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 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:

$$Areaoftriangle=\frac{Base\times Height}{2}$$

Whereas the area of a rectangle is mathematically equal to:

$$Areaofrectangle=Width\times Length$$

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:

$$=\frac{x}{4}$$

Now the square area is:

$$=(\frac{x}{4}{)}^2$$

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

The area of an equilateral triangle is:

$$=\frac{\sqrt{3}}{4}{a}^2$$

Where $a$ is the triangle length.

Thus:

$$=\frac{10–x}{3}$$

$$=\frac{\sqrt{3}}{4}(\frac{10–x}{3}{)}^2$$

$$=\frac{\sqrt{3}(10-x{)}^2}{36}$$

Now the total area is:

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

Now differentiating  $A^{\prime }(x)=0$

$$=\frac{x}{8}–\sqrt{3}(10–x)18=0$$

$$\frac{x}{8}=\sqrt{3}(10–x)18$$

By cross multiplication, we get:

$$18x=8\sqrt{(}3)(10–x)$$

$$18x=80\sqrt{(}3)–8\sqrt{(}3x)$$

$$(18+8\sqrt{(}3)x)=80\sqrt{(}3)$$

By simplifying, we get:

$$x=4.35$$

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:

$$=\frac{x}{4}$$

Now the square area is:

$$=(\frac{x}{4}{)}^2$$

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

The area of an equilateral triangle is:

$$=\frac{\sqrt{3}}{4}{a}^2$$

Where $a$ is the triangle length.

Thus:

$$=\frac{10–x}{3}$$

$$=\frac{\sqrt{3}}{4}(\frac{20–x}{3}{)}^2$$

$$=\frac{\sqrt{3}(20-x{)}^2}{36}$$

Now the total area is:

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

Now differentiating  $A^{\prime }(x)=0$

$$=\frac{x}{8}–\sqrt{3}(20–x)18=0$$

$$\frac{x}{8}=\sqrt{3}(20–x)18$$

By cross multiplication, we get:

$$18x=8\sqrt{(}3)(20–x)$$

$$18x=160\sqrt{(}3)–8\sqrt{(}3x)$$

$$(18+8\sqrt{(}3)x)=160\sqrt{(}3)$$

By simplifying, we get:

$$x=8.699$$

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