Lời giải:
\( V = (a - 2x)( b - 2x)x \, \text{với} \, 0 < x < \frac{b}{2} \)
\( = 4x^3 - 2(a + b)x^2 + abx \)
V' = 12x^2 - 4(a + b)x + ab = 0
\( \Leftrightarrow \left[ \begin{array}{} x = \frac{a + b - \sqrt{a^2 + b^2 - ab}}{6} < \frac{b}{2} \quad (x > 0) \\ x = \frac{a + b + \sqrt{a^2 + b^2 - ab}}{6} > \frac{b}{2} \end{array} \right. \)
\( V_{\text{đạt Max}} \quad \Leftrightarrow \quad x = \frac{a + b - \sqrt{a^2 + b^2 - ab}}{6} \)
**Nhớ:** \( V_{\text{đạt Max}} \quad \Leftrightarrow \quad x = \frac{a + b - \sqrt{a^2 + b^2 - ab}}{6} \)