Michael P. answered • 10/11/19

Enthusiastic Math Tutor, Mechanical Engineer, and Mentor

Part A:

We are basically given f(x) and asked to solve for x:

f(x)=15 seconds = 0.25 minutes = (x-5)/(x^2-10x)

Multiply both sides by that denominator:

.25x^2-2.5x=x-5

Combine like terms:

.25x^2-3.5x+5=0

Use the quadratic equation to solve for x

x=[-(-3.5)+- sqrt((-3.5^2)-4(.25)(5))]/2(.25)

x= 1.61 or x = 12.39

This equation is only good for x>10, so x = 12.39 cars through the gate per minute

Part B:

We just solved for where f(x) = .25 minutes, and found that at least 12.39 cars need to be let through the gate per minute to keep f(x) at or below 0.25 minutes, so now, to reframe the question, how do we get at least 12.39 cars through the gate per minute? With attendants. Because each attendant can get 5 cars through the gate per minute, we need:

(12.39 cars/minute)/(5 cars/minute) = at least 2.47 attendants, so at least 3 attendants