A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square an

Q: Why do we use $x$ for the triangle and $22-x$ for the square? Could we swap them?

Absolutely! Mathematics is indifferent to your labeling. If you let the square be $x$, you would simply solve for the square's side first. The final result for the triangle's side would remain identical because the physics of the problem—the trade-off between the two shapes—is invariant.

Q: How do we know this critical point is a minimum and not a maximum?

Great question! In optimization, we check the second derivative. Here, $\frac{d^2A}{dx^2} = \frac{\sqrt{3}}{18} + \frac{1}{8}$, which is clearly positive. A positive second derivative confirms that our critical point is indeed a local minimum.

Q: Why did we divide by 3 at the very end?

The question asks for the side length of the equilateral triangle, not the length of the wire piece. We defined the triangle's perimeter as $x$, so its side is $x/3$. Always re-read the final question to ensure you are answering what is asked!

Question

A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :

Options:

(A) $\frac{22}{9 + 4\sqrt{3}}$

(B) $\frac{66}{9 + 4\sqrt{3}}$

(C) $\frac{22}{4 + 9\sqrt{3}}$

(D) $\frac{66}{4 + 9\sqrt{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a wire of length $22 	ext{ m}$. You hold a pair of shears, ready to make a single cut. This is a choice that dictates the geometry of two distinct worlds: a square and an equilateral triangle.

We define our cut at length $x$. This leaves us with two pieces: one of length $x$ and another of length $22 - x$.

We bend the first piece ($x$) into an equilateral triangle and the second piece ($22 - x$) into a square. For the triangle, the perimeter is $3a = x$, which implies $a = \frac{x}{3}$. For the square, the perimeter is $4b = 22 - x$, which implies $b = \frac{22 - x}{4}$.

### The Area Function

We must now quantify the total area $A(x)$ of these shapes. The area of the equilateral triangle is $A_T = \frac{\sqrt{3}}{4} a^2$, and the area of the square is $A_S = b^2$.

Substituting our expressions for $a$ and $b$, we obtain:

$$A_T = \frac{\sqrt{3}}{4} \left( \frac{x}{3} ight)^2 = \frac{\sqrt{3} x^2}{36}$$

$$A_S = \left( \frac{22 - x}{4} ight)^2 = \frac{(22 - x)^2}{16}$$

The total area function, $A(x) = A_T + A_S$, serves as our master equation:

$$A(x) = \frac{\sqrt{3} x^2}{36} + \frac{(22 - x)^2}{16}$$

### The Calculus of Change

To find the minimum area, we locate the critical point where the derivative of the area with respect to $x$ is zero. Differentiating $A(x)$ yields:

$$\frac{dA}{dx} = \frac{\sqrt{3}}{36} \cdot 2x + \frac{1}{16} \cdot 2(22 - x) \cdot (-1)$$

Simplifying this expression, we get:

$$\frac{dA}{dx} = \frac{\sqrt{3} x}{18} - \frac{22 - x}{8}$$

Setting $\frac{dA}{dx} = 0$ identifies the equilibrium point. We equate the two terms:

$$\frac{\sqrt{3} x}{18} = \frac{22 - x}{8}$$

### Final Calculation

Multiplying by $72$ to clear the denominators, we arrive at:

$$4\sqrt{3} x = 9(22 - x)$$

Expanding and isolating $x$:

$$4\sqrt{3} x = 198 - 9x$$

$$x(9 + 4\sqrt{3}) = 198 \implies x = \frac{198}{9 + 4\sqrt{3}}$$

Since the side length of the triangle is $a = \frac{x}{3}$, we divide our result by $3$:

$$a = \frac{1}{3} \left( \frac{198}{9 + 4\sqrt{3}} ight) = \frac{66}{9 + 4\sqrt{3}}$$

The side length of the triangle that minimizes the total area is **$a = \frac{66}{9 + 4\sqrt{3}} 	ext{ m}$**.

Select Answer:

Visualized Solution

Visualizing the Wire

Defining the Cut

Forming the Shapes

Calculating Side Lengths

Area of the Triangle

Area of the Square

The Total Area Function

Condition for Minima

Differentiating the Function

Setting Derivative to Zero

Solving for x (Cross Multiplication)

Solving for x (Grouping terms)

Finding x

Finding the Side of the Triangle

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Area Function

The Calculus of Change

Final Calculation

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