Let's denote:

• The total length of the wire: $L = 6666$
• The length of the wire used to form the circle: $x$
• The length of the wire used to form the square: $6666 - x$

Step 1: Express the areas of the circle and the square

• For the circle: The circumference of the circle is $x$, so we have: $2 \pi r = x \quad \Rightarrow \quad r = \frac{x}{2\pi}$ The area of the circle is: $A_{\text{circle}} = \pi r^2 = \pi \left( \frac{x}{2\pi} \right)^2 = \frac{x^2}{4\pi}$

• For the square: The perimeter of the square is $6666 - x$, so each side of the square is: $s = \frac{6666 - x}{4}$ The area of the square is: $A_{\text{square}} = s^2 = \left( \frac{6666 - x}{4} \right)^2 = \frac{(6666 - x)^2}{16}$

Step 2: Define the total area

The total area $A_{\text{total}}$ is the sum of the areas of the circle and the square: $A_{\text{total}}(x) = \frac{x^2}{4\pi} + \frac{(6666 - x)^2}{16}$

Step 3: Minimize the total area

To minimize the total area, we need to find the value of $x$ that minimizes the function $A_{\text{total}}(x)$. We do this by finding the derivative of $A_{\text{total}}(x)$ and setting it to zero:

$A_{\text{total}}'(x) = \frac{2x}{4\pi} - \frac{2(6666 - x)}{16}$ Simplifying: $A_{\text{total}}'(x) = \frac{x}{2\pi} - \frac{6666 - x}{8}$ Set the derivative equal to zero: $\frac{x}{2\pi} = \frac{6666 - x}{8}$

Multiply both sides by $16\pi$ to eliminate fractions: $8x = 2\pi(6666 - x)$ Expanding: $8x = 13332\pi - 2\pi x$ Now combine the $x$-terms: $8x + 2\pi x = 13332\pi$ Factor out $x$: $x(8 + 2\pi) = 13332\pi$ Solve for $x$: $x = \frac{13332\pi}{8 + 2\pi}$ Using $\pi \approx 3.1416$, this gives: $x \approx \frac{13332 \times 3.1416}{8 + 2 \times 3.1416} \approx \frac{41865.43}{14.2832} \approx 2932.619$

Step 4: Answer for Part (a)

Thus, to minimize the combined area:

• The length used for the circle is approximately 2932.619.
• The length used for the square is $6666 - 2932.619 \approx 3733.381$.

Would you like details on part (b) or more explanations?

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Here are 5 questions that build upon this:

1. How do you find the maximum of a function after finding the minimum?
2. What role does calculus play in optimization problems like this?
3. How would the areas change if the total wire length was different?
4. Can you solve the same problem for a triangle instead of a square?
5. How does the derivative help in finding where a function's critical points occur?

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