Q. 88 - Precalculus Enhanced with Graphing Utilities

Question

Bending Wire: A wire 60 feet long is cut into two pieces. Is it possible to bend one piece into the shape of a square and the other into the shape of a circle so that the total area enclosed by the two pieces is 100 square feet? If this is possible, find the length of the side of the square and the radius of the circle.

Step-by-Step Solution

Answer

It is not possible to cut the wire into square and circle.

Step 1. Given information

Let x denotes the length of side of the square and r denotes the radius of circle.

There are two conditions:

1. A wire 60 feet long is cut into square and circle means 

$4x+2\pi r=60$

2. The total area enclosed by the two pieces is $100$ square feet means 

$x^2+\pi r^2=100$

Step 2. Solve the system of equations.

From first equation,

$$\begin{gathered} 4x+2\mathrm{\pi r}=60 \\ 4\mathrm{x}=60-2\mathrm{\pi r} \\ \mathrm{x}=\frac{60-2\mathrm{\pi r}}{4} \\ \mathrm{x}=\frac{30-\mathrm{\pi r}}{2} \end{gathered}$$

Substitute $x=\frac{30-\mathrm{\pi r}}{2}$ in $x^2+{\mathrm{\pi r}}^2=100$,

$$\begin{gathered} {\left(\frac{30-\mathrm{\pi r}}{2}\right)}^2+{\mathrm{\pi r}}^2=100 \\ {(30-\mathrm{\pi r})}^2+4{\mathrm{\pi r}}^2=400 \\ 900-60\mathrm{\pi r}+{\mathrm{\pi }}^2{\mathrm{r}}^2+4{\mathrm{\pi r}}^2-400=0 \\ {\mathrm{\pi }}^2{\mathrm{r}}^2+4{\mathrm{\pi r}}^2-60\mathrm{\pi r}+500=0 \\ {\mathrm{r}}^2(4\mathrm{\pi }+{\mathrm{\pi }}^2)-60\mathrm{\pi r}+500=0 \end{gathered}$$

Using quadratic equation formula,

$$\begin{gathered} r=\frac{60\mathrm{\pi }\pm \sqrt{{\left(60\mathrm{\pi }\right)}^2-4(4\mathrm{\pi }+{\mathrm{\pi }}^2)\left(500\right)}}{2(4\mathrm{\pi }+{\mathrm{\pi }}^2)} \\ =\frac{60\mathrm{\pi }\pm \sqrt{3600{\mathrm{\pi }}^2-8000\mathrm{\pi }-2000{\mathrm{\pi }}^2}}{2(4\mathrm{\pi }+{\mathrm{\pi }}^2)} \\ =\frac{60\mathrm{\pi }\pm (40\mathrm{\pi }-40\sqrt{5\mathrm{\pi }})}{2(4\mathrm{\pi }+{\mathrm{\pi }}^2)} \\ =\frac{30\mathrm{\pi }\pm 20(\mathrm{\pi }-\sqrt{5\mathrm{\pi }})}{4\mathrm{\pi }+{\mathrm{\pi }}^2} \end{gathered}$$

It is imaginary value.

So it is not possible to cut the wire into square and circle.