6. 6. 6. The average speed of a vehicle on a stretch of Route 134 between 6 a.m.

Question

6. 6. 6. The average speed of a vehicle on a stretch of Route 134 between 6 a.m. and 12 p.m. on a typical weekday is approximated by the function f(t) = 14t - 28 t + 38 (0 t 6) where f (t) is measured in miles per hour, and t is measured in hours, with t = 0 corresponding to 6 a.m. At what time of the morning commute is the traffic moving at the slowest rate? What is the average speed of a vehicle at that time? Lynbrook West, an apartment complex, has 100 two-bedroom units. The monthly profit (in dollars) realized from renting out x apartments is given by the following function. P(x) = -11x2 + 1958x - 40,000 To maximize the monthly rental profit, how many units should be rented out? What is the maximum monthly profit realizable?

Explanation / Answer

Slowest rate is similar as to finding the minimum. So finding the slowest rate is the same as finding the minimum value of the function f(t) = 10t - 20?(t) + 30 with domain 0 ? t ? 6. The minimum can occur at

1. Critical points (where f ' (t) = 0)
2. Points where f isn't differentiable (where f ' (t) does not exist)
3. End points of the domain

1. Critical Points.

Let's find f ' (t). Let's rewrite ?(t) as t ^ (1/2).

f (t) = 14t - 28t^(1/2) + 30
f ' (t) = 14 - 28(1/2)*t^(-1/2) + 0

Now set f ' (t) = 0 and solve for t.

0 = 14 - 14t^(-1/2)

Add 10t^(-1/2) to both sides.

14t^(-1/2) = 14

Divide both sides by 10

t^(-1/2) = 1

Mulitply both sides by t^(1/2)

t^(1/2) * t^(-1/2) = t^(1/2) * 1
t^(1/2 + -1/2) = t^(1/2)
t^(0) = t^(1/2)
1 = t^(1/2)

Square both sides

1^2 = (t^(1/2))^2
1 = t^((1/2)*2)
1 = t^1
t = 1 and this is in our domain.

2. Points where f ' isn't differentialbe.

Since f ' (t) = 14 - 28t^(-1/2) = 14 - 28 / t^(1/2), we know that t cannot be negative nor 0. Only 0 is in our domain.

3. End points of the domain

The end points are 0 and 6.

Therefore, our only possibities of acheiving a minimum are t = 0, t = 1, and t = 6.

Our minimum is the min {f(0), f(1), f(6)}

f(0) = 14*0 - 28?(0) + 30 = 30
f(1) = 14*1 - 28?(1) + 30 = 16
f(6) = 14*6 - 28?(6) + 30 = 90 - 20?(6) > 30 because
90 - 20?(6) > 90 - 20?(9) = 90 - 20*3 = 30

So when t = 1 this corresponds to 7:00 a.m. which is the time when traffic is moving at the slowest rate. The average speed at this time is f(1) = 16 mph.

Thus our answers are

a) 7:00 a.m.
b) 16

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The max will be when dP/dx = 0

dP/dx = -22x + 1958 = 0
x = 1958/22 = 89

So 89 units need to be rented out to maximize profits.

Just plug x = 88 into the eqution for P to get the actual profit.
P = -11(88)^2 + 1958(88) - 40000
P = 47120 dollars

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