Key Concepts and Formulas

• Optimization using Derivatives: Finding the minimum or maximum value of a function by finding critical points (where the derivative is zero or undefined) and using the first or second derivative test.
• Area of a Square: $A = s^2$, where $s$ is the side length.
• Area of a Circle: $A = \pi r^2$, where $r$ is the radius.
• Circumference of a Circle: $C = 2\pi r$, where $r$ is the radius.

Step-by-Step Solution

Step 1: Define Variables and Formulate the Constraint

We are given a wire of length 36 m, which is cut into two pieces. One piece is used to form a square, and the other to form a circle. Let $x$ be the length of the wire used for the square, and $y$ be the length used for the circle. The total length is the sum of these two lengths:

$x + y = 36$

This is our constraint. We also know that $x \ge 0$ and $y \ge 0$.

Why this step? Defining variables and the constraint equation is crucial for setting up the optimization problem.

Step 2: Express the Areas in Terms of the Variables

• Square:

  • Perimeter of the square is $x$.
  • Side length, $s = \frac{x}{4}$.
  • Area of the square, $A_S = s^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}$.

• Circle:

  • Circumference of the circle is $y$.
  • Radius of the circle, $r = \frac{y}{2\pi}$.
  • Area of the circle, $A_C = \pi r^2 = \pi \left(\frac{y}{2\pi}\right)^2 = \frac{y^2}{4\pi}$.

Why this step? We need to express the areas in terms of the wire lengths $x$ and $y$ to formulate the total area function.

Step 3: Formulate the Total Area Function

The total area, $A$, is the sum of the area of the square and the area of the circle:

$A = A_S + A_C = \frac{x^2}{16} + \frac{y^2}{4\pi}$

We need to express $A$ as a function of a single variable. Using the constraint $x + y = 36$, we can write $y = 36 - x$. Substituting this into the equation for $A$, we get:

$A(x) = \frac{x^2}{16} + \frac{(36 - x)^2}{4\pi}$

Why this step? We create a function of a single variable that represents the total area, enabling us to use calculus to find the minimum.

Step 4: Find Critical Points using Differentiation

To find the critical points, we differentiate $A(x)$ with respect to $x$ and set the derivative equal to zero:

$A'(x) = \frac{d}{dx} \left( \frac{x^2}{16} + \frac{(36 - x)^2}{4\pi} \right) = \frac{2x}{16} + \frac{2(36 - x)(-1)}{4\pi} = \frac{x}{8} - \frac{36 - x}{2\pi}$

Setting $A'(x) = 0$:

$\frac{x}{8} - \frac{36 - x}{2\pi} = 0$ $\frac{x}{8} = \frac{36 - x}{2\pi}$ $2\pi x = 8(36 - x)$ $\pi x = 4(36 - x)$ $\pi x = 144 - 4x$ $(\pi + 4)x = 144$ $x = \frac{144}{\pi + 4}$

Why this step? We find the critical points, which are potential locations for minimum or maximum area.

Step 5: Verify Minimum (Second Derivative Test)

To confirm that $x = \frac{144}{\pi + 4}$ minimizes the area, we find the second derivative of $A(x)$:

$A''(x) = \frac{d}{dx} \left( \frac{x}{8} - \frac{36 - x}{2\pi} \right) = \frac{1}{8} + \frac{1}{2\pi}$

Since $A''(x) > 0$ for all $x$, the function $A(x)$ is concave up, and the critical point we found corresponds to a minimum.

Why this step? We use the second derivative test to confirm that the critical point corresponds to a local minimum.

Step 6: Calculate $k$ and the Final Expression

We have $x = \frac{144}{\pi + 4}$. Now we find $y$:

$y = 36 - x = 36 - \frac{144}{\pi + 4} = \frac{36(\pi + 4) - 144}{\pi + 4} = \frac{36\pi + 144 - 144}{\pi + 4} = \frac{36\pi}{\pi + 4}$

The circumference of the circle is $k = y = \frac{36\pi}{\pi + 4}$. We want to find the value of $\left(\frac{4}{\pi} + 1\right)k$:

$\left(\frac{4}{\pi} + 1\right)k = \left(\frac{4 + \pi}{\pi}\right) \left(\frac{36\pi}{\pi + 4}\right) = \frac{4 + \pi}{\pi} \cdot \frac{36\pi}{\pi + 4} = 36$

Why this step? We calculate the circumference of the circle, $k$, and then evaluate the final expression to arrive at the answer.

Common Mistakes & Tips

• Units: Ensure consistency in units throughout the problem.
• Second Derivative Test: Always verify that the critical point corresponds to a minimum (or maximum) using the second derivative test.
• Algebraic Manipulation: Be careful with algebraic manipulations, especially when dealing with fractions and the constant $\pi$.

Summary

We minimized the sum of the areas of a square and a circle formed from a wire of length 36 m. By expressing the total area as a function of a single variable, differentiating, and using the second derivative test, we found the optimal length for each shape. Finally, we calculated the value of the given expression, which is 36.

Final Answer

The final answer is \boxed{36}.