Đáp án đúng: C

Phương pháp giải:

Sử dụng công thức \({V_{BDA'M}} = \dfrac{1}{6}\left| {\left[ {\overrightarrow {BA'} ,\overrightarrow {BD} } \right].\overrightarrow {BM} } \right|\).Giải chi tiết:Ta có
\(\overrightarrow {AB}  = \overrightarrow {DC} \) \( \Rightarrow \left( {m;0;0} \right) = \left( {{x_C};{y_C} - m;{z_C}} \right)\) \( \Rightarrow \left\{ \begin{array}{l}{x_C} = m\\{y_C} - m = 0\\{z_C} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_C} = m\\{y_C} = m\\{z_C} = 0\end{array} \right. \Rightarrow C\left( {m;m;0} \right)\).
\(\overrightarrow {AA'}  = \overrightarrow {CC'} \) \( \Rightarrow \left( {0;0;n} \right) = \left( {{x_{C'}} - m;{y_{C'}} - m;{z_{C'}}} \right)\) \( \Rightarrow \left\{ \begin{array}{l}{x_{C'}} - m = 0\\{y_{C'}} - m = 0\\{z_{C'}} = n\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_{C'}} = m\\{y_{C'}} = m\\{z_{C'}} = n\end{array} \right. \Rightarrow C'\left( {m;m;n} \right)\).
\(M\) là trung điểm của cạnh \(CC'\)\( \Rightarrow M\left( {m;m;\dfrac{n}{2}} \right)\).
Ta có:
\(\begin{array}{l}\overrightarrow {BA'}  = \left( { - m;0;n} \right)\\\overrightarrow {BD}  = \left( { - m;m;0} \right)\\\overrightarrow {BM}  = \left( {0;m;\dfrac{n}{2}} \right)\\ \Rightarrow \left[ {\overrightarrow {BA'} ,\overrightarrow {BD} } \right] = \left( { - mn; - mn; - {m^2}} \right)\\ \Rightarrow \left[ {\overrightarrow {BA'} ,\overrightarrow {BD} } \right].\overrightarrow {BM}  =  - {m^2}n - \dfrac{{{m^2}n}}{2} =  - \dfrac{{3{m^2}n}}{2}\\ \Rightarrow {V_{BDA'M}} = \dfrac{1}{6}\left| {\left[ {\overrightarrow {BA'} ,\overrightarrow {BD} } \right].\overrightarrow {BM} } \right| = \dfrac{1}{6}.\dfrac{{3{m^2}n}}{2} = \dfrac{{{m^2}n}}{4}\,\,\left( {do\,\,m,\,\,n > 0} \right)\end{array}\)
Áp dụng BĐT Cô-si ta có \({m^2}n = \dfrac{1}{2}m.m.2n \le \dfrac{1}{2}{\left( {\dfrac{{m + m + 2n}}{3}} \right)^3} = \dfrac{4}{{27}}{\left( {m + n} \right)^3} = \dfrac{4}{{27}}{.4^3} = \dfrac{{256}}{{27}}\).
\({V_{BDA'M}} \le \dfrac{{64}}{{27}}\). Dấu “=” xảy ra khi \(\left\{ \begin{array}{l}m = 2n\\m + n = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m = \dfrac{8}{3}\\n = \dfrac{4}{3}\end{array} \right.\).
Vậy \({\left( {{V_{BDA'M}}} \right)_{\max }} = \dfrac{{64}}{{27}} \Leftrightarrow m = \dfrac{8}{3},\,\,n = \dfrac{4}{3}\).
Chọn C.