Let there be x operations at Refinery 1 and y operations at the Refinery 2.
(b). The total amount of A produced is 2x+5y.
( c). The total amount of B produced is x+y.
(d). The total amount of C produced is 5x+2y
(e). In view of the requirements of the consumer, we have 2x+5y ≥ 1040…(1), x+y ≥ 340…(2) and 5x+2y ≥ 890…(3).
A graph of the lines 2x+5y = 1040 (in red), x+y = 340 (in blue) and 5x+2y = 890 (in green) is attached. The feasible region is on and above all these 3 lines.
The blue and the green lines intersect at A: ( 70,270) and the blue and the red lines intersect at B: (220,120). Thus, the only feasible basic solutions are (70,270) and (220,120).
(f). At A, the cost of operation of the 2 refineries is 70*200+270*350 = 14000+94500 = $ 108500.
At B, the cost of operation of the 2 refineries is 220*200+120*350 = 44000+42000 = $ 86000.
The cheapest cost is $ 86000 from 220 operations at Refinery 1 and 120 operations at the Refinery 2.
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