# A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is a maximum?

This question aims to find the total area enclosed by a wire when it is cut down into two piecesThis 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:

$Area \space of \space rectangle \space = \space Width \space \times \space Length$

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:

$x \space = \space 4.35$

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:

$x \space = \space 8.699$