Chapter 4.1, Problem 41PS

Solution

a.

To write: A model giving the price $p$ (in dollars) of a ticket $t$ years after 2000.

  $p=48.28{\left(1.06\right)}^t$ .

Given:

The average price of a ticket is $48.28.

The price is increased by 6% per year for the next 4 years.

Calculation:

Here, the initial price is

  $a=48.28$

The growth rate is,

  $\begin{array}{c}r=\frac{6}{100}\\ =0.06\end{array}$

Now the model is,

  $\begin{array}{c}p=a{\left(1+r\right)}^t\\ p=48.28{\left(1+0.06\right)}^t\\ p=48.28{\left(1.06\right)}^t\end{array}$

b.

To graph: The model and to estimate the year in which the price of a ticket is $60.

  

The price of the ticket is $60 in the year 2004.

Given:

  $p=48.28{\left(1.06\right)}^t$

Calculation:

The graph of the model passes through the points $\left(0,48.28\right),\left(1,51.1768\right)$ .

Now the graph is,

  

From the graph, it can be seen that the price of ticket is $60 in about four years after 2000.

Hence, the price of the ticket is $60 in the year 2004.

c.

To graph: The model and to estimate the year in which the price of a ticket is $60.

The minimum value is $48.28 and the maximum value is $61.

Given:

  $p=48.28{\left(1.06\right)}^t$

Calculation:

From the graph in subpart (b), it can be seen that $t=0$ represents the year 2000, and $t=4$ represents the year 2004.

So, the domain is $\left[0,4\right]$ .

In this domain values, the price of the ticket ranges from $48.28 to $61.

So, the minimum value is 48.28 and the maximum value is 61.