Overall, this problem consists of selecting where 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.

Wire Image

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 to 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 guess-and-check method by trying out different values xx. A table of values for xx and its corresponding areas can be constructed.

Using the Grapher app, input functions to compute the area of the circle, the area of the square, and the total area. Then view the Table tab and input possible values for xx.

Functions for the area of the circle and the area of the square

Function for the total area

Table representing values of x

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 f(x)=x24π+(40x4)2f(x)=\frac{x^2}{4\pi}+\left( \frac{40-x}{4} \right) ^2 and look for maximums.

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

Function definition

Before viewing the Graph, open the Function options by clicking on the three dots to the right of the expression. Adjust the plot restriction in the Function options to match the practical domain for this scenario.

Function domain

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

Function graph

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