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


vertex=(5000, 500000) and axis of symmetry - x=500



a= (1000)(500)


*2bb = 2b squared

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Congratulations! Your answer is correct!

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.


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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial