Chapter 1.7, Problem 10CFU

To determine

To find: the less expensive parking lot for the given situation.

Answer to Problem 10CFU

Shuttle parking

Explanation of Solution

Given information:

The fee of parking for Long term lot: $1 per hour or fraction thereof, or $6 max per day

The fee of parking for Shuttle’s lot: $4 a day or fraction thereof

Duration of car parked $=2\text{ days 3 hours}=51\text{ hours}=2\frac{3}{24}\text{ days}$

Calculation:

Let the Long term parking lot be represented as L, and Shuttle’s parking lot be represented by S.

Case 1: Long term parking

The function would be as follows:

  $L\left(h\right)=\{\begin{array}{l}1〚h+\frac{3599}{3600}〛,\text{ if }0<h\le 6\\ 6〚\frac{h}{24}+\frac{3599}{3600}〛,\text{ if }h>6\end{array}$

Put $h=51$ and calculate $L\left(51\right)$ as follows:

  $\begin{array}{c}L\left(51\right)=6〚\frac{51}{24}+\frac{3599}{3600}〛\\ =6〚2\frac{450}{3600}+\frac{3599}{3600}〛\left[\text{Taking common denominator}\right]\\ =6〚3\frac{449}{3600}〛\\ =6\left(3\right)\\ =18\end{array}$

The 2 days 3 hour parking fee for Long-term lot is $18.

Case 2: Shuttle parking

The function would be as follows:

  $S\left(d\right)=4〚d+\frac{86399}{86400}〛$

Put $d=2\frac{3}{24}$ and calculate $S\left(2\frac{3}{24}\right)$ as follows:

  $\begin{array}{c}S\left(2\frac{3}{24}\right)=4〚2\frac{3}{24}+\frac{86399}{86400}〛\\ =4〚2\frac{10800}{86400}+\frac{86399}{86400}〛\left[\text{Taking common denominator}\right]\\ =4〚3\frac{10879}{86400}〛\\ =4\left(3\right)\\ =12\end{array}$

The 2 days 3 hour parking fee for Shuttle’s lot is $12.

Therefore, Shuttle parking lot is less expensive for 2 days 3 hours.