Author Question: What is the maximum area for this question? (Quadratics)? (Read 1786 times)


  • Sr. Member
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So i got this math question, and I have a lousy teacher, so I don't know exactly how to find the maximum area for this question.
A farmer has a 200m of fencing material to enclose a rectangular feild adjacent to a river. No fencing is required along the river.
The image has a rectangle with the widths (smaller sides of rectangle) labeled with 'w' and the length is 200-w. The function I came up with is a=200w-2w^2. I did graph it on the calculator. Vertex is (50, 5000). X-intercepts are (0,0) and (100,0). Y intercept is (0,0).
That is all I can get from what I have self-learned and I still need to find the maximum area of the field and the dimensions of it.
If possible, please explain each step so that I can learn it. :)
Heres a picture of the diagram if my description didn't make sense:


  • Hero Member
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l = 200 - 2w

A = w * l
A = w * (200 - 2w)
A = 2 * (100w - w^2)
A = -2 * (w^2 - 100w)
A = -2 * (w^2 - 2 * 50w + 50^2 - 50^2)
A = -2 * (w - 50)^2 + 2 * 50^2
A = -2 * (w - 50)^2 + 2 * 2500
A = 5000 - 2 * (w - 50)^2

The vertex is at (50 , 5000)

w = 50
A = 5000
l = 100

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