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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?
Find the dimensions of the rectangular field of maximum area which can be enclosed with 400 feet of fence.
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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
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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
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From the above, we see that the vertex is:
(h,k) = (100, 10000)
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This says that when the width=100 feet, the area will be maximized at 10000 square feet.
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Solution:
width = 100 feet
length = 200-w = 200-100 = 100 feet
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