Task B-2.5. (20 points) The length of the side of a square is $6 \mathrm{~cm}$. On the sides $\overline{A B}$ and $\overline{A D}$, points $K$ and $L$ are given such that $|A K|=2 \mathrm{~cm}$ and $|A L|=3 \mathrm{~cm}$. A trapezoid is inscribed in the square with the base $\overline{K L}$. What is the maximum possible area of the inscribed trapezoid?
Solution. Let's denote the vertices of the trapezoid base as $M$ and $N$. Triangles $A K L$ and $C N M$ are similar $(\angle A K L = \angle C N M)$.

If we denote the lengths of the sides $\overline{C N}$ and $\overline{C M}$ as $2x$ and $3x$, the area $P$ of the trapezoid can be represented as the difference between the area of the square $A B C D$ and the areas of the four triangles $A K L, B M K, C N M, D L N$.

![](https://cdn.mathpix.com/cropped/2024_05_30_9c56249806d755c5106ag-08.jpg?height=560&width=585&top_left_y=220&top_left_x=744)

\[
\begin{aligned}
P & =36-\frac{3 \cdot 2}{2}-\frac{4 \cdot(6-3 x)}{2}-\frac{3 x \cdot 2 x}{2}-\frac{3 \cdot(6-2 x)}{2} \\
& =-3 x^{2}+9 x+12 .
\end{aligned}
\]

(8 points)

The quadratic function $f(x)=-3 x^{2}+9 x+12$ reaches its maximum at $x=\frac{-9}{2 \cdot(-3)}=\frac{3}{2}$.

(6 points)

Thus, $P_{\max }=-3 \cdot \frac{9}{4}+9 \cdot \frac{3}{2}+12=\frac{75}{4}$.

(2 points)

(Or from $f(x)=-3\left(x-\frac{3}{2}\right)^{2}+\frac{75}{4}$ it follows that $P_{\max }=\frac{75}{4}$.)

## COUNTY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B category

23rd February 2009.