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:
\[x \space = \space 8.699 \]