To solve the problem, let’s break it down:

Definitions:

1. A wire of 16 m is divided into two pieces.
2. One piece is bent into a square, and the other into an equilateral triangle.
3. Let x be the length of the wire used for the square, and 16 - x be the length used for the triangle.

Formulas:

• Square Area: If the perimeter of the square is $x$, the side length is $\frac{x}{4}$, and the area is: $A_{\text{square}} = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}.$

• Equilateral Triangle Area: If the perimeter of the triangle is $16 - x$, each side is $\frac{16 - x}{3}$, and the area is: $A_{\text{triangle}} = \frac{\sqrt{3}}{4} \left(\frac{16 - x}{3}\right)^2 = \frac{\sqrt{3}}{36} (16 - x)^2.$

• Total Area: The total area $A$ is the sum of the two: $A(x) = \frac{x^2}{16} + \frac{\sqrt{3}}{36} (16 - x)^2.$

Steps to Solve:

(a) Maximize Total Area:

We need to find $x$ that maximizes $A(x)$. To do this:

1. Compute the derivative $A'(x)$.
2. Solve $A'(x) = 0$ to find critical points.
3. Use the second derivative test or boundary values ($x = 0$ or $x = 16$) to confirm the maximum.

(b) Minimize Total Area:

To minimize $A(x)$, follow the same steps as above but identify the $x$ corresponding to the minimum.

Let me calculate these values for you. sympy as sp

Define variables

x = sp.symbols('x') total_length = 16

Area formulas

A_square = (x**2) / 16 # Square area A_triangle = (sp.sqrt(3) / 36) * (total_length - x)**2 # Triangle area A_total = A_square + A_triangle # Total area

Derivative of the total area

A_prime = sp.diff(A_total, x)

Solve critical points

critical_points = sp.solve(A_prime, x)

Evaluate second derivative to classify critical points

A_double_prime = sp.diff(A_prime, x) critical_values = [(cp, A_total.subs(x, cp), A_double_prime.subs(x, cp)) for cp in critical_points]

Boundary points for total area

boundary_values = [(0, A_total.subs(x, 0)), (total_length, A_total.subs(x, total_length))]

critical_values, boundary_valuesThe critical points and boundary values have been analyzed. Here's the result:

1. Critical point: $x = -\frac{256}{11} + \frac{192\sqrt{3}}{11}$ meters for the square.
   • Corresponding total area: $\frac{(-\frac{256}{11} + \frac{192\sqrt{3}}{11})^2}{16} + \frac{\sqrt{3}}{36} (16 - (-\frac{256}{11} + \frac{192\sqrt{3}}{11}))^2$.
   • Nature (from second derivative): Positive, indicating **minimize area}.

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