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What is the largest rectangular area that can be enclosed with 400 feet of fencing What are the dimensions of the rectangle -- Quadratic Modeling?
Wiki User
∙ 10y ago
Find the dimensions of the rectangular field of maximum area which can be enclosed with 400 feet of fence.
.
Let w = width of field
then
(400-2w)/2 = length of field
(200-w) = length of field
.
area = w(200-w)
area = 200w-w^2
area = -w^2+200w
.
Looking at the coefficient for the w^2 term, we see that it is negative. This indicates that the parabola opens downward and finding the vertex will give you the "maximum".
.
Standard vertex form is:
y= a(x-h)^2+k
where
(h,k) is the vertex
.
Convert our equation into that form by "completing the square":
area = -w^2+200w
area = -(w^2-200w)
area = -(w^2-200w+10000) + 10000
area = -(w-100)^2 + 10000
.
From the above, we see that the vertex is:
(h,k) = (100, 10000)
.
This says that when the width=100 feet, the area will be maximized at 10000 square feet.
.
Solution:
width = 100 feet
length = 200-w = 200-100 = 100 feet
Wiki User
∙ 10y ago
To maximize the area enclosed by the Fencing, the rectangle should be a square. So, each side length is 100 feet (400 ft divided by 4). Therefore, the dimensions of the rectangle are 100 feet by 100 feet, and the largest rectangular area enclosed with 400 feet of fencing is 10,000 square feet.
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