2 w + L = 40 so L = 40-2w
A = w L = w (40-2w)
-2 w^2 + 40 w = A
w^2 - 20 w = -A/2
w - 20 + 100 = -(1/2)(A-200)
(w - 10)^2 = -(1/2)(A-200)
w = 10, L = 20, A = 200
A. 6 ft
B. 8 ft
C. 10 ft
D. 12 ft
A = w L = w (40-2w)
-2 w^2 + 40 w = A
w^2 - 20 w = -A/2
w - 20 + 100 = -(1/2)(A-200)
(w - 10)^2 = -(1/2)(A-200)
w = 10, L = 20, A = 200
The perimeter of a rectangle is given by the formula:
Perimeter = 2(length + width)
In this case, the perimeter is given as 40 feet. We can plug that into the formula and solve for length in terms of width:
40 = 2(length + width)
20 = length + width
length = 20 - width
The area of a rectangle is given by the formula:
Area = length * width
We can substitute the expression for length into the formula for area:
Area = (20 - width) * width
To find the width that results in the maximum area, we can calculate the area for each of the given options and see which one is the largest.
For option A:
Width = 6 ft
Length = 20 - 6 = 14 ft
Area = 14 * 6 = 84 square feet
For option B:
Width = 8 ft
Length = 20 - 8 = 12 ft
Area = 12 * 8 = 96 square feet
For option C:
Width = 10 ft
Length = 20 - 10 = 10 ft
Area = 10 * 10 = 100 square feet
For option D:
Width = 12 ft
Length = 20 - 12 = 8 ft
Area = 8 * 12 = 96 square feet
From the calculations, we can see that option C, with a width of 10 ft, results in the maximum area of 100 square feet. Therefore, the correct answer is C.