Question

a wire b units long is cut into two pieces. one piece is bent into an equilateral triangle 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? the length of the piece bent into the shape of an equilateral triangle is units. (type an exact answer, using π as needed.)

Explanation:

Step1: Let the length of the wire used for the equilateral triangle be \(x\) units. Then the length of the wire used for the circle is \((b - x)\) units.

The side length of the equilateral triangle \(s=\frac{x}{3}\), and its area \(A_{t}=\frac{\sqrt{3}}{4}s^{2}=\frac{\sqrt{3}}{4}(\frac{x}{3})^{2}=\frac{\sqrt{3}}{36}x^{2}\).
The circumference of the circle \(C = b - x\), and since \(C = 2\pi r\) (where \(r\) is the radius of the circle), \(r=\frac{b - x}{2\pi}\). The area of the circle \(A_{c}=\pi r^{2}=\pi(\frac{b - x}{2\pi})^{2}=\frac{(b - x)^{2}}{4\pi}\).

Step2: The total area \(A(x)=\frac{\sqrt{3}}{36}x^{2}+\frac{(b - x)^{2}}{4\pi}\)

Differentiate \(A(x)\) with respect to \(x\):
\(A^{\prime}(x)=\frac{\sqrt{3}}{18}x-\frac{1}{2\pi}(b - x)\)
Set \(A^{\prime}(x) = 0\) for critical points:
\(\frac{\sqrt{3}}{18}x-\frac{1}{2\pi}(b - x)=0\)
\(\frac{\sqrt{3}}{18}x+\frac{1}{2\pi}x=\frac{b}{2\pi}\)
\(x(\frac{\sqrt{3}\pi+9}{9\pi})=\frac{b}{2\pi}\)
\(x=\frac{9b}{2(\sqrt{3}\pi + 9)}\)

Answer:

The length of the wire used for the equilateral triangle is \(\frac{9b}{2(\sqrt{3}\pi + 9)}\) units, and the length of the wire used for the circle is \(b-\frac{9b}{2(\sqrt{3}\pi + 9)}=\frac{b(2\sqrt{3}\pi+ 18 - 9)}{2(\sqrt{3}\pi + 9)}=\frac{b(2\sqrt{3}\pi + 9)}{2(\sqrt{3}\pi + 9)}\) units.