A piece of wire 18 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 wire should be used for the square in order to maximize the total area?
BACON!!!!
true
To maximize the total area, we need to find the length of wire that should be used for the square.
Let's assume the length of wire used for the square is x. Since the total length of the wire is 18m, the length of the wire used for the circle would be (18 - x) meters.
The perimeter of a square is given by 4 * side length, so the side length of the square would be x/4.
The circumference of a circle is given by 2 * π * radius. Since the length of wire used for the circle is (18 - x) meters, we can set up the equation 2 * π * radius = (18 - x). Solving for the radius, we get radius = (18 - x) / (2 * π).
The area of the square is given by side length squared, which is (x/4)^2 = x^2/16.
The area of the circle is given by π times the radius squared, which is π * [(18 - x) / (2 * π)]^2 = (18 - x)^2 / (4 * π).
To find the total area, we add the area of the square and the area of the circle: total area = x^2/16 + (18 - x)^2 / (4 * π).
To maximize the total area, we can take the derivative of the total area function with respect to x, set it equal to zero, and solve for x.
If x is allocated to the square, that leaves 18-x for the circle. So,
r = (18-x)/2π
a = π((18-x)/2π)^2 + (x/4)^2
= (1/4π + 1/16)x^2 - 9/π x + 81/π
now just set da/dx=0 and solve for x