Đáp án
\( 2x^2 + 6x y = 5 \quad (x > 0) \)
\( y = \frac{5 - 2x^2}{6x} > 0 \Leftrightarrow 0 < x < \frac{\sqrt{5}}{2} \)
\( V = 2x^2 y = 2x^2 \left( \frac{5 - 2x^2}{6x} \right) \)
\( = \frac{1}{3} \left( 5x - 2x^3 \right) \)
\( V' = \frac{1}{3} (5 - 6x^2) = 0 \quad \Leftrightarrow \quad x = \pm \frac{\sqrt{5}}{6} \)
Max \( V = f \left( \frac{\sqrt{5}}{6} \right) = \frac{\sqrt{30}}{27} \approx 1.014301... \dots \quad \Rightarrow \boxed{B} \)