Answer:
80,00[tex]ft^{2}[/tex]
Step-by-step explanation:
According to my research, the formula for the Area of a rectangle is the following,
[tex]A = L*W[/tex]
Where
Since the building wall is acting as one side length of the rectangle. We are left with 1 length and 2 width sides. To maximize the Area of the parking lot we will need to equally divide the 800 ft of fencing between the Length and Width.
800 / 2 = 400ft
So We have 400 ft for the length and 400 ft for the width. Since the width has 2 sides we need to divide 60 by 2.
400/2 = 200 ft
Now we can calculate the maximum Area using the values above.
[tex]A = 400ft*200ft[/tex]
[tex]A = 80,000ft^{2}[/tex]
So the Maximum area we are able to create with 800 ft of fencing is 80,00[tex]ft^{2}[/tex]
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The dimensions of the largest possible parking lot is 200 feet by 200 feet.
Let x represent the length and y represent the width, hence:
Since there is 800 feet of fence;
2(x + y) = 800
x + y = 400
y = 400 - x
Also:
xy = Area (A)
A = x(400 - x)
A = 400x - x²
The maximum area is at A' = 0, hence:
A' = 400 - 2x
400 - 2x = 0
x = 200 feet
y = 400 - x = 400 - 200 = 200
The dimensions of the largest possible parking lot is 200 feet by 200 feet.
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