This question aims to find the **total area** enclosed by a wire when it is **cut down** into **two pieces**. This question uses the concept of the **area of a rectangle** and **an equilateral triangle**. The area of a triangle is mathematically equal to:

\[Area \space of \space triangle \space = \space \frac{Base \space \times \space Height}{2} \]

**Whereas** the area of a **rectangle** is **mathematically** equal to:

\[Area \space of \space rectangle \space = \space Width \space \times \space Length \]

## Expert Answer

Let $ x $ be the amount to be **clipped** from the** square**.

The **sum remaining** for such an **equilateral triangle** would be $ 10 – x $.

We **know** that the **square length** is:

\[= \space \frac{x}{4} \]

Now the **square area** is:

\[= \space (\frac{x}{4})^2 \]

\[= \space \frac{x^2}{16} \]

The area of an **equilateral triangle** is:

\[= \space \frac{\sqrt 3}{4} a^2 \]

Where $ a $ is the **triangle length**.

**Thus**:

\[= \space \frac{10 – x}{3} \]

\[= \space \frac{\sqrt 3}{4} (\frac{10 – x}{3})^2 \]

\[= \space \frac{\sqrt 3(10-x)^2}{36} \]

Now the **total area** is:

\[A(x) \space = \space \frac{x^2}{16} \space + \space \frac{\sqrt 3(10-x)^2}{36}\]

Now **differentiating ** $ A'(x) = 0 $

\[= \space \frac{x}{8} \space – \space {\sqrt 3(10 – x)}{18} \space = \space 0 \]

\[ \frac{x}{8} \space =\space {\sqrt 3(10 – x)}{18} \]

By **cross multiplication**, we get:

\[18x \space = \space 8 \sqrt(3) (10 – x) \]

\[18x \space = \space 80 \sqrt(3) \space – \space 8 \sqrt(3x) \]

\[(18 \space + \space 8 \sqrt(3) x) = \space 80 \sqrt(3) \]

By **simplifying**, we get:

\[x \space = \space 4.35 \]

## Numerical Answer

The value of $ x = 4.35 $ is where we can obtain the **maximum** area **enclosed** by this wire.

## Example

A 20 m **long piece** of wire is **divided** into two parts. Both **pieces** are bent, with one** becoming** a square and the other an **equilateral triangle**. And how would the wire be **spliced** to ensure that the** covered area** is as large as **possible**?

Let $ x $ be the amount to be** clipped** from the square.

The **sum remaining** for such an **equilateral triangle** would be $ 20 – x $.

We **know** that the **square length** is:

\[= \space \frac{x}{4} \]

Now the **square area** is:

\[= \space (\frac{x}{4})^2 \]

\[= \space \frac{x^2}{16} \]

The area of an **equilateral triangle** is:

\[= \space \frac{\sqrt 3}{4} a^2 \]

**Where** $ a $ is the **triangle length**.

**Thus**:

\[= \space \frac{10 – x}{3} \]

\[= \space \frac{\sqrt 3}{4} (\frac{20 – x}{3})^2 \]

\[= \space \frac{\sqrt 3(20-x)^2}{36} \]

Now the **total area** is:

\[A(x) \space = \space \frac{x^2}{16} \space + \space \frac{\sqrt 3(20-x)^2}{36}\]

Now **differentiating** $ A'(x) = 0 $

\[= \space \frac{x}{8} \space – \space {\sqrt 3(20 – x)}{18} \space = \space 0 \]

\[ \frac{x}{8} \space =\space {\sqrt 3(20 – x)}{18} \]

By **cross multiplication**, we get:

\[18x \space = \space 8 \sqrt(3) (20 – x) \]

\[18x \space = \space 160 \sqrt(3) \space – \space 8 \sqrt(3x) \]

\[(18 \space + \space 8 \sqrt(3) x) = \space 160 \sqrt(3) \]

By **simplifying**, we get:

\[x \space = \space 8.699 \]