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Problem 17 - Entrance Test

A wire of length 28 m 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 lengths of the two pieces so that the combined area of the square and the circle is minimum?

A. (28π)/(π+4), 112/(π+4)B. (112)/(π+4), (28*4)/(π+4)C. (28π)/(π+4), (112)/(π+4)D. 28/(π+4), 112/(π+4)

Correct: C

Let the length of one piece be x, which is used to make a square. Then the length of the other piece is 28 - x, which is used to make a circle. Side of the square = x/4. Radius of the circle = (28 - x) / (2π). Area of the square = (x/4)² = x²/16. Area of the circle = π * ((28 - x) / (2π))² = (28 - x)² / (4π). Total area, A = x²/16 + (28 - x)² / (4π). To minimize A, we find dA/dx = 0. dA/dx = (2x/16) + (2(28 - x)(-1)) / (4π) = x/8 - (28 - x) / (2π) = 0. x/8 = (28 - x) / (2π) => xπ = 4(28 - x) => xπ = 112 - 4x => x(π + 4) = 112 => x = 112 / (π + 4). Length of the piece for the circle = 28 - x = 28 - 112 / (π + 4) = (28π + 112 - 112) / (π + 4) = (28π) / (π + 4). So (28π)/(π+4), 112/(π+4).