A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the oth

Q: Why do we use $x$ and $36-x$ instead of just $x$ and $y$?

Using two variables $x$ and $y$ makes the calculus harder. By using the constraint $x+y=36$, we can express $y$ as $36-x$, which turns the problem into a single-variable optimization task, which is much easier to solve.

Q: How do I know if this is a minimum or a maximum?

You can check the second derivative. The second derivative of $A(x) = \frac{x^2}{16} + \frac{(36-x)^2}{4\pi}$ is $A''(x) = \frac{1}{8} + \frac{1}{2\pi}$. Since this is positive, the function is concave up, meaning our critical point is indeed a minimum.

Q: Can I use $r$ as the variable instead of $x$?

Absolutely! You could express everything in terms of the radius $r$ of the circle. The circumference would be $2\pi r$, and the square's perimeter would be $36 - 2\pi r$. The math would lead to the same result, just with a different variable.

Question

A wire of length $36	ext{ m}$ is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is $k$ (meter), then $(\frac{4}{\pi} + 1) k$ is equal to .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a thirty-six-meter wire, a simple, straight line of potential. Your task is to partition this wire into two segments, one to be bent into a square and the other into a circle.

We start with the constraint: if the first piece has length $x$ and the second has length $y$, then $x + y = 36$. This is our anchor.

### The Architecture of Shapes

Now, let us consider the geometry. A square with perimeter $x$ has a side length of $s = \frac{x}{4}$. Its area is:

$$A_s = s^2 = \frac{x^2}{16}$$

A circle with circumference $y$ has a radius $r = \frac{y}{2\pi}$. Its area is:

$$A_c = \pi r^2 = \pi \left( \frac{y}{2\pi} \right)^2 = \frac{y^2}{4\pi}$$

By substituting $y = 36 - x$, we transform our area function into a single-variable expression:

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

This function represents the total area for any given cut $x$.

### The Calculus of Change

To find the minimum area, we must look for the point where the rate of change of the area with respect to $x$ is zero. We differentiate $A(x)$ with respect to $x$:

$$\frac{dA}{dx} = \frac{2x}{16} + \frac{2(36 - x)(-1)}{4\pi}$$

Simplifying this, we get:

$$\frac{dA}{dx} = \frac{x}{8} - \frac{36 - x}{2\pi}$$

Setting this derivative to zero, we find the critical point:

$$\frac{x}{8} = \frac{36 - x}{2\pi}$$

Multiplying by 8, we get $x = \frac{4(36 - x)}{\pi}$, which rearranges to $\pi x = 144 - 4x$. Solving for $x$, we find:

$$x = \frac{144}{\pi + 4}$$

### The Elegant Cancellation

The problem asks for the value of $(\frac{4}{\pi} + 1)k$, where $k$ is the circumference of the circle, $k = y = 36 - x$. Substituting our value of $x$, we get:

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

Now, we evaluate the final expression:

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

Notice the beautiful symmetry: the $(\pi + 4)$ terms cancel out, and the $\pi$ terms cancel out. This leaves us with the final result of **36**.

Enter Numerical Value:

Visualized Solution

Visualizing the Wire Split

Area of the Square

Area of the Circle

Total Area Function $A (x)$

Differentiating the Area Function

Finding the Critical Point

Solving for $x$

Finding the Circumference $k$

Final Calculation

Key Takeaways

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Architecture of Shapes

The Calculus of Change

The Elegant Cancellation

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