# A piece of wire 20 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How much wire should be used for the square in order to maximize the total area?

Apr 30, 2015

Let $x$ metres be the side of the square, so $4 x$ metres are the total amount of wire used for the square (with $0 \le 4 x \le 20 \Rightarrow 0 \le x \le 5$). For the equilater triangle $20 - 4 x$ metres of wire remain, and the side will be $\frac{20 - 4 x}{3}$.

Given the side $l$ of an equilatrer triangle, the height is, using the Pitagora's theorem, $h = \frac{l}{2} \sqrt{3}$, so the area of the triangle is:

$A = l \cdot \frac{l}{2} \sqrt{3} \cdot \frac{1}{2} = {l}^{2} \frac{\sqrt{3}}{4}$

($A = \frac{b \cdot h}{2}$)

So the total area is:

$A = {x}^{2} + {\left(\frac{20 - 4 x}{3}\right)}^{2} \frac{\sqrt{3}}{4}$.

$A ' = 2 x + \frac{\sqrt{3}}{4} \cdot \frac{1}{9} \cdot 2 \left(20 - 4 x\right) \cdot \left(- 4\right) =$

$= 2 x - 2 \frac{\sqrt{3}}{9} \left(20 - 4 x\right)$.

$A ' > 0$ if

$2 x - 2 \frac{\sqrt{3}}{9} \left(20 - 4 x\right) > 0 \Rightarrow 18 x - 40 \sqrt{3} + 8 \sqrt{3} x > 0 \Rightarrow$

$2 x \left(9 + 4 \sqrt{3}\right) > 40 \sqrt{3} \Rightarrow x > \frac{40 \sqrt{3}}{2 \left(9 + 4 \sqrt{3}\right)} \Rightarrow$

$x > \frac{20 \sqrt{3}}{9 + 4 \sqrt{3}} \cdot \frac{9 - 4 \sqrt{3}}{9 - 4 \sqrt{3}} \Rightarrow$

$x > \frac{20 \cdot 3 \left(3 \sqrt{3} - 4\right)}{81 - 48} \Rightarrow x > \frac{20}{11} \left(3 \sqrt{3} - 4\right) = \overline{x}$.

So our function Area is decreasing in $\left[0 , \overline{x}\right)$ and growing in $\left(\overline{x} , 5\right]$ and so the point $\left(\overline{x} , A \left(\overline{x}\right)\right)$ is a minimum of the function. The conclusion is that maximum (absolute maximum) is in one of the two sides of the interval of definition of $x$, $0 \mathmr{and} 5$.

If $x = 0$ the value of area will be: $A = 100 \frac{\sqrt{3}}{9} \cong 19 , 2$.

If $x = 5$ the value of the area will be: $A = 25$, that is greater than the other value.

In conclusion the maximum of the area will be in the case in which all the wire will be used for the square!