A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square an

Q: Wait, why do we reject the negative root or boundary values?

In this optimization problem, we are looking for a physical length $h$. Since $h$ represents the side of a hexagon, it must be greater than $0$. Furthermore, the total length of the wire is $20\text{ m}$, so $6h$ cannot exceed $20$. Our critical point falls within this physical domain, so it is the valid minimum.

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure optimization problem, not a dynamics problem. While physics often uses energy minimization (like a soap film forming a shape to minimize surface energy), here we are using calculus to minimize a geometric area function. The logic is identical: we are finding the state of 'least cost' or 'least area'.

Q: Is there a geometric shortcut without using calculus?

You could potentially use the AM-GM inequality if the function were simpler, but because we have a quadratic function of $h$, calculus is the most robust and standard tool. It guarantees we find the exact point where the slope of the area function is zero, which is the hallmark of a minimum.

Question

A wire of length $20	ext{ m}$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is:

Options:

(A) $\frac{5}{2+\sqrt{3}}$

(B) $\frac{10}{2+3\sqrt{3}}$

(C) $\frac{5}{3+\sqrt{3}}$

(D) $\frac{10}{3+2\sqrt{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a workshop with a single, pristine piece of wire, exactly $20	ext{ m}$ long. You have a pair of wire cutters in your hand and a challenge: cut this wire into two pieces to form a square and a regular hexagon such that the total area enclosed by both shapes is as small as possible.

This is not just a math problem; it is a lesson in trade-offs. Every centimeter of wire you give to the square is a centimeter you cannot give to the hexagon. This tension is the heart of optimization.

### Defining the Boundaries

Let $s$ be the side length of the square and $h$ be the side length of the regular hexagon. The wire is our total resource. The perimeter of the square is $4s$, and the perimeter of the hexagon is $6h$.

Since the total length is $20	ext{ m}$, we have the constraint equation:
$$4s + 6h = 20$$

To make our lives easier, we simplify this by dividing by $2$, giving us $2s + 3h = 10$. This is our anchor. It tells us that $s$ and $h$ are locked in a dance; if $h$ grows, $s$ must shrink. We can express $s$ explicitly as:
$$s = \frac{10 - 3h}{2}$$

### Building the Area Model

Now, we construct the total area function $A(h)$. The area of the square is $A_s = s^2$. Substituting our expression for $s$, we get:
$$A_s = \left(\frac{10 - 3h}{2}ight)^2$$

Next, the hexagon. A regular hexagon is just six equilateral triangles glued together. The area of one such triangle with side $h$ is $\frac{\sqrt{3}}{4}h^2$. Multiplying by $6$, we get:
$$A_h = \frac{3\sqrt{3}}{2}h^2$$

The total area is the sum:
$$A(h) = \left(\frac{10 - 3h}{2}ight)^2 + \frac{3\sqrt{3}}{2}h^2$$

### Finding the Minimum

To find the minimum, we look for the critical point where the rate of change of the area is zero. We differentiate $A(h)$ with respect to $h$:
$$\frac{dA}{dh} = 2\left(\frac{10 - 3h}{2}ight)\left(-\frac{3}{2}ight) + 2\left(\frac{3\sqrt{3}}{2}ight)h$$

Simplifying this, we get:
$$\frac{dA}{dh} = -\frac{3}{2}(10 - 3h) + 3\sqrt{3}h$$

Setting this to zero, we have:
$$-\frac{3}{2}(10 - 3h) + 3\sqrt{3}h = 0$$

Multiplying by $2$ to clear the fraction, we get $-3(10 - 3h) + 6\sqrt{3}h = 0$, which expands to $-30 + 9h + 6\sqrt{3}h = 0$. Factoring out $h$, we find:
$$h(9 + 6\sqrt{3}) = 30$$

### Final Calculation

Finally, isolating $h$, we get:
$$h = \frac{30}{9 + 6\sqrt{3}}$$

Dividing numerator and denominator by $3$, we arrive at the elegant result:
**$$h = \frac{10}{3 + 2\sqrt{3}}$$**

This is the precise length of the hexagon side that minimizes the combined area. It is a beautiful intersection of geometry and calculus, proving that even with a simple piece of wire, the mathematics of optimization reveals a perfect, singular solution.

Select Answer:

Visualized Solution

Visualizing the Problem

The Perimeter Constraint

Expressing $s$ in terms of $h$

Area of the Square

Area of the Regular Hexagon

Total Area Function $A (h)$

Differentiating the Area

Simplifying the Derivative

Solving for $h$ - Part 1

The Final Result

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Defining the Boundaries

Building the Area Model

Finding the Minimum

Final Calculation

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