To tackle the problem, we first need to derive the combined area function $A(x)$ based on the wire lengths allocated to the circle and square.

Part 1: Combined Area Function

1. Circle: Let $x$ be the length of wire used for the circle. The circumference $C$ of the circle is $x$, so the radius $r$ is: $r = \frac{x}{2\pi}$ The area $A_c$ of the circle is: $A_c = \pi r^2 = \pi \left(\frac{x}{2\pi}\right)^2 = \frac{x^2}{4\pi}$

2. Square: The remaining length of wire for the square is $59 - x$. The perimeter $P$ of the square is $59 - x$, so the side length $s$ is: $s = \frac{59 - x}{4}$ The area $A_s$ of the square is: $A_s = s^2 = \left(\frac{59 - x}{4}\right)^2 = \frac{(59 - x)^2}{16}$

3. Combined Area: Therefore, the total area $A(x)$ is: $A(x) = A_c + A_s = \frac{x^2}{4\pi} + \frac{(59 - x)^2}{16}$

Part 2: Interval of Interest

The interval for $x$ (the wire used for the circle) is from 0 to 59: $[0, 59]$

Part 3: Minimizing the Combined Area

To minimize the area, we can take the derivative of $A(x)$, set it to zero, and solve for $x$:

1. Find the derivative: $A'(x) = \frac{1}{2\pi} x - \frac{59 - x}{8}$ Setting $A'(x) = 0$: $\frac{x}{2\pi} = \frac{59 - x}{8}$

2. Solve for $x$: $8x = 2\pi(59 - x)$ $8x + 2\pi x = 118\pi$ $x(8 + 2\pi) = 118\pi$ $x = \frac{118\pi}{8 + 2\pi} \approx 25.954$

Thus, the lengths for minimizing the area are:

• Circle: $x \approx 25.954$
• Square: $59 - x \approx 33.046$

Part 4: Maximizing the Combined Area

To maximize the area, we note that the areas of both shapes increase with more wire allocated. Hence, to maximize the area, allocate all wire to one shape:

1. All wire to the circle:

   • Circle: $x = 59$
   • Square: $0$

2. All wire to the square:

   • Circle: $0$
   • Square: $59$

However, typically, we want a balanced allocation for optimal areas. The critical points found through the first derivative will be evaluated to check for a maximum.

Summary:

• For minimizing combined area: Circle = $25.954$, Square = $33.046$
• For maximizing combined area: Circle = $59$, Square = $0$ or Circle = $0$, Square = $59$

If you have any questions or need further details, feel free to ask!

Further Exploration Questions:

1. How does changing the total length of the wire affect the area?
2. What are the implications of different shapes on the area calculations?
3. Can we derive a general formula for any lengths of wire?
4. How would this change if we introduced different shapes, like rectangles?
5. What are practical applications of maximizing area with constrained lengths?

Tip: Consider using calculus to find optimal solutions in problems involving areas and perimeters.