-grade oil each day. The company has orders totaling 25,000 barrels of high-grade oil, 27,000 barrels of medium-grade oil, and 30,000 barrels of low-grade oil. How many days should it run each refinery to minimize its costs and still refine enough oil to meet its orders?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x = the number of days to run refinery 1
let y = the number of days to run refinery 2
your objective function is 20,000 * x + 25,000 * y which equals the total cost to provide the necessary amount of oil.
your constraint functions are:
400x + 300y >= 25000 (hi grade oil requirements)
300x + 400y >= 27000 (medium grade oil requirements)
200x +500y >= 30000 (lo grade oil requirements)
x,y >= 0 (number of days can't be negative).
using the desmos.com calculator, you would graph the opposite of these inequalities.
the area of the graph that is not shaded is your region of feasibility.
the minimum cost will be found by evaluating the objective function at each corner point of the region of feasibility.
the graph looks like this:
the corner points are at (0,83.3333),(25,50),(150,0)
at (0,83.3333), the cost is 0 * 20,000 + 83.33 * 25,000 = 2,083,250.
at (25,50), the cost is 25 * 20,000 + 50 * 25,000 = 1,750,000.
at (150,0), the cost is 150 * 20,000 + 0 * 25,000 = 3,000,000
the minimum cost is achieved when refinery 1 is operated for 25 days and refinery 2 is operated for 50 days at a cost of 1,750,000.
the desmos.com calculator can be found at https://www.desmos.com/calculator
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