## For real numbers (x,y) satisfying (0 le x,y le 1) and ({log _2}frac{{x + y}}{{2 – xy}} + 2left( {x + 1} right) )left( {y + 1} right) – 6 = 0). Find the maximum value of (P = 3x + y). – Math book

For real numbers $$x,y$$ satisfying $$0 \le x,y \le 1$$ and $${\log _2}\frac{{x + y}}{{2 – xy}} + 2\left( {x + 1} \right)\left( {y + 1} \right) – 6 = 0$$. Find the maximum value of $$P = 3x + y$$.

A. $$3$$. B. $$\frac{7}{2}$$. C. $$4$$. D. $$\frac{5}{2}$$.
With the condition $$\left\{ \begin{array}{l}0 \le x,y \le 1\\x + y > 0\end{array} \right.$$ we have: $${$$ log _2}\frac{{x + y}}{{2 – xy}} + 2\left( {x + 1} \right)\left( {y + 1} \right) – 6 = 0\)
$$\Leftrightarrow {\log _2}\left( {x + y} \right) – {\log _2}\left( {2 – xy} \right) + 2xy + 2x + 2y – 4 = 0$$
$$\Leftrightarrow {\log _2}\left( {x + y} \right) + 2\left( {x + y} \right) = {\log _2}\left( {2 – xy} \right) + 2\left( {2 – xy} \right)\left( * \right)$$.
Consider the function $$f\left( t \right) = {\log _2}t + 2t$$ on $$\left( {0;2} \right)$$.
$$f’\left( t \right) = \frac{1}{{t\ln 2}} + 2 > 0,\forall t \in \left( {0;2} \right)$$ so the function the number $$f\left( t \right)$$ covariates on the interval $$\left( {0;2} \right)$$.
So from $$\left( * \right)$$ we have $$x + y = 2 – xy \Leftrightarrow y\left( {1 + x} \right) = 2 – x \Leftrightarrow y = \frac{ {2 – x}}{{1 + x}}$$.
$$P = 3x + y = 3x + \frac{{2 – x}}{{1 + x}}$$. Consider the function $$g\left( x \right) = 3x + \frac{{2 – x}}{{1 + x}},\,\,x \in \left[ {0;1} \right]$$.
$$g’\left( x \right) = 3 – \frac{3}{{{{\left( {1 + x} \right)}^2}}} = \frac{{3x\left( { x + 2} \right)}}{{{\left( {1 + x} \right)}^2}}} \ge 0,\forall x \in \left[ {0;1} \right]$$. It follows that $$MaxP = \frac{7}{2}$$ is obtained when $$x = 1,y = \frac{1}{2}$$.