In studying of traffic flow at a highway toll booth over the course of 60 minutes, it is determined that the arrival and departure rates are deterministic, but not uniform. The arrival rate is found to vary according to the function A(t) = 1.8 + 0.25t - 0.0030t^2. The departure rate function is D(t) = 1.4 + 0.11t. In both of these functions, t is in minutes after the beginning of the observation and A(t) and D(t) are in vehicles per minute. At what time does the maximum queue length occur?
a. 2.7 min
b. 9.4 min
c. 49.4 min
d. 60.0 min

Question

contestada

Respuesta :

DeniceSandidge DeniceSandidge · Answer 1 · 2020-01-23T11:17:25+01:00

Answer:

correct answer is c. 49.4 min

Explanation:

given data

course time = 60 minutes

arrive rate A(t) = 1.8 + 0.25t - 0.0030t²

departure rate function D(t) = 1.4 + 0.11t

to find out

maximum queue length occur

solution

first we get here total vehicle arrive after time = t is

total vehicle = [tex]\int_{0}^{t}1.8 + 0.25t - 0.0030t^2 dt[/tex]

total vehicle = 1.8t + [tex]\frac{0.25t^2}{2} +\frac{0.003t^3}{3}[/tex] .............1

and

now we get here here total vehicle departed after time = t is

total vehicle departed = [tex]\int_{0}^{t}1.4 + 0.11t dt[/tex]

total vehicle departed = 1.4 t + [tex]\frac{0.11t^2}{2}[/tex] ..............2

so length at timer t is equation 1 - equation 2

so length at timer t = 1.8t + [tex]\frac{0.25t^2}{2} +\frac{0.003t^3}{3}[/tex] - 1.4 t + [tex]\frac{0.11t^2}{2}[/tex]

and here for maximum length [tex]\frac{dQ}{dt}[/tex] = 0

so put here equation = 0 we get t

1.8t + [tex]\frac{0.25t^2}{2} +\frac{0.003t^3}{3}[/tex] - 1.4 t + [tex]\frac{0.11t^2}{2}[/tex] = 0

solve it we get

t = 49.40 min

so correct answer is c. 49.4 min