Then its length is:

(50 – 2w) / 2

= 25 – w.

a = w(25 – w)

a – 25w + w^2 = 0

– 25w + w^2 = – a

Adding the square (156.25) of 12.5 to complete the square:

156.25 – 25w + w^2 = 156.25 – a

(12.5 – w)^2 = 156.25 – a

a = 156.25 – (12.5 – w)^2.

As (12.5 – w)^2 is never negative, this shows that the maximum area is 156.25 units^2, when the width is 12.5. This is when the rectangle is a square.

The domain of possible widths is [0, 25], as when the length of the rectangle is 25, its width is 0, and vice versa.

The range of areas is [0, 156.25], as the maximum area is achieved when the rectangle is a square with side 12.5.