Task B-2.4. 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_06_03_ade005c4283eff439485g-09.jpg?height=580&width=334&top_left_y=580&top_left_x=863)

Let $x, y$ be the lengths of the sides of the rectangle. Since $y \| d_{1}$ and $x \| d_{2}$, from the similarity of triangles we have

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

Now we have

$$
P=P(x)=x \cdot(24-3 x)=24 x-3 x^{2}
$$

The area is maximized for

$$
\begin{aligned}
x_{M} & =-\frac{24}{2 \cdot(-3)}=4 \text{ cm} \\
y_{M} & =24-3 \cdot 4=12 \text{ cm}
\end{aligned}
$$

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