9.3. The numbers x and y satisfy the equation:

$$
\sqrt{x y}+\sqrt{(1-x)(1-y)}=\sqrt{7 x(1-y)}+\frac{\sqrt{y(1-x)}}{\sqrt{7}}
$$

Find the maximum value of the expression $x+7 y$. Justify your answer.
# Solution:

Method 1. Perform equivalent transformations:

$$
\begin{aligned}
& \sqrt{x y}+\sqrt{(1-x)(1-y)}=\sqrt{7 x(1-y)}+\frac{\sqrt{y(1-x)}}{\sqrt{7}} \\
& \sqrt{7}(\sqrt{x y}-\sqrt{7 x(1-y)})+(\sqrt{7(1-x)(1-y)}-\sqrt{y(1-x)})=0 \\
& \sqrt{7}\left(\sqrt{x}(\sqrt{y}-\sqrt{7(1-y)})+\frac{\sqrt{1-x}}{\sqrt{7}}(\sqrt{7(1-y)}-\sqrt{y})\right)=0 \\
& \sqrt{7}(\sqrt{y}-\sqrt{7(1-y)})\left(\sqrt{x}-\frac{\sqrt{1-x}}{\sqrt{7}}\right)=0 \\
& {\left[\begin{array}{l}
\left\{\begin{array}{r}
\sqrt{y}-\sqrt{7(1-y)}=0 \\
0 \leqslant x \leqslant 1
\end{array}\right. \\
\left\{\begin{array}{r}
\sqrt{x}-\frac{\sqrt{1-x}}{\sqrt{7}}=0 \\
0 \leqslant y \leqslant 1
\end{array}\right.
\end{array}\right.} \\
& {\left[\begin{array}{r}
y=7 / 8 \\
0 \leqslant x \leqslant 1 \\
\left\{\begin{array}{r}
x=1 / 8 \\
0 \leqslant y \leqslant 1
\end{array}\right.
\end{array}\right.}
\end{aligned}
$$

Then the maximum value of the expression $x+7 y$ is $\frac{57}{8}$.

Method 2. The numbers $x$ and $y$ have the same sign (otherwise $\sqrt{x y}$ does not exist). They cannot both be negative (otherwise the square roots on the right side of the equation do not exist). If $x>0$, then $y \leqslant 1$ (otherwise $\sqrt{7 x(1-y)}$ does not exist) and similarly from the inequality $y>0$ it follows that $x \leqslant 1$. Therefore, $0 \leqslant x \leqslant 1$ and $0 \leqslant y \leqslant 1$. Under these conditions, square both sides of the equation (an equivalent transition, since both sides of the equation are non-negative):

$$
\begin{gathered}
x y+2 \sqrt{x y(1-x)(1-y)}+(1-x)(1-y)= \\
=7 x(1-y)+2 \sqrt{x y(1-x)(1-y)}+\frac{y(1-x)}{7} \\
7 x y+7(1-x)(1-y)=49 x(1-y)+y(1-x) \\
64 x y-56 x=8 y-7 \\
8 x(8 y-7)=8 y-7
\end{gathered}
$$

from which either $y=\frac{7}{8}$ and $0 \leqslant x \leqslant 1$, or $x=\frac{1}{8}$ and $0 \leqslant y \leqslant 1$. In the first case, the maximum value of the expression $x+7 y$ is achieved when $y=\frac{7}{8}, x=1$ and is $\frac{57}{8}$, in the second case - when $x=\frac{1}{8}, y=1$ and is also $\frac{57}{8}$.

Answer: $\frac{57}{8}$.

Recommendations for checking:

| included in the work | points |
| :--- | :--- |
| Correct and justified answer with the numbers $x, y$ indicated, at which it is achieved | 7 points |
| Correct and justified answer without indicating the numbers $x, y$ at which it is achieved | 6 points |
| In the correct solution process, there is an arithmetic error, possibly leading to an incorrect answer | 6 points |
| The original equation is correctly reduced to an equation of the form $f(x, y) \cdot g(x, y)=0$ | 3 points |
| Correct answer without justification | 0 points |