Question

a piece of wire of length 61 is cut into two pieces. one piece is bent into a square and the other is bent into a circle. if the sum of the areas enclosed by each part is a minimum, what is the length of each part? to minimize the combined area, the wire should be cut so that a length of □ is used for the circle and a length of □ is used for the square. (round to the nearest thousandth as needed.)

Explanation:

Step1: Set up variables

Let the length of wire used for the circle be \(x\), then the length used for the square is \(61 - x\).
The radius of the circle \(r=\frac{x}{2\pi}\), and the side of the square \(s = \frac{61 - x}{4}\).

Step2: Write the area formula

The area of the circle \(A_{1}=\pi r^{2}=\pi(\frac{x}{2\pi})^{2}=\frac{x^{2}}{4\pi}\).
The area of the square \(A_{2}=s^{2}=(\frac{61 - x}{4})^{2}=\frac{(61 - x)^{2}}{16}\).
The combined area \(A(x)=\frac{x^{2}}{4\pi}+\frac{(61 - x)^{2}}{16}\).

Step3: Differentiate the area function

Differentiate \(A(x)\) with respect to \(x\):
\(A^{\prime}(x)=\frac{2x}{4\pi}+\frac{2(61 - x)(- 1)}{16}=\frac{x}{2\pi}-\frac{61 - x}{8}\).
Set \(A^{\prime}(x) = 0\) for critical points:
\(\frac{x}{2\pi}-\frac{61 - x}{8}=0\).
Multiply through by \(8\pi\) to clear the fractions:
\(4x-\pi(61 - x)=0\).
\(4x-61\pi+\pi x = 0\).
\(x(4 + \pi)=61\pi\).
\(x=\frac{61\pi}{4+\pi}\approx\frac{61\times3.1416}{4 + 3.1416}=\frac{191.6376}{7.1416}\approx26.834\).
Then \(61 - x=61-26.834 = 34.166\).

Answer:

The length used for the circle is approximately \(26.834\) and for the square is approximately \(34.166\).