Answer:
C(t) = 5t for 0 < t ≤ 3
C(t) = (8t-9) for 3 < t ≤ 6
C(t) = 44 for 6 < t ≤ 10
C is given in thousands of dollars and t is given in months.
Total lump sum to be paid at the end of the 6 months for the total 10 months that the construction company works on the parking garage = $44000
Step-by-step explanation:
We will do a piecewise analysis for this cost function.
How to obtain the total cost changes with each time interval.
In the first 3 months,
C(t) = 5t for 0 < t ≤ 3
Note that C is given in thousands of dollars
In the next 3 months,
C(t) = (8t-9) for 3 < t ≤ 6
- Here, it gets a little complex, we use the upper limit of the previous interval and the lower limit of this new interval to get the constant to be subtracted from the normal 8t that characterizes this interval.
8(3) = 24
5(3) = 15
constant = 24 - 15 = 9
In the last 4 months,
C(t) = 44 for 6 < t ≤ 10
- For the last four months, it is a single sum of $5000 plus the [8(6) - 9] from the previous inteval
C = [8(6) - 9] + 5 = 39 + 5 = 44
So, the cost of parking, tracked month after month gives
T | Cost (in thousands of dollars)
1 | 5
2 | 10
3 | 15
4 | 23
5 | 31
6 | 39
7 | 44
8 | 44
9 | 44
10 | 44
Total lump sum to be paid at the end of the 6 months for the total 10 months that the construction company works on the parking garage = $44000
Hope this Helps!!!
