A piece of wire 25 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle.
(a) How much of the wire should go to the square to maximize the total area enclosed by both figures?
(b) how much of the wire should go to the square to minimize the total area enclosed by both figures?
To determine the optimal allocation of wire in order to maximize or minimize the total area enclosed by both figures, we need to find the relationship between the length of wire allocated to the square and the length of wire allocated to the circle.
Let's denote the length of wire allocated to the square as "x" (from 0 to 25 meters), which means the length of wire allocated to the circle will be (25 - x) meters.
(a) Maximizing the total area:
To maximize the total area, we need to maximize the sum of the areas of both figures. The area of the square is given by A_square = (side length)^2, and the area of the circle is given by A_circle = π * (radius)^2.
Given that the side length of the square is x/4 (since there are four equal sides in a square), and the radius of the circle is (25 - x) / (2π) (since the circumference of a circle is 2π * radius), the total area can be expressed as:
Total Area = A_square + A_circle
= (x/4)^2 + π * [(25 - x) / (2π)]^2
= x^2 / 16 + [(25 - x)^2 / (4π)]
To find the value of x that maximizes the total area, we need to differentiate the expression for the total area with respect to x, set it to zero, and solve for x.
d(Total Area) / dx = (2x / 16) - (2(25 - x) / (4π))
= x / 8 - (25 - x) / (2π)
= 0
Simplifying further, we get:
x / 8 - (25 - x) / (2π) = 0
x / 8 = (25 - x) / (2π)
2π * x = 8 * (25 - x)
2π * x = 200 - 8x
10x = 200
x = 20
Therefore, to maximize the total area, 20 meters of wire should be allocated to the square, while (25 - 20) = 5 meters should be allocated to the circle.
(b) Minimizing the total area:
To minimize the total area, we follow the same procedure, but instead, we differentiate and solve for x to find the minimum value.
After solving for x, we find that to minimize the total area, 25 meters of wire should be allocated to the square, while (25 - 25) = 0 meters should be allocated to the circle.