Question

A school is constructing a rectangular play area against an exterior wall of the school

building. It uses 120 feet of fencing material to enclose three sides of the play area.
b
10
30
40
W
W
100 un gesa
е
24 for
a. Complete the table by giving the length and area for each width. (The width is
the side perpendicular to the building.)
school building
width (feet) length (feet) area (square feet)
160
60
40
(120-w)²
Perimeter = 112D
1/20
10
who nt
1,000
esh an
STA

Answer 1

I have no clue the the answer is ma boi

Answer 2

The table should be completed with the length and area for each width as follows;

width (feet) length (feet) area (square feet)

10 100 1000

30 60 1800

40 40 1600

w 120 - 2w
`120W-2W^2`

In Mathematics and Geometry, the perimeter of a rectangle can be calculated by using this mathematical equation (formula);

P = 2(L + W)

Where:

• P represent the perimeter of a rectangle.
• W represent the width of a rectangle.
• L represent the length of a rectangle.

Since 120 feet of fencing material was used to enclose three sides of the play area, it implies there are only 3 sides;

120 = L + 2W

L = 120 - 2W

In order to maximize the area, the width must be 30 feet and the length must be 60 feet;

120 = L + 2W

120 = 60 + 2(30)

120 = 120

For the area, we have;

Area of rectangle, A = LW

A = (120 - 2W)W

`A=120W-2W^2`

By taking the first derivative of the area function and equating to zero, we have;

A' = 120 - 4W

0 = 120 - 4W

4W = 120

W = 120/4 = 30 feet.

When W is 30 feet, the length and area are given by;

L = 120 - 2(30)

L = 60 feet.

A = 30 × 60

A = 1800 sq. ft.

When W is 10 feet, the length and area are given by;

L = 120 - 2(10)

L = 100 feet.

A = 100 × 10

A = 1000 sq. ft.

When W is 40 feet, the length and area are given by;

L = 120 - 2(40)

L = 40 feet.

A = 40 × 40

A = 1600 sq. ft.

When W is w feet, the length and area are given by;

L = 120 - 2(w)

L = (120 - 2w) feet.

A = (120 - 2w) × w

A =
`120W-2W^2` sq. ft.