## Let two positive real numbers (x,y) vary and satisfy the relation(4 + ln frac{{2x + 2y + 1}}{{5xy}} = 20xy – left( {8x + 8y} right)). Minimum value of expression (P = xy + 9) is equal to – Math Book

Let two positive real numbers $$x,y$$ change and satisfy the relation
$$4 + \ln \frac{{2x + 2y + 1}}{{5xy}} = 20xy – \left( {8x + 8y} \right)$$.
The minimum value of the expression $$P = xy + 9$$ is equal to
A. $$m = 11.$$ B. $$m = 10.$$ C. $$m = 12 \cdot$$ D. $$m = \frac{{19}}{2} \cdot \ ) The answer: adsense We have \(4 + \ln \frac{{2x + 2y + 1}}{{5xy}} = 20xy – 8x – 8y\,\,\, \Leftrightarrow \ln \left( {2x + 2y + 1} \right ) – \ln \left( {5xy} \right)\, = 4\left( {5xy} \right) – 4\left( {2x + 2y + 1} \right)$$$$\Leftrightarrow \ln \ left( {2x + 2y + 1} \right) + 4\left( {2x + 2y + 1} \right) = \ln \left( {5xy} \right)\, + 4\left( {5xy} \ right)\,\,\,\,\left( 1 \right).$$
Consider the function $$f\left( t \right) = 4t + \ln t$$ with $$t > 0$$.
Since $$f’\left( t \right) = 4 + \frac{1}{t} > 0,\,\,\forall t > 0$$ \(f