Chapter 11.1, Problem 36PPS

a.

To determine

To calculate: The equation of the perimeter and the area of rectangle with the given information.

Answer to Problem 36PPS

The equation of the perimeter of the rectangle is $x+y=6$ and the equation of the area of the rectangle is $x\times y=A$ .

Explanation of Solution

Given information: The perimeter of a rectangle is 12 units and length of the rectangle is x and the width of the rectangle is y.

Formula used:

The perimeter and area of a rectangle is given by

  $\begin{array}{l}P=2\left(l+b\right)\\ A=l\times b\\ l\text{is}\text{the}\text{length}\text{of}\text{the}\text{rectangle}\text{.}\\ b\text{is}\text{the}\text{width}\text{of}\text{the}\text{rectangle}\text{.}\end{array}$

Calculation:

The given information is:

  $\begin{array}{l}l=\text{length}\text{of}\text{the}\text{rectangle}=x\\ b=\text{width}\text{of}\text{the}\text{rectangle}=y\\ \text{Perimeter}\text{of}\text{rectangle}=12\end{array}$

Hence, the equation of the perimeter of the given rectangle is

  $\begin{array}{l}P=2\left(l+b\right)=12\\ 2\left(x+y\right)=12\\ x+y=6\mathrm{.....}\left(1\right)\end{array}$

The equation of the area of the given rectangle is

  $\begin{array}{l}A=l\times b\\ x\times y=A\mathrm{.....}\left(2\right)\end{array}$

Therefore, the equation of the perimeter of the rectangle is $x+y=6$ and the equation of the area of the rectangle is $x\times y=A$ .

b.

To determine

To calculate: All the possible whole-number values that satisfies as the length and width of the rectangle and to find the area of the rectangle for each pair.

Answer to Problem 36PPS

The possible values of x & y and the area associated with them is given below in the tabular form.

$\left(x,y\right)$	$Area$
$\left(1,5\right)$	$A=5$
$\left(2,4\right)$	$A=8$
$\left(3,3\right)$	$A=9$
$\left(4,2\right)$	$A=8$
$\left(5,1\right)$	$A=5$

Explanation of Solution

Given information: The perimeter of a rectangle is 12 units and length of the rectangle is x and the width of the rectangle is y.

Formula used:

The perimeter and area of a rectangle is given by

  $\begin{array}{l}P=2\left(l+b\right)\\ A=l\times b\\ l\text{is}\text{the}\text{length}\text{of}\text{the}\text{rectangle}\text{.}\\ b\text{is}\text{the}\text{width}\text{of}\text{the}\text{rectangle}\text{.}\end{array}$

Calculation:

By using the results of equation (1) & (2)

The equation of the perimeter of the rectangle is $x+y=6$ .

The equation of the area of the rectangle is $x\times y=A$ .

Here for different values of x & y satisfying equation (1), there will be different value of the area of the rectangle.

Also since x is the length of the rectangle so it cannot be less than 1 and it cannot be more than 5 because then y will be zero which is also not possible as it is the width of the rectangle, hence $1\le x\le 5$ .

Therefore the possible values of x & y and the area associated with them is given below in the tabular form.

$\begin{array}{l}\text{For}x=1\\ \text{From}\text{equation}\left(1\right)\\ x+y=6\\ y=6-1=5\\ \\ \text{here}\left(x,y\right)=\left(1,5\right)\end{array}$	$\begin{array}{l}\text{For}\left(x,y\right)=\left(1,5\right)\\ \text{From}\text{equation}\left(2\right)\\ A=x\times y\\ A=1\times 5=5\\ \\ \text{here}A=5unit{s}^2\end{array}$
$\begin{array}{l}\text{For}x=2\\ \text{From}\text{equation}\left(1\right)\\ x+y=6\\ y=6-2=4\\ \\ \text{here}\left(x,y\right)=\left(2,4\right)\end{array}$	$\begin{array}{l}\text{For}\left(x,y\right)=\left(2,4\right)\\ \text{From}\text{equation}\left(2\right)\\ A=x\times y\\ A=2\times 4=8\\ \\ \text{here}A=8unit{s}^2\end{array}$
$\begin{array}{l}\text{For}x=3\\ \text{From}\text{equation}\left(1\right)\\ x+y=6\\ y=6-3=3\\ \\ \text{here}\left(x,y\right)=\left(3,3\right)\end{array}$	$\begin{array}{l}\text{For}\left(x,y\right)=\left(3,3\right)\\ \text{From}\text{equation}\left(2\right)\\ A=x\times y\\ A=3\times 3=9\\ \\ \text{here}A=9unit{s}^2\end{array}$
$\begin{array}{l}\text{For}x=4\\ \text{From}\text{equation}\left(1\right)\\ x+y=6\\ y=6-4=2\\ \\ \text{here}\left(x,y\right)=\left(4,2\right)\end{array}$	$\begin{array}{l}\text{For}\left(x,y\right)=\left(4,2\right)\\ \text{From}\text{equation}\left(2\right)\\ A=x\times y\\ A=4\times 2=8\\ \\ \text{here}A=8unit{s}^2\end{array}$
$\begin{array}{l}\text{For}x=5\\ \text{From}\text{equation}\left(1\right)\\ x+y=6\\ y=6-5=1\\ \\ \text{here}\left(x,y\right)=\left(5,1\right)\end{array}$	$\begin{array}{l}\text{For}\left(x,y\right)=\left(5,1\right)\\ \text{From}\text{equation}\left(2\right)\\ A=x\times y\\ A=5\times 1=5\\ \\ \text{here}A=5unit{s}^2\end{array}$

c.

To determine

To graph: The area of the rectangle with respect to the length of the rectangle.

Explanation of Solution

Given information: By using the result of part-b, the variation of area with the value of x & y is given below.

$\left(x,y\right)$	$Area$
$\left(1,5\right)$	$A=5$
$\left(2,4\right)$	$A=8$
$\left(3,3\right)$	$A=9$
$\left(4,2\right)$	$A=8$
$\left(5,1\right)$	$A=5$

Graph:

The graph of area with respect to the length of the rectangle is given below.

  

d.

To determine

To interpret: The variation of area of the rectangle with respect to the length of the rectangle

Explanation of Solution

Given information: By using the result of part-c, the variation of area with the value of length is given below.

  

From the given information it can be observed that the area of the rectangle is continuously varying with respect to the length of the rectangle. As the length of the rectangle increases the area first increases and reaches a maximum value and then starts decreasing as the length is further increased.

The maximum value of the area of the rectangle is 9 sq. units.

The minimum value of the area of the rectangle is 5 sq. units.

e.

To determine

To find: The length and width of the rectangle for which the area of the rectangle is the greatest and the least.

Answer to Problem 36PPS

The area is maximum (9) at $\left(x,y\right)=\left(3,3\right)$ and the area is minimum (5) at $\left(x,y\right)=\left(1,5\right)$ .

Explanation of Solution

Given information: By using the result of part-c, the variation of area with the value of length is given below.

  

From the given information it can be observed that

The maximum value of the area is

  $\begin{array}{l}A=9unit{s}^{2}atx=3\\ \text{Also}\text{from}\text{equation}\left(1\right)\text{of}\text{part}\text{a}\\ x+y=6\\ 3+y=6\\ y=3\end{array}$

Hence the area is maximum at $\left(x,y\right)=\left(3,3\right)$ .

The minimum value of the area is

  $\begin{array}{l}A=5unit{s}^{2}atx=1\\ \text{Also}\text{from}\text{equation}\left(1\right)\text{of}\text{part}\text{a}\\ x+y=6\\ 1+y=6\\ y=5\end{array}$

Hence the area is minimum at $\left(x,y\right)=\left(1,5\right)$ .