Twenty feet wire is used to make two figures? What is the maximum areas of enclosed figures.

max area for a given perimeter is a circle. Is the wire evenly divided?

If so, c = p/2, so r = p/4π, and the total area is

a = 2(πr^2) = 2π(p^2/16π^2) = p^2/8π

If not, then if we have a tiny circle, (area effectively zero), then r = p/2π and

a = πr^2 = πp^2/4π^2 = p^2/4π

I suspect there is something missing here from the problem statement.

To find the maximum area of the enclosed figures using a 20 feet wire, we need to determine which shapes can be created and which combination will maximize the total area.

We have a couple of different options for figures that can be made with a 20 feet wire. Let's consider two possible figures: a rectangle and a circle.

1. Rectangle:
To find the maximum area of a rectangle, we need to determine its dimensions. Let's assume the length of the wire is used as the perimeter of the rectangle. Since a rectangle has two sets of equal sides, we can divide the 20 feet wire into four equal parts, each representing one side of the rectangle.

Let's say the wire is divided into 4 equal parts of 5 feet each. To form a rectangle, we need two pieces of wire for the length and two pieces for the width, each measuring 5 feet.
Therefore, the dimensions of the rectangle would be: length = 5 feet and width = 5 feet.

The formula to calculate the area of a rectangle is: Area = length x width.
Substituting the values, we get: Area = 5 feet x 5 feet = 25 square feet.

2. Circle:
To find the maximum area of a circle, we need to determine its radius. The circumference of a circle is equal to the wire used to form it.
We know the circumference of the circle is 20 feet. The formula for the circumference of a circle is: Circumference = 2πr, where π (pi) is a constant approximately equal to 3.14.

Let's solve for the radius:
20 feet = 2πr
Dividing both sides of the equation by 2π, we get:
r = 20 feet / (2π) = 10 feet / π

Now we can calculate the area of the circle using the formula: Area = πr^2
Substituting the value of the radius, we get:
Area = 3.14 x (10 feet / π)^2

Calculating this expression, we get an approximate area of 31.42 square feet.

Therefore, between the rectangle and the circle, the maximum area of the enclosed figure when using a 20 feet wire is approximately 31.42 square feet, which is achieved when creating a circle.