6. What is the maximum area that a rectangle can have, the coordinates of whose vertices satisfy the equation

$$
|y+1|\left(y^{2}+2 y+28\right)+|x-2|=9\left(y^{2}+2 y+4\right)
$$

and whose sides are parallel to the coordinate axes?

(12 points)

#
# Solution.

$$
|y+1|\left((y+1)^{2}+27\right)+|x-2|=9(y+1)^{2}+27
$$

Substitution: $x_{1}=y+1, y_{1}=x-2$.

Then $\quad\left|y_{1}\right|=-\left(\left|x_{1}\right|-3\right)^{3}$

The area of the rectangle does not change with this substitution. The area is calculated using the formula $S(x)=-4 x(x-3)^{3}, x \in[0 ; 3]$.

We have

$$
\begin{aligned}
& S^{\prime}(x)=-4\left((x-3)^{3}+3 x(x-3)^{2}\right) \\
&=-4(x-3)^{2}(4 x-3)
\end{aligned}
$$

At the point $x=\frac{3}{4}$, the function $S(x)$ attains its maximum value. $S_{\max }=S(0.75)=34.171875$.

## Answer: 34.171875

![](https://cdn.mathpix.com/cropped/2024_05_06_fb302ee711d03f314b6cg-04.jpg?height=1668&width=665&top_left_y=191&top_left_x=1255)