Augusto has a wire that is $10 \mathrm{~m}$ long. He makes a cut at a point on the wire, obtaining two pieces. One piece has a length of $x$ and the other has a length of $10-x$ as shown in the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-05.jpg?height=645&width=1166&top_left_y=568&top_left_x=527)

Augusto uses the two pieces of wire to make two squares.

a) What is the length of the side of each of the squares? What is the area of each?

b) What is the length of each of the two pieces of wire so that the sum of the areas of the squares obtained is minimized?

c) Suppose Augusto cuts the wire into ten pieces and uses each one to make a square. What should be the size of each piece so that the sum of the areas of the squares obtained is minimized?
Solution

In this problem, all lengths are given in meters and areas in square meters.

a) A piece of rope has length $x$ and another piece of rope has length $10-x$. Since a square has four sides of equal length, one square will have a side length of $\frac{x}{4}$ and the other square will have a side length of $\frac{10-x}{4}$.

The area of a square with side length $\ell$ is $\ell^{2}$. Therefore, one square will have an area of $\left(\frac{x}{4}\right)^{2}=\frac{x^{2}}{16}$, while the other square will have an area of $\left(\frac{10-x}{4}\right)^{2}=\frac{100-20 x+x^{2}}{16}$.

b) Let $S(x)$ be the sum of the areas of the two squares. From the previous part, we have

$$
S(x)=\frac{x^{2}}{16}+\frac{100-20 x+x^{2}}{16}=\frac{100-20 x+2 x^{2}}{16}=\frac{1}{8} x^{2}-\frac{5}{4} x+\frac{25}{4}
$$

which is a quadratic function. The minimum of a function of the form

$$
f(x)=a x^{2}+b x+c
$$

with $a>0$ is achieved at $x=\frac{-b}{2 a}$. Thus, the minimum area will be achieved if

$$
x=-\frac{\left(-\frac{5}{4}\right)}{2 \frac{1}{8}}=5
$$

In other words, if the rope is cut exactly in the middle!

c) From the previous part, we know that to minimize the sum of the areas, it is necessary to cut the rope exactly in the middle. Well, we claim that to minimize the area with nine cuts (i.e., creating ten squares), it is necessary that all pieces of rope be equal. To show this, consider the following argument: if two of the ten pieces of rope were different, it would be possible to reduce the area by cutting the pieces of rope so that these two were equal (we are using the previous part). Therefore, any two pieces of rope must be equal. Hence, all must be equal!