Đáp án đúng: C
Đặt $x=\frac{{SM}}{{SB}};\,y=\frac{{SN}}{{SD}}\,\Rightarrow \,V'={{V}_{{S.AMK}}}+{{V}_{{S.ANK}}}=\frac{V}{4}(x+y)\,\,\,\,(1)$
Mặt khác$V'={{V}_{{S.AMN}}}+{{V}_{{S.MNK}}}=\frac{{3xy}}{4}V\,\,\,\,\,(2)$
Từ (1) và (2) ta có$x+y=3xy$
$\Rightarrow \,y=\frac{x}{{3x-1}},\left( {y>0\,\Rightarrow \,x>\frac{1}{3}} \right);\,y=\frac{{SN}}{{SD}}\le 1\,\Rightarrow \,\frac{x}{{3x-1}}\le 1$
$\begin{array}{l}\Rightarrow \,x\ge \frac{1}{2}\,\Rightarrow \,\frac{1}{2}\le x\le 1\\\frac{{V'}}{V}=\frac{{3{{x}^{2}}}}{{4(3x-1)}},\left( {\frac{1}{2}\le x\le 1} \right)\end{array}$
Xét hàm số$f(x)=\frac{{3{{x}^{2}}}}{{4(3x-1)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{1}{2}\le x\le 1} \right)$.$f(x)$ đạt GTNN$=\frac{1}{3}$
Đáp án C