## Task A-2.3.

A wire of length $10 \mathrm{~m}$ needs to be cut into two parts, and from one part a square should be made, and from the other part an equilateral triangle. At what point should the wire be cut so that the total area of the square and the equilateral triangle is minimized?
## Solution.

Let $x$ be the point at which the wire should be cut, i.e., let $x$ be the perimeter of the square, and $10-x$ be the perimeter of the triangle.

Then the area of the square is $\frac{x^{2}}{4}$, while the area of the equilateral triangle is $\frac{\sqrt{3}}{4}\left(\frac{10-x}{3}\right)^{2}$. 1 point

The total area of the shapes is

$$
\frac{x^{2}}{4}+\frac{\sqrt{3}}{4}\left(\frac{10-x}{3}\right)^{2}=\frac{1}{144}\left[(9+4 \sqrt{3}) x^{2}-80 \sqrt{3} x+400 \sqrt{3}\right]
$$

We have obtained a quadratic function of the form $a x^{2}+b x+c$. Since the leading coefficient $a$ is positive, the expression achieves its minimum value at the vertex of the quadratic function.

We find its value using the formula $x_{0}=-\frac{b}{2 a}$. Therefore,

$$
x_{0}=\frac{40 \sqrt{3}}{9+4 \sqrt{3}}=\frac{40 \sqrt{3}(9-4 \sqrt{3})}{(9-4 \sqrt{3})(9+4 \sqrt{3})}=\frac{40(3 \sqrt{3}-4)}{11}
$$

We conclude that the wire should be cut at a distance of $x_{0}=\frac{40(3 \sqrt{3}-4)}{11}$ from its edge.

Note: The solution is accepted if the expression is simplified in any other way, such as $x_{0}=$ $\frac{120 \sqrt{3}-160}{11}$.

The problem can also be solved by taking some other measure from the problem as the unknown. If $y$ is taken as the perimeter of the equilateral triangle, the solution is $y_{0}=10-x_{0}=$ $\frac{270-120 \sqrt{3}}{11}=\frac{30(9-4 \sqrt{3})}{11}$. In this case, the correct solution is also to cut the wire at a distance of $y_{0}$ from the edge. If the student expresses the solution in terms of $x_{0}$ or $y_{0}$, it is not necessary to write the final solution in a sentence.

If, for example, the length of the side of the square or the length of the side of the equilateral triangle is taken as the unknown, the solution is $\frac{x_{0}}{4}$ or $\frac{y_{0}}{3}$, respectively. In these cases, if the student does not express the point at which the wire should be cut (which is the question in the problem), the solution is worth a maximum of 5 points.