Question

A construction company is building a new parking garage and is charging the following rates: $5000 a month for the first 3 months; $8000 a month for the next 3 months; $5000 in total for the last 4 months when the garage will be completed. This amount will be paid in a lump sum at the end of the 6th month. Express the cost C( in thousands of dollars) as a function of time t (in months) that the construction company works on the parking garage.

Answer 1

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!!!