Answer:
$56,000
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:
![\left[\begin{array}{ccc}x&y&C\\0&9&\$90,000\\1&6&\$72,000\\2&4&\$64,000\\3&2&\$56,000\\4&1&\$58,000\\5&0&\$60,000\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/yyxczuygl44cmrjh2i3umjjwdcmlnner8a.png)
The minimum cost is $56,000.