You have a 64 foot of fencing. what are the dimensions of the rectangle of the greatest area you could enclose?

In a rectangle, opposite sides are congruent.
Let one side have length x.
The opposite side also has length x.
The lengths of these two sides add to 2x.
The lengths of all 4 sides add to 64, so the lengths of the other 2 sides
add up to 64 - 2x. Each side measures 32 - x.

The rectangle has sides of length x and 32 - x.

The area of the rectangle is

A = LW

A = x(32 - x)

A = 32x - x^2

y = 32x - x^2 is a parabola that opens downward.
The maximum value of the parabola is the vertex on top.

32x - x^2 = 0

(32 - x)x = 0

32 - x = 0 or x = 0

x = 32 or x = 0

Since the parabola is symmetric with respect to the vertical axis, the vertex has x-coordinate 16.

At x = 16, you get maximum area.

Two opposite sides measure 16 ft each.
32 - x = 32 - 16 = 16
The other two opposite sides also measure 16 ft.

Since all sides turned out to have length 16 ft, the rectangle is a square.

The maximum area is enclosed by a square with side 16 ft.