6. What is the greatest area that a rectangle can have, the coordinates of whose vertices satisfy the equation $|y-x|=(y+x+1)(5-x-y)$, and whose sides are parallel to the lines $y=x$ and $y=-x$? Write the square of the found area in your answer. $\quad(12$ points $)$

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# Solution.

Substitution: $x_{1}=x+y, y_{1}=y-x$. This substitution increases all dimensions by a factor of $\sqrt{2}$. We have $\left|y_{1}\right|=\left(x_{1}+1\right)\left(5-x_{1}\right), \quad S\left(x_{1}\right)=4\left(x_{1}-2\right)\left(x_{1}+1\right)\left(5-x_{1}\right), x_{1} \in(2 ; 5)$. $S^{\prime}\left(x_{1}\right)=4\left(\left(x_{1}+1\right)\left(5-x_{1}\right)+\left(x_{1}-2\right)\left(5-x_{1}\right)-\left(x_{1}-2\right)\left(x_{1}+1\right)\right)=$ $=-12\left(x_{1}^{2}-4 x_{1}+1\right)=-12\left(x_{1}-2-\sqrt{3}\right)\left(x_{1}-2+\sqrt{3}\right)=0$.

The maximum area is achieved at the point $x_{1}=2+\sqrt{3}$. In the coordinate system $\mathrm{Ox}_{1} \mathrm{y}_{1}$, the maximum area is

$$
S(2+\sqrt{3})=4 \sqrt{3}(3+\sqrt{3})(3-\sqrt{3})=24 \sqrt{3}
$$

In the original coordinate system, $S_{\max }=12 \sqrt{3}$. $\left(S_{\max }\right)^{2}=432$. Answer: 432

Answer: 432

![](https://cdn.mathpix.com/cropped/2024_05_06_fb302ee711d03f314b6cg-12.jpg?height=1034&width=1433&top_left_y=151&top_left_x=317)