Question

How do you find the maximum area of rectangle with 80ft perimeter?

Answer 1

Algebraically, we can show that the maximum area of the rectangle is achieved when it is a square. In this case, the maximum area is `"400 ft"^2`

Explanation:

Since this is not in calculus, I'll provide a non-calculus answer.

We know the rectangle always has a perimeter of `80`, so `2l+2w=80`, which simplifies to be `l+w=40`.

We also know that the area of the rectangle is `A=lw`, but we can express this as a function of a single variable.

Use the perimeter expression `l+w=40` to say that `l=40-w`. Because this will always be true in the rectangle, we can substitute `40-w` for `l` in `A=lw`.

`A=lw`

`A=(40-w)w`

`A=-w^2+40w`

This quadratic function can be graphed, and the highest point will be the spot where the rectangle's area is maximized.

graph{-x^2+40x [-5, 45, -145, 460]}

The highest point, the vertex of the parabola, is `(20,400)`. When `w=20`, we also know that `l=20` (which forms a square) and the maximum area of the rectangle (square) is `"400 ft"^2`.