A piece of wire of length 2 m. is cut into two parts, one of which is bent into the shape of a square and the other into a shape of a circle. How should the wire be cut so that sum of the enclosed areas is minimum.
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To find the point of minimum sum of the enclosed areas, we need to optimize the problem and consider the constraints. Let's denote the length of the wire used for the square as "x," and the remaining length of the wire used for the circle as "2 - x".
Now, we can determine the formulas for the area of the square and the circle, respectively:
1. Area of the square = side length ^ 2 = (x / 4) ^ 2 = x ^ 2 / 16
2. Area of the circle = π * radius ^ 2 = π * ((2 - x) / (2 * π)) ^ 2 = (2 - x) ^ 2 / (4 * π)
To find the point of minimum sum of the enclosed areas, we need to take the derivative of the sum of the areas with respect to x and set it equal to zero:
d(sum of areas) / dx = d(square area) / dx + d(circle area) / dx = 0
Now, let's solve for the value of x:
d(square area) / dx = (1/16) * 2x = x / 8
d(circle area) / dx = (2 - x) / (2 * π)
Setting these derivatives equal to zero:
x / 8 + (2 - x) / (2 * π) = 0
Simplifying and solving for x:
(x + 4 - x) / 8 = (2 - x) / (2 * π)
4 / 8 = (2 - x) / (2 * π)
1 / 2 = (2 - x) / (2 * π)
π = 2 - x
x = 2 - π
Therefore, to minimize the sum of the enclosed areas, the wire should be cut into two parts, with the length of the wire used for the square as (2 - π) meters and the remaining length used for the circle as (π - 2) meters.