Key Concepts and Formulas

• Area of a Square: If $s$ is the side length of a square, its area is $A = s^2$.
• Area of an Equilateral Triangle: If $a$ is the side length of an equilateral triangle, its area is $A = \frac{\sqrt{3}}{4}a^2$.
• Optimization using Derivatives: To find the minimum (or maximum) value of a function $f(x)$, find the critical points by solving $f'(x) = 0$. Check the second derivative $f''(x)$ to confirm if it's a minimum ($f''(x) > 0$) or maximum ($f''(x) < 0$).

Step-by-Step Solution

Step 1: Define Variables and Formulate the Constraint Equation

Let $L = 22$ m be the total length of the wire. Let $x$ be the length of the wire used to form the square, and $y$ be the length of the wire used to form the equilateral triangle. Then, we have the constraint: $x + y = 22$ We want to express $y$ in terms of $x$: $y = 22 - x$

Step 2: Express Side Lengths in Terms of x and y

• Square: The perimeter of the square is $x$, so each side has length $s = \frac{x}{4}$. The area of the square is $A_s = s^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}$.
• Equilateral Triangle: The perimeter of the equilateral triangle is $y$, so each side has length $a = \frac{y}{3}$. The area of the equilateral triangle is $A_t = \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}\left(\frac{y}{3}\right)^2 = \frac{\sqrt{3}y^2}{36}$.

Step 3: Express the Total Area in Terms of a Single Variable

The total area $A$ is the sum of the area of the square and the area of the equilateral triangle: $A = A_s + A_t = \frac{x^2}{16} + \frac{\sqrt{3}y^2}{36}$ Substitute $y = 22 - x$ into the equation for $A$: $A(x) = \frac{x^2}{16} + \frac{\sqrt{3}(22 - x)^2}{36}$

Step 4: Find the Critical Points by Taking the Derivative

To minimize the total area, we need to find the critical points by taking the derivative of $A(x)$ with respect to $x$ and setting it equal to zero: $A'(x) = \frac{2x}{16} + \frac{2\sqrt{3}(22 - x)(-1)}{36} = \frac{x}{8} - \frac{\sqrt{3}(22 - x)}{18}$ Set $A'(x) = 0$: $\frac{x}{8} = \frac{\sqrt{3}(22 - x)}{18}$ $18x = 8\sqrt{3}(22 - x)$ $18x = 176\sqrt{3} - 8\sqrt{3}x$ $18x + 8\sqrt{3}x = 176\sqrt{3}$ $x(18 + 8\sqrt{3}) = 176\sqrt{3}$ $x = \frac{176\sqrt{3}}{18 + 8\sqrt{3}} = \frac{88\sqrt{3}}{9 + 4\sqrt{3}}$

Step 5: Solve for y (the side length of the triangle)

Now, we find the value of $y = 22 - x$: $y = 22 - \frac{88\sqrt{3}}{9 + 4\sqrt{3}} = \frac{22(9 + 4\sqrt{3}) - 88\sqrt{3}}{9 + 4\sqrt{3}} = \frac{198 + 88\sqrt{3} - 88\sqrt{3}}{9 + 4\sqrt{3}} = \frac{198}{9 + 4\sqrt{3}}$ Since $y$ is the perimeter of the equilateral triangle, and $a$ is the side length of the triangle, $a = \frac{y}{3}$. Thus: $a = \frac{198}{3(9 + 4\sqrt{3})} = \frac{66}{9 + 4\sqrt{3}}$

Step 6: Verify that this is a Minimum (Optional, but good practice)

Take the second derivative of A(x): $A''(x) = \frac{1}{8} + \frac{\sqrt{3}}{18}$ Since $A''(x) > 0$, we have found a minimum.

Step 7: State the Final Answer

The length of the side of the equilateral triangle that minimizes the combined area is $\frac{66}{9 + 4\sqrt{3}}$.

Common Mistakes & Tips

• Units: Always keep track of the units. In this case, the units are meters.
• Simplification: Be careful when simplifying expressions, especially with square roots.
• Check for Minimum/Maximum: While not strictly necessary for this problem, it's good practice to check the second derivative to ensure you've found a minimum (or maximum) and not a saddle point.

Summary

We minimized the combined area of a square and an equilateral triangle formed from a wire of length 22m. By expressing the areas in terms of a single variable (the length of the wire used for the square), taking the derivative, and setting it equal to zero, we found the optimal side length of the equilateral triangle to be $\frac{66}{9 + 4\sqrt{3}}$.

Final Answer

The final answer is \boxed{\frac{66}{9 + 4\sqrt 3}}, which corresponds to option (B).