Problema Solution
To prevent traffic jams, a city funds a courtesy patrol to aid stranded drivers on local roads. The patrol can repair a flat tire, provide the motorist with 2 gallons of gas, or call a tow truck for more serious problems. It takes 15 minutes to help a driver who is out of gas and 45 minutes to help a driver with a flat tire. The courtesy patrol driver carries 28 gallons of gas. What is the maximum number of stops for flat tires or empty gas tanks that the courtesy patrol can make in an 8-hour shift?
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let
g = the number of empty gas tanks, g>=0
t = the number of flat tires, t>=0
the courtesy patrol driver carries 28 gallons of and provides the motorist with 2 gallons of gas thus
2g <= 28 divide both sides by 2
g <= 14
It takes 15 minutes to help a driver who is out of gas and 45 minutes to help a driver with a flat tire and the maximum total time is 8 hours shift that is 8*60 = 480 min
15g + 45t <=480 divide both sides by 15
g + 3t <= 32
thus the constrains are
g <= 14
g + 3t <= 32
g >=0
t >= 0
the objective function is
F(g , t) = g + t
we need to find the maximum of F(g, t)
by graphing the inequalities
g <= 14
g + 3t <= 32
g >=0
t >= 0
click here to see the graph
we find the critical points (14, 0), (0, 32/3) and (6,14)
in the point (14, 0) we have F(14, 0) = 14
in the point (0, 32/3) we have F(0, 32/3) = 32/3 <= 11
in the point (6, 14) we have F(6, 14) = 20
the maximum number of stops for flat tires or empty gas tanks that the courtesy patrol can make in an 8-hour shift is 20.