Enrollment in a particular class for the last four semesters has been 122, 128, 100, and 155 (listed from oldest to most recent)
1 answer:
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Answer:
the lowest operating cost is achieved when mine 1 is operated for 0 hours and mine 2 is operated for 2 hours.
Explanation:
Mining hours of mine 1: M1
mining hours of mine 2: M2
Objective function: Minimize operating cost < 200M1 + 160 M2
Explicit constraints:
6M1 + 2M2 ≥ 12
2M1+ 2M2≥ 8
4M1+ 12M2 ≥ 24
Implicit constraints:
M1> 0
M2>0
See the attachment for feasible region
The coordinates of feasible region are (0,0), (2,0),(0,2),(1.5,1.5)
To maximize cost substitute each of these coordinates in objective function
for(0,0)
cost=0 (neglect this cost since M1>0 and M2 >0)
for(2,0)
cost= 320
for (0,2)
cost= 400
for (1.5,1.5)
cost= 540
the lowest operating cost is achieved when mine 1 is operated for 0 hours and mine 2 is operated for 2 hours
