Optimizing Wire Length for Minimum Combined Area

We are given a wire of total length $L$. This wire is cut into two pieces. One piece is used to form a circle, and the other piece is used to form an equilateral triangle. Our goal is to find the length of the wire used for the circle that minimizes the total combined area of the two shapes.

Mathematical Formulation

Let $x$ be the length of the wire used for the circle. Then, the length of the wire used for the equilateral triangle is $L-x$.

Circle Calculations

• Circumference of the circle, $C = x$.
• The radius $r$ is given by $C = 2\pi r$, so $r = \frac{x}{2\pi}$.
• Area of the circle, $A_c = \pi r^2 = \pi \left(\frac{x}{2\pi}\right)^2 = \frac{\pi x^2}{4\pi^2} = \frac{x^2}{4\pi}$.

Equilateral Triangle Calculations

• Perimeter of the triangle, $P = L-x$.
• The side length $s$ is given by $P = 3s$, so $s = \frac{L-x}{3}$.
• Area of the equilateral triangle, $A_t = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \left(\frac{L-x}{3}\right)^2 = \frac{\sqrt{3}}{4} \frac{(L-x)^2}{9} = \frac{\sqrt{3}(L-x)^2}{36}$.

Total Area and Minimization

The total combined area $A(x)$ is the sum of the areas of the circle and the triangle: $ A(x) = A_c + A_t = \frac{x^2}{4\pi} + \frac{\sqrt{3}(L-x)^2}{36} $

To find the minimum area, we need to find the value of $x$ where the derivative of $A(x)$ with respect to $x$ is zero.

Calculate the derivative $A'(x)$: $ A'(x) = \frac{d}{dx} \left( \frac{x^2}{4\pi} + \frac{\sqrt{3}(L-x)^2}{36} \right) $ $ A'(x) = \frac{2x}{4\pi} + \frac{\sqrt{3}}{36} \cdot 2(L-x)(-1) $ $ A'(x) = \frac{x}{2\pi} - \frac{\sqrt{3}(L-x)}{18} $

Set $A'(x) = 0$ to find the critical point: $ \frac{x}{2\pi} - \frac{\sqrt{3}(L-x)}{18} = 0 $ $ \frac{x}{2\pi} = \frac{\sqrt{3}(L-x)}{18} $

Now, solve for $x$: $ 18x = 2\pi\sqrt{3}(L-x) $ $ 9x = \pi\sqrt{3}(L-x) $ $ 9x = \pi\sqrt{3}L - \pi\sqrt{3}x $ $ 9x + \pi\sqrt{3}x = \pi\sqrt{3}L $ $ x(9 + \pi\sqrt{3}) = \pi\sqrt{3}L $ $ x = \frac{\pi\sqrt{3}L}{9 + \pi\sqrt{3}} $

To confirm this is a minimum, we check the second derivative: $ A''(x) = \frac{d}{dx} \left( \frac{x}{2\pi} - \frac{\sqrt{3}(L-x)}{18} \right) = \frac{1}{2\pi} - \frac{\sqrt{3}}{18}(-1) = \frac{1}{2\pi} + \frac{\sqrt{3}}{18} $ Since $A''(x)$ is positive, the value of $x$ found corresponds to a minimum area.

Conclusion

The length of the wire used to construct the circle that minimizes the combined area is $ x = \frac{\sqrt{3}\pi L}{9+\sqrt{3}\pi} $.