We aim to find where to cut a 6-meter wire so that the total area of an equilateral triangle and a circle formed from the two resulting pieces is minimized. Here's the detailed breakdown:

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Step 1: Define variables

Let:

• $x$ = length of the piece used to form the equilateral triangle.
• $6 - x$ = length of the piece used to form the circle.

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Step 2: Express the areas

Equilateral triangle

The perimeter of the triangle is $x$, so the side length of the triangle is: $s = \frac{x}{3}$ The area of an equilateral triangle is: $A_{\text{triangle}} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \left(\frac{x}{3}\right)^2 = \frac{\sqrt{3}}{36} x^2$

Circle

The circumference of the circle is $6 - x$, so the radius of the circle is: $r = \frac{6 - x}{2\pi}$ The area of the circle is: $A_{\text{circle}} = \pi r^2 = \pi \left(\frac{6 - x}{2\pi}\right)^2 = \frac{(6 - x)^2}{4\pi}$

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Step 3: Total area

The total area $A_{\text{total}}$ is the sum of the areas of the triangle and the circle: $A_{\text{total}} = A_{\text{triangle}} + A_{\text{circle}}$ $A_{\text{total}} = \frac{\sqrt{3}}{36} x^2 + \frac{(6 - x)^2}{4\pi}$

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Step 4: Minimize the total area

To minimize $A_{\text{total}}$, we differentiate with respect to $x$ and set the derivative to zero: $\frac{dA_{\text{total}}}{dx} = \frac{d}{dx} \left( \frac{\sqrt{3}}{36} x^2 + \frac{(6 - x)^2}{4\pi} \right)$ $\frac{dA_{\text{total}}}{dx} = \frac{\sqrt{3}}{18} x - \frac{2(6 - x)}{4\pi}$ Simplify: $\frac{\sqrt{3}}{18} x - \frac{3 - \frac{x}{2}}{\pi} = 0$

Let’s solve this equation for $x$. sympy import symbols, diff, Eq, solve, pi, sqrt

Define variable

x = symbols('x')

Define total area function

A_triangle = (sqrt(3)/36) * x**2 A_circle = ((6 - x)**2) / (4 * pi) A_total = A_triangle + A_circle

Differentiate total area with respect to x

dA_dx = diff(A_total, x)

Solve for critical points where derivative is zero

critical_points = solve(Eq(dA_dx, 0), x) critical_pointsThe critical point for $x$ is:

$x = \frac{-54\sqrt{3}\pi + 486}{81 - 3\pi^2}$

This can be evaluated numerically for practical use. Let me calculate the approximate value. Numerical evaluation of the critical point critical_points_numeric = [point.evalf() for point in critical_points] critical_points_numericThe critical point is approximately $x \approx 3.74$ meters. This means:

• Use 3.74 meters of the wire for the equilateral triangle.
• Use the remaining 6 - 3.74 = 2.26 meters for the circle.

This allocation minimizes the total area of both figures.

Would you like a detailed explanation of the derivation or verification of the solution? Here are some follow-up questions to consider:

1. Why does minimizing the total area reduce storage space?
2. How does the geometry of the triangle and circle affect the optimization?
3. What happens if we use the entire wire for one shape?
4. How sensitive is the result to small changes in wire length?
5. How can this approach be generalized to other shapes?

Tip: Always check the second derivative to confirm whether a critical point is a minimum or maximum.