Let's solve this one in general, that is for any given perimeter.
Let P represent the fixed perimeter.
Let w represent the width of the rectangle.
Let l represent the length of the rectangle.
The perimeter of a rectangle is:
The area of a rectangle is the length times the width so a function for the area in terms of the width is:
The area function is a parabola, opening downward, with vertex at:
Since the parabola opens downward, the vertex represents a maximum value of the area function. The value of the width that gives this maximum value is one-fourth of the given perimeter. Therefore, the shape must be a square, and the area is the width squared.
The area function is continuous and twice differentiable over its domain, therefore there will be a local extrema wherever the first derivative is equal to zero and that extreme point will be a maximum if the second derivative is negative at that point.
Therefore the maximum area is obtained when
And that maximum area is:
My calculator said it, I believe it, that settles it