## Task B-2.5.

In a kite where the lengths of the diagonals are $d_{1}=24 \mathrm{~cm}$ and $d_{2}=8 \mathrm{~cm}$, a rectangle is inscribed such that its sides are parallel to the diagonals of the kite. Determine the dimensions of the inscribed rectangle that has the maximum area.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_7713364b1a15726862ddg-08.jpg?height=599&width=503&top_left_y=1828&top_left_x=708)

Let $x$ and $y$ be the sides of the rectangle, and let $y \parallel d_{1}, x \parallel d_{2}$. Notice triangles $B C A$ and $B F E$. According to the K-K theorem,

$$
\triangle B C A \sim \triangle B F E
$$

Therefore, we have:

$$
\begin{aligned}
& d_{1}: y = \frac{d_{2}}{2} : \left(\frac{d_{2}}{2} - \frac{x}{2}\right) \\
& \Rightarrow 24: y = 4: \left(4 - \frac{x}{2}\right) \\
& \Rightarrow y = 24 - 3x .
\end{aligned}
$$

The area of the rectangle is a quadratic function of the side $x$:

$$
\begin{aligned}
& P = P(x) = x \cdot (24 - 3x) \\
& P(x) = 24x - 3x^2
\end{aligned}
$$

This quadratic function achieves its maximum value for

$$
x_{M} = -\frac{24}{2 \cdot (-3)} = 4 \text{ cm}
$$

It follows that $y_{M} = 24 - 3 \cdot 4 = 12 \text{ cm}$. 1 point

Thus, the rectangle with the largest possible area has sides of length $x = 4 \text{ cm}$ and $y = 12 \text{ cm}$.