# If a rectangular field's enclosed area is the maximum, find the demensions of the field. There are 3 sides of the rectangular field with 2000 ft. of fencing. The other side of the rectangle will be a river. (addition info. before the question is asked) I worked this problem and keep getting 500,000 a=lw.... a= (2000-2b).... 2000b-2bb...-2bb+2000b vertex: a= (-2bb + 2000b)+0 ....-2(bb-1000b+250,000)+0+500,000....-2(b-500)squared + 500,000 so... vertex=(5000, 500000) and axis of symmetry - x=500 Dimensions: a=lw a= (1000)(500) a=500,000*2bb = 2b squared 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.

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