C = 12000 p + 10000 q
40 p + 80 q >/= 2000
40 p + 30 q >/= 1500
12 p + 40 q >/= 2400
or
p + 2 q >/= 50
4 p + 3 q >/= 150
3 p + 10 q >/= 60
Graph those lines
You will see that the First two lines are way up there to the right and above the third. Therefore only the first two need be investigated
They intersect at:
4 p + 8 q = 200
4 p + 3 q = 150
-----------------
5 q = 50
q = 10
p = 30 so (p,q) at (30,10)
also axis intersections at (0,50) and (50,0)
test those three for cost
12p + 10 q or 6 p + 5 q
at intersection (30,10)
6*30+5*10 =230
at (0,50)
6*0 + 5*50 = 250
at (50,0)
6*50 + 0 = 300
so min cost at (30,10)
An airline will provide accommodations for a minimum of 2000 first-class, 1500 tourist, and 2400 economy-class passengers. Airplane P-1 costs $12,000 per miles to operate and can accommodate 40 first-class, 40 tourist, and 12 economy-class passengers. Airplane Q-2 costs $10,000 per miles to operate and can accommodate 80 first-class, 30tourist, and 40 economy-class passengers. How many of each type of plane should be used to minimize the operating cost?
1 answer
1 answer