Chapter 4.6, Problem 62PPSE

Solution

a.

To Plot: The points. And find the model for the data set.

Given:

Width	Length	Area $$
$1$	$49$	$49$
$2$	$48$
$3$
$4$
$5$

Explanation:

The completed table is shown below:

If $l$ is length and $w$ is width

Then perimeter of rectangle is $2\left(l+w\right)$ and area of the rectangle is $l\times w$:

Width	Length	Area
$1$	$49$	$49$
$2$	$48$	$96$
$3$	$47$	$141$
$4$	$46$	$184$
$5$	$45$	$225$

The points $\left(w,a\right)$ are plotted below:

  $a=$ Area of the rectangle [Vertical axis]

  $w=$ Width of the rectangle [Horizontal axis]

The given data can be modelled by $a=44w+7$:

  

b.

To Find: The another point in the data set. Verify the model by using it.

The another point can be $\left(6,264\right)$ .

Given:

Width	Length	Area $$
$1$	$49$	$49$
$2$	$48$
$3$
$4$
$5$

Explanation:

In the data set, another point can be $\left(6,264\right)$

Since when width is $6$ , length is $50-6=44$ :

Calculate the area: $6\times 44=264$

Verify by using the model $a=44w+7$ :

  $\begin{array}{c}264=44\times 6+7\\ 264\approx 271\end{array}$

Approximately the values are equal.

c.

To Find: The reasonable domain for this function.

The reasonable domain for this function is $1\le w\le 49$ .

Given:

Width	Length	Area $$
$1$	$49$	$49$
$2$	$48$
$3$
$4$
$5$

Explanation:

When width is $w$ , the length is $50-w$ :

Hence, the domain conditions are $50-w\ge 1$ and $w\ge 1$ :

  $50-w\ge 1$

Add $-50$ on both sides:

  $-w\ge -49$

Rewrite it as:

  $w\le 49$

Domain of the function $a\left(w\right)=50w-w^2$ is $1\le w\le 49$ .

d.

To Find: The maximum possible area. Find the dimensions that yield the area.

The maximum possible area is $625$ . The dimensions are width $w=25$ and length $l=25$ .

Given:

Width	Length	Area $$
$1$	$49$	$49$
$2$	$48$
$3$
$4$
$5$

Explanation:

Maximum possible area is the maximum value of $a\left(w\right)=50w-w^2$ :

  $\begin{array}{c}a\left(w\right)=50w-w^2\\ =-w^2+50w\\ =-\left[w^2-50w+\left({25}^2-{25}^2\right)\right]\\ =-\left[w^2-50w+{25}^2\right]+{25}^2\\ =-{\left[w-25\right]}^2+{25}^2\end{array}$

Maximum area of the rectangle is ${25}^2=625$ .

Find the Dimension:

When width $w=25$ and length $l=25$ is the dimension for maximum area.

e.

To Find: The function for area in terms of width without using the table.

The function for area in terms of width without using the table is $a=50w-w^2$ .

Given:

Width	Length	Area $$
$1$	$49$	$49$
$2$	$48$
$3$
$4$
$5$

Explanation:

When width is $w$ , Length is $50-w$ since the perimeter is $100$ .

Area of the rectangle:

  $\begin{array}{c}=Width\times Length\\ =w\left(50-w\right)\\ =50w-w^2\end{array}$

So, the model is $a=50w-w^2$ which varies from model obtained from part $\left(a\right)$ $a=44w+7$ which is approximate.