Answer :
Answer: Let's go through each part of the problem:
A) To find the time
ℎ
h when the number of vehicles in the parking lot is at a minimum, we need to find the critical points of the function
(
ℎ
)
=
9
ℎ
3
−
45
ℎ
2
+
44
ℎ
+
59
v(h)=9h
3
−45h
2
+44h+59 and determine whether they correspond to a minimum. We'll do this by taking the derivative of
(
ℎ
)
v(h) and setting it equal to zero.
(
ℎ
)
=
9
ℎ
3
−
45
ℎ
2
+
44
ℎ
+
59
v(h)=9h
3
−45h
2
+44h+59
′
(
ℎ
)
=
27
ℎ
2
−
90
ℎ
+
44
v
′
(h)=27h
2
−90h+44
Setting
′
(
ℎ
)
v
′
(h) equal to zero and solving for
ℎ
h:
27
ℎ
2
−
90
ℎ
+
44
=
0
27h
2
−90h+44=0
Using the quadratic formula:
ℎ
=
90
±
9
0
2
−
4
⋅
27
⋅
44
2
⋅
27
h=
2⋅27
90±
90
2
−4⋅27⋅44
ℎ
=
90
±
8100
−
4752
54
h=
54
90±
8100−4752
ℎ
=
90
±
3348
54
h=
54
90±
3348
ℎ
≈
90
±
57.86
54
h≈
54
90±57.86
So, the critical points are approximately
ℎ
≈
2.35
h≈2.35 and
ℎ
≈
0.61
h≈0.61.
To determine whether these critical points correspond to a minimum, we can use the second derivative test. Calculating the second derivative:
′
′
(
ℎ
)
=
54
ℎ
−
90
v
′′
(h)=54h−90
At
ℎ
=
2.35
h=2.35:
′
′
(
2.35
)
=
(
54
×
2.35
)
−
90
≈
71.9
v
′′
(2.35)=(54×2.35)−90≈71.9
Since
′
′
(
2.35
)
>
0
v
′′
(2.35)>0, this critical point corresponds to a local minimum.
Similarly, at
ℎ
=
0.61
h=0.61:
′
′
(
0.61
)
=
(
54
×
0.61
)
−
90
≈
−
56.3
v
′′
(0.61)=(54×0.61)−90≈−56.3
Since
′
′
(
0.61
)
<
0
v
′′
(0.61)<0, this critical point corresponds to a local maximum.
Therefore, the time after opening when the number of vehicles in the parking lot is at a minimum is approximately
ℎ
≈
2.35
h≈2.35 hours. To find the number of vehicles at that time, we substitute
ℎ
≈
2.35
h≈2.35 into the function
(
ℎ
)
v(h):
(
2.35
)
≈
9
(
2.35
)
3
−
45
(
2.35
)
2
+
44
(
2.35
)
+
59
v(2.35)≈9(2.35)
3
−45(2.35)
2
+44(2.35)+59
(
2.35
)
≈
9
(
13.61
)
−
45
(
5.52
)
+
44
(
2.35
)
+
59
v(2.35)≈9(13.61)−45(5.52)+44(2.35)+59
(
2.35
)
≈
122.49
−
248.4
+
103.6
+
59
v(2.35)≈122.49−248.4+103.6+59
(
2.35
)
≈
−
3.31
v(2.35)≈−3.31
Rounded to the nearest vehicle, there are approximately 3 vehicles in the lot at that time.
B) To find
(
−
0.6
)
v(−0.6), we substitute
ℎ
=
−
0.6
h=−0.6 into the function
(
ℎ
)
v(h):
(
−
0.6
)
=
9
(
−
0.6
)
3
−
45
(
−
0.6
)
2
+
44
(
−
0.6
)
+
59
v(−0.6)=9(−0.6)
3
−45(−0.6)
2
+44(−0.6)+59
(
−
0.6
)
=
9
(
−
0.216
)
−
45
(
0.36
)
−
26.4
+
59
v(−0.6)=9(−0.216)−45(0.36)−26.4+59
(
−
0.6
)
=
−
1.944
−
16.2
−
26.4
+
59
v(−0.6)=−1.944−16.2−26.4+59
(
−
0.6
)
=
14.456
v(−0.6)=14.456
This means that, approximately 14 vehicles were in the lot 0.6 hours before the store opened.
C) To find when the amount of vehicles in the parking lot is decreasing at an increasing rate, we need to analyze the sign of the rate of change
(
ℎ
)
=
27
ℎ
2
−
90
ℎ
+
44
r(h)=27h
2
−90h+44.
First, let's find the critical points by setting
(
ℎ
)
=
0
r(h)=0:
27
ℎ
2
−
90
ℎ
+
44
=
0
27h
2
−90h+44=0
Solving this quadratic equation will give us the critical points.
ℎ
=
90
±
9
0
2
−
4
⋅
27
⋅
44
2
⋅
27
h=
2⋅27
90±
90
2
−4⋅27⋅44
ℎ
=
90
±
8100
−
4752
54
h=
54
90±
8100−4752
ℎ
=
90
±
3348
54
h=
54
90±
3348
ℎ
≈
90
±
57.86
54
h≈
54
90±57.86
So, the critical points are approximately
ℎ
≈
2.35
h≈2.35 and
ℎ
≈
0.61
h≈0.61.
Now, let's analyze the intervals:
For
ℎ
<
0.61
h<0.61,
(
ℎ
)
>
0
r(h)>0, meaning the rate of change is positive, and the number of vehicles is increasing.
For
0.61
<
ℎ
<
2.35
0.61<h<2.35,
(
ℎ
)
<
0
r(h)<0, meaning the rate of change is negative, and the number of vehicles is decreasing.
For
ℎ
>
2.35
h>2.35,
(
ℎ
)
>
0
r(h)>0, meaning the rate of change is positive again, and the number of vehicles is increasing.
Thus, the amount of vehicles in the parking lot is decreasing at an increasing rate on the interval
0.61
<
ℎ
<
2.35
0.61<h<2.35.
D) The model
(
ℎ
)
v(h) is only consistent with the data on the domain
−
0.725
≤
ℎ
≤
72
−0.725≤h≤72, as the store cannot have a negative time or a time greater than 72 hours.
