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A wire of length 28m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

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The length of the wire used for making the square is approximately 14.71 meters

Let the length of the wire used for making the square be x meters. Then, the length of the wire used for making the circle is (28 - x) meters.

Let the side of the square be s-meter. Then, the perimeter of the square is 4s = x.

Solving for s, we get s = x/4.

The area of the square is s^2 = (x/4)^2 = x^2/16.

Let the radius of the circle be r meters. Then, the circumference of the circle is 2πr = (28 - x)

Solving for r, we get r = (14 - x/2π).

The area of the circle is πr^2 = π((14 - x/2π)^2)

To minimize the combined area of the square and the circle, we need to minimize the expression:

A(x) = x^2/16 + π((14 - x/2π)^2)

Expanding and simplifying the expression, we get:

A(x) = x^2/16 + 196π - 28πx + x^2/4π

To find the minimum value of A(x), we can take the derivative of A(x) concerning x and set it equal to zero:

dA(x)/dx = x/8 - 14π + x/4π = 0

Solving for x, we get x ≈ 14.71.

Therefore, the length of the wire used for making the square is approximately 14.71 meters, and the length of the wire used for making the circle is approximately 13.29 meters.

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