Objective

The objective of this activity is to optimize the area formed when a wire of a given length is cut and used to create two shapes.

A wire of length 40 cm will be used to form a circle, a square, or both. The wire can be cut so that one section is used to create a circle and the other section is used to create a square.

How should the wire be used to form shapes with a maximum combined area? How much of the wire should be used to create the circle? How much should be used to create the square?

There are many ways that students may approach this problem. Consider selecting and sequencing techniques when having students present their work. In precalculus, a graphing approach is ideal.

Overall, this problem consists of selecting a location to cut the wire, constructing a square from one piece and a circle from the other, and computing the area of each shape.

Let xx be the distance from the left end of the wire as shown below.

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Cutting here will result in two pieces of wire, one with length xx and the other with length 40x40-x. Let the piece of wire with length xx be used to construct the circle and the piece of wire with length 40x40-x be used to construct the square.

For the circle, the length of the wire xx corresponds to the circumference of the circle. That is, x=2πrx=2\pi r. To determine the area of the circle, we must first solve for rr

x=2πrx = 2\pi r

x2π=r\frac{x}{2\pi} = r

Now we find the area of the circle using A=πr2A=\pi r^2

Acircle=π(x2π)2A_{circle}=\pi \left( \frac{x}{2\pi} \right) ^2

Acircle=x24πA_{circle}=\frac{x^2}{4\pi}

For the square, the length of the wire 40x40-x corresponds its perimeter. That is, P=40xP=40-x. To calculate the area of the square, we must first determine the side length of the square. Since a square has 4 congruent sides, we have:

4s=40x4s = 40-x

s=40x4s=\frac{40-x}{4}

Now we find the area of the square using s2s^2

Asquare=(40x4)2A_{square}=\left( \frac{40-x}{4} \right) ^2

The total combined area can be found by adding the two areas

A=x24π+(40x4)2A=\frac{x^2}{4\pi}+\left( \frac{40-x}{4} \right) ^2

Creating a Table

Using the area formula above, students may consider a table of values for the location xx to cut the wire.

Using the Regression app, input values for xx in X1. Then use formulas to compute the area of the circle, the area of the square, and the total area.

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x Area of Circle Area of Square Total Area
0 0 100 100
5 1.989 76.563 78.552
10 7.958 56.25 64.208
15 17.905 39.063 56.967
20 31.831 25 56.831
25 49.736 14.063 63.794
30 71.62 6.25 77.87
35 97.482 1.563 99.045
40 127.324 0 127.324

Using this approach, students will find the maximum at x=40x = 40. That is, when the entire piece of wire is used to construct a circle, the area is maximized at 127.324cm2127.324 \textup{cm}^2

Graphing

Using the area formula above, students may graph the function y=x24π+(40x4)2y=\frac{x^2}{4\pi}+\left( \frac{40-x}{4} \right) ^2 and look for maximums.

Using the Functions app, enter the function f(x)=x24π+(40x4)2f(x) = \frac{x^2}{4\pi}+\left( \frac{40-x}{4} \right) ^2.

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Before viewing the Graph, adjust the plot range in the Function options to match the practical domain for this scenario.

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Graphing the function, we see the maximum occurs at the right endpoint (40,127.324)\left( 40,127.324 \right).

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That is, when the entire piece of wire is used to construct a circle, the area is maximized at 127.324cm2127.324 \textup{cm}^2


Alternative Prompts

The following alterations can be used depending on your learning objectives:

  1. Change the length of the wire for each student or group of students.
  2. Replace the length of the wire with a parameter LL.
  3. Instruct students to find the minimum combined area instead of the maximum area.
  4. Use the wire to form a circle and an equilateral triangle with minimized combined area.
  5. Use the wire to form a square and an equilateral triangle with a minimized combined area.
  6. Use the wire to form a circle and another regular polygon with minimized combined area.