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Let x be the width of the rectangle and y the length.

The perimeter of the rectangle sides that will be fenced is 2x+y. Since the length of fencing is 2000ft we have 2x+y=2000.

The area of the rectangle is A=xy.

Now 2x+y=2000 ==> y=2000-2x

Substituting for y in the area equation we get:

A=x(2000-2x). The graph of this function is a parabola opening down. The maximum will be at the vertex.

Written is standard form we get -2x^2+2000x. The vertex is on the axis of symmetry, which is located at `x=-b/(2a)` so the vertex has x-coordinate `-2000/-4=500`

If x=500, y=2000-2(500)=1000.

**The maximum area is 500(1000)=500,000 sq ft.**

** Note that the length of fencing is 2(500)+1000=2000ft as required.

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Using calculus the maximum occurs at a critical point. If `f(x)=x(2000-2x)=-2x^2+2000x` then `f'(x)=-4x+2000` and the critical point is located at x=500 as above.