College Algebra
posted by Brooklyn on .
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:
1002L=2W
Divide by two: 50L=W
Substitute in original equation:
A=LW A=L(50L)
A= 50L  L squared Now I am stuck.

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.