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The post by @boreal represents the solution in a very inaccurately way, and is incorrect in many places.
I came to bring the correct solution.
sides are x and 200-x (where 200 = 400/2 is half of the perimeter)
The area is x(200-x) = -x^2 + 200x
the vertex for this quadratic is a maximum and will have x value of
-b/2a = (-200)/(-2) = 100 yds
The maximum area for this 100*100 SQUARE (!) is 10000 yds^2.
The formula for the area is = -W^2 + 200*W for W= width.
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The answer is very well known fact: under given condition, the maximum area is provided by the square,
and when the perimeter of a rectangle is given (= the fence length), the side of this square is one fourth of the perimeter.
See the lesson
- A rectangle with a given perimeter which has the maximal area is a square
- A farmer planning to fence a rectangular garden to enclose the maximal area
in this site.
On finding the maximum/minimum of a quadratic function see the lessons
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
- HOW TO complete the square to find the vertex of a parabola
- Briefly on finding the vertex of a parabola
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the topic "Finding minimum/maximum of quadratic functions".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.