A wire 7 meters long is cut into two pieces. One piece is bent into a square for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each:

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To minimize the total area of both figures, we need to find the optimal length of wire to be used for each figure. Let's solve this problem step by step:

1. Let's assume that the length of wire used for the square frame is x (in meters).
2. The remaining length of wire for the TV antenna will be 7 - x (since the total length of wire is 7 meters).
3. The perimeter of a square can be calculated by multiplying the length of one side by 4, so the side length of the square frame will be x/4.
4. The perimeter of a circle is given by the formula 2πr, but since we're using wire instead of a string, the wire will form the circumference. Therefore, the circumference of the circle will be 7 - x.
5. Solving for the radius (r) of the circle, we can use the formula r = (circumference)/(2π), so the radius will be (7 - x) / (2π).
6. The area of the square is given by the formula A = side^2, so the area of the square frame will be (x/4)^2 = x^2/16.
7. The area of the circle is given by the formula A = πr^2, so the area of the circle for the TV antenna will be π((7 - x)/(2π))^2 = (7 - x)^2 / (4π).

Now, we can express the total area, Atotal, as the sum of the two areas:

Atotal = x^2/16 + (7 - x)^2 / (4π)

To find the minimum total area, we can take the derivative of Atotal with respect to x and set it equal to zero:

dAtotal/dx = (2x/16) - (2(7 - x))/(4π) = 0

Simplifying, we get:

x/8 - (7 - x)/(2π) = 0

Multiplying through by 8π, we have:

πx - 4(7 - x) = 0

Simplifying further:

πx - 28 + 4x = 0

Combining like terms:

(π + 4)x - 28 = 0

Solving for x, we get:

x = 28 / (π + 4)

Therefore, the optimal length of wire to be used for the square frame is approximately 2.81 meters (rounded to two decimal places), and the remaining length of wire for the TV antenna is approximately 4.19 meters (since 7 - 2.81 = 4.19).

To minimize the total area and reduce storage space, the wire should be cut at approximately 2.81 meters to allocate for the square frame and 4.19 meters to allocate for the TV antenna.

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