An oil company has two refineries. Refinery A produces 200 barrels of high-grade oil, 300 barrels of medium-grade oil, and 200 barrels of low-grade oil and costs $12,000 to operate. Each day, Refinery B produces 100 barrels of high-grade oil, 100 barrels of medium-grade oil, and 200 barrels of low-grade oil and costs $10,000 to operate. The company must produce at least 800 barrels of high-grade oil, 900 barrels of medium-grade oil, and 1,000 barrels of low-grade oil. What is the minimum cost to operate both refineries?

$56,000

Step-by-step explanation:

If x is the number of days that Refinery A is in operation, and y is the number of days that Refinery B is in operation, then:

200x + 100y ≥ 800

300x + 100y ≥ 900

200x + 200y ≥ 1000

The total cost is:

C = 12,000x + 10,000y

x and y must be integers.  Possible combinations are:

x = 0, y = 9

x = 1, y = 6

x = 2, y = 4

x = 3, y = 2

x = 4, y = 1

x = 5, y = 0

The costs of these combinations are:

The minimum cost is $56,000.

1/4 and 8/3 because:

the reciprocal if 4 is 1/4 because 4 is used as 4/1 and the opposite of that is 1/4. the opposite of 3/8 is 8/3 so that is the answer.