A parking garage is changing its parking fees for a week long festival. The function below represents the cost to park in the garage as a function of time (days into the festival).
Piece wise
c(t) =
10; 0 <= t <= 2
10+ t; 2 < t <= 5
7; 5 < t <= 7
a) What would be the best days to park in the garage?
b) calculate c(4) - c(2)
c(10) = 10 +4 = 14
c(2) = 10
14 -10 = 4
To determine the best days to park in the garage, we need to analyze the function piece by piece:
a) For days 0 to 2 (inclusive), the cost to park is a constant $10 per day.
b) For days 2 to 5 (inclusive), the cost to park increases linearly with time, starting from $10 and increasing by $1 for each additional day.
c) For days 5 to 7 (inclusive), the cost to park is a constant $7 per day.
Now, let's determine the best days to park in the garage:
- Days 0 to 2: The cost is $10 per day, which is the lowest the cost will ever be during the week-long festival. So, parking on any of these days would be a good choice.
- Days 2 to 5: The cost increases by $1 each day during this period, so parking on day 2 would be the cheapest option within this time frame.
- Days 5 to 7: The cost is a constant $7 per day, which is lower than the cost during days 2 to 5. Therefore, parking on any of these days would be more cost-effective than days 2 to 5.
In conclusion, the best days to park in the garage would be on days 0 to 2 and days 5 to 7.
To calculate c(4) - c(2), we need to find the cost on day 4 and subtract the cost on day 2.
c(4) = 10 + 4 = 14
c(2) = 10
Therefore, c(4) - c(2) = 14 - 10 = 4.