# 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 maximum?

Jun 28, 2018

The triangle should have $3.022$ metres of wire, which means the square should have 6.978 meters

#### Explanation:

We know the perimeter of the two shapes combined will be the amount of wire, 10 m.

Call the two pieces $a$ and $b$. Then $P = 10 = a + b$. Now we need to find an expression for area.

A_“square” = (a/4)^2
A_“triangle” = (b/3)sqrt(b^2 - (b/2)^2)(1/2)

Since $a + b = 10$, $a = 10 - b$

A_“square” = ((10 -b)/4)^2

A_“combined” = ((10-b)/4)^2+ b^2/12sqrt(3)

A_“combined” = (100 - 20b +b^2)/16 + b^2/12sqrt(3)

A’_“combined” = -20/16+ b/8 + b/6sqrt(3)

Look for critical values.

$0 = - \frac{20}{16} + \frac{3 b + 4 b \sqrt{3}}{24}$

$\frac{20}{16} \left(24\right) = b \left(3 + 4 \sqrt{3}\right)$

$b = \frac{30}{3 + 4 \sqrt{3}} \approx 3.022$ m

It follows that $a = 10 - \frac{30}{3 + 4 \sqrt{3}} \approx 6.978$.

Hopefully this helps!