Chapter 11.3, Problem 67E

(a)

To determine

To find: A quadratic model using a graphic utility.

Answer to Problem 67E

A quadratic model is $y\left(t\right)=0.248t^2-3.56t+20.6614$

Explanation of Solution

Given:

The data set is:

Year	Revenue
2007	7.67
2008	8.54
2009	8.73
2010	9.16
2011	11.65
2012	14.07
2013	16.05

Calculation:

Follow the provided steps of Ti-83 plus calculator to compute a quadraticas:

1. Turn on the calculator.
2. Click on STAT > Edit > Enter.
3. Enter the data of x in L1 and data of y in L2.
4. Click om STAT > CALC >QuadReg> ENTER.
5. Hit 2nd and 1 to choose L1 and Hit 2nd and 2 to choose L2.
6. Hit Enter.

The obtained output is:

  

From the above output, the quadratic model is $y\left(t\right)=0.248t^2-3.56t+20.6614$

(b)

To determine

To construct: The model using the graphing utility. Also, estimate the slope of the graph if t is 12 and interpret the result.

Answer to Problem 67E

The slope is 13.6534.

Explanation of Solution

Graph:

Follow the provided steps of Ti-83 plus calculator to construct the figure:

1. Turn on the calculator.
2. Click on STAT > Edit > Enter.
3. Hit Y= and enter the regression.
4. Press 2nd key and Y and press twice and chose the mark of rectangle.
5. Hit zoom button and 9th key.

The obtained output is:

  

Calculation:

From the above part, the quadratic model is:

  $\begin{array}{c}y\left(t\right)=0.248t^2-3.56x+20.6614\\ =0.248{\left(12\right)}^2-3.56\left(12\right)+20.6614\\ =13.6534\end{array}$

Then the slope is approximately 13.6534.

It implies that the population is increasing by the rate of 1365.34 per year

(c)

To determine

To find: The derivative of the model computed in part (a) and evaluate it for t = 12

Answer to Problem 67E

The value is 2.392.

Explanation of Solution

Calculation:

The quadratic model is:

  $y\left(t\right)=0.248t^2-3.56t+20.6614$

Now, take the derivative of above function with respect to t as:

  $\begin{array}{c}y\left(t\right)=0.248t^2-3.56t+20.6614\\ \frac{d\left(y\left(t\right)\right)}{dt}=0.496t-3.56\end{array}$

Now, plug in t = 12 in above expression as:

  $\begin{array}{c}{\frac{d\left(y\left(t\right)\right)}{dt}\}}_{t=12}=0.496t-3.56\\ =2.392\end{array}$

Thus, the required value is 2.392.

(d)

To determine

To draw:The conclusion regarding the results the previous parts.

Explanation of Solution

Since, the quadratic model is:

  $y\left(t\right)=0.248t^2-3.56t+20.6614$

And, its derivative is:

  $\frac{d\left(y\left(t\right)\right)}{dt}=0.496t-3.56$

Because it is depending on the time. Thus, it could be said that it is a good model