## Task A-2.6. (20 points)

A wire of length $1 \mathrm{~m}$ needs to be cut and from the resulting pieces, one equilateral triangle and one square should be made. The entire wire must be used. Determine the length of the side of the triangle and the length of the side of the square so that the sum of the areas of the triangle and the square is minimized.
## Solution.

Let $a$ be the length of the side of the triangle and $b$ be the length of the side of the square.

According to the conditions of the problem, we have $3a + 4b = 1$, or $b = \frac{1 - 3a}{4}$.

The total area is $P = \frac{a^2 \sqrt{3}}{4} + b^2$,

or

\[
\begin{aligned}
P & = \frac{a^2 \sqrt{3}}{4} + \left(\frac{1 - 3a}{4}\right)^2 = \frac{a^2 \sqrt{3}}{4} + \frac{1 - 6a + 9a^2}{16} \\
& = \frac{1}{16}\left((4\sqrt{3} + 9)a^2 - 6a + 1\right)
\end{aligned}
\]

The expression in parentheses is a quadratic function of the variable $a$, and it has its minimum value for

\[
a = \frac{6}{2(4\sqrt{3} + 9)} = \frac{9 - 4\sqrt{3}}{11}
\]

Then, $b = \frac{1 - 3 \cdot \frac{9 - 4\sqrt{3}}{11}}{4} = \frac{3\sqrt{3} - 4}{11}$.