Posted by **Brooklyn** on Tuesday, March 15, 2011 at 1:05pm.

You have 100yds. of fencing to enclose a rectangular area. Find the dimensions which maximize the area. What is the maximum area?

I know that A=L times W.

I know that 100 = 2L+2w

Subtract 2L from both sides:

100-2L=2W

Divide by two: 50-L=W

Substitute in original equation:

A=LW A=L(50-L)

A= 50L - L squared Now I am stuck.

- College Algebra -
**tchrwill**, Wednesday, March 16, 2011 at 11:50am
Considering all rectangles with the same perimeter, the square encloses the greatest area.

Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

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