Chapter 8.1, Problem 60PPS

a.

To determine

To draw: Three rectangles that each have a perimeter of 400 feet.

Explanation of Solution

Given:

The perimeter of the rectangle.

Draw:

Consider three rectangles with lengths 100 ft, 75ft, and 50ft such that each these have perimeter 400ft and draw these rectangles as shown below.

  

b.

To determine

To find: The area of each rectangle by recording there width and length.

Explanation of Solution

Given:

The table as shown below.

Rectangle	Length	Width	Area
1	100 ft
2	75 ft
3	50 ft
4	X ft

Calculation:

Width of rectangle is given by $w=\frac{P-2l}{2}$ .

Area of rectangle is given by $A=lw$ .

Fill the table assuming the perimeter to 400 ft as shown below.

Rectangle	Length	Width	Area
1	100 ft	100 ft	10000 sq. ft
2	75 ft	125 ft	9375 sq. ft
3	50 ft	150 ft	7500 sq. ft
4	$x$ ft	$\frac{400-2x}{2}=200-x$	$x\left(200-x\right)$ sq. ft

c.

To determine

To graph: The area of the rectangle 4 found in part (b) and then find the largest area.

Answer to Problem 60PPS

1000 sq. ft.

Explanation of Solution

Given:

The area of the rectangle 4 is given by $A\left(x\right)=x\left(200-x\right)$ .

Graph:

Graph the function $A\left(x\right)=x\left(200-x\right)$ as shown below.

  

From the graph it can be observed that the largest area is 1000 sq. ft.

d.

To determine

To graph: The length and width of the rectangle giving the largest area.

Answer to Problem 60PPS

The area is largest when length is 100 ft and width is 100 ft.

Explanation of Solution

Given:

The area of the rectangle 4 is given by $A\left(x\right)=x\left(200-x\right)$ .

Graph:

Graph the function $A\left(x\right)=x\left(200-x\right)$ as shown below.

  

From the graph, it can be observed that the largest area is obtained when $x=100$ . That means, the length of the rectangle is 100 ft the area is largest. The corresponding width of the rectangle is $w=\frac{400-2\cdot 100}{2}=100\text{ft}$ .