Bending an Iron Wire

Geometry Level 3

An iron wire is bent into a circle with a certain radius R. If the wire is now bent into a rectangle, what is the maximum area of that rectangle?

( 2 R π 2 ) 2 (\frac{2Rπ}{2})^{2} R 2 π 2 4 \frac{R^{2}π^{2}}{4} R 2 π 4 \frac{R^{2}π}{4} R 2 π 2 16 \frac{R^{2}π^{2}}{16}

Ir J by IR J , 21, Philippines

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1 solution

Ir J
Jun 12, 2018

=>Given that the radius of the circle was R. It's circumference will be 2Rπ

=>2Rπ will be the perimeter of the formed rectangle.

=>With this we can say that the sum of the consecutive sides of the rectangle will be 2 R π 2 \frac{2Rπ}{2} = Rπ

Let x be one of the side of the rectangle and f(x) be the function for the area.

f(x) = x (Rπ-x) = -x^{2} + Rπx (Parabola)

To know the maximum value of f(x), just find the vertex. You can use the Vertex Formula " − b 2 a \frac{-b}{2a} " and find out that this would result to − R π 2 ( − 1 ) \frac{-Rπ}{2(-1)} = R π 2 \frac{Rπ}{2} . This means that f(x) is maximum when x = R π 2 \frac{Rπ}{2} .

The maximum area is ( R π 2 ) 2 (\frac{Rπ}{2})^{2} = R 2 π 2 4 \frac{R^{2}π^{2}}{4}

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