Chọn D
Gắn hệ trục Axyz như hình vẽ.
Đặt \(x=AM,y=AN(0\le x;y\le 2);\)
\(\[\begin{array}{l} {S_{AMCN} =S_{ABCD} -S_{BMC} -S_{DNC} } \\ {S_{AMCN} =4-(2-x)-(2-y)} \\ {S_{AMCN} =x+y} \end{array}\] \)
\(\[\begin{array}{l} {S(0;0;2),M(x;0;0),C(2;2;0),N(0;y;0)} \\ {=>\overrightarrow{SM}=(x;0;-2),\overrightarrow{SC}=(2;2;-2),\overrightarrow{SN}=(0;y;-2)} \end{array}\] \)\(
\[\begin{array}{l} {[\overrightarrow{SM},\overrightarrow{SC}]=(4;2x-4;2x)} \\ {[\overrightarrow{SN},\overrightarrow{SC}]=(4-2y;-4;-2y)} \\ {(SMC)\bot (SNC)<=>[\overrightarrow{SM},\overrightarrow{SC}].[\overrightarrow{SN},\overrightarrow{SC}]=0} \end{array}\]
\[\begin{array}{l} {\Leftrightarrow 4(4-2y)-4(2x-4)-4xy=0} \\ {\Leftrightarrow 4-2y-2x+4-xy=0} \\ {\Leftrightarrow xy+2x+2y-8=0} \end{array}\] \)
\(\[\Leftrightarrow y=\frac{8-2x}{x+2} .1\le x\le 2\] \)
\(\[V_{S.AMCN} =\frac{1}{3} .SA.S_{AMCN} =\frac{2}{3} (x+y)=\frac{2}{3} (x+\frac{8-2x}{x+2} )\] \)
Xét \(f(x)=x+\frac{8-2x}{x+2} trên [1;2]\)
\(\[f'(x)=1-\frac{12}{(x+2)^{2} } =\frac{(x+2)^{2} -12}{(x+2)^{2} } \] \)
\(\[f'(x)=0\Leftrightarrow (x+2)^{2} =12\Leftrightarrow x=\pm 2\sqrt{3} -2\] \)
\(\[\begin{array}{l} {f(1)=f(2)=3} \\ {f(2\sqrt{3} -2)=4\sqrt{3} -4} \\ {=>\max V_{SABCD} =\frac{2}{3} .3=2} \end{array}\]
\)
Dấu "=" xảy ra \(\Leftrightarrow \left\{\begin{array}{l} {x=1} \\ {y=2} \end{array}\right. hoặc \left\{\begin{array}{l} {x=2} \\ {y=1} \end{array}\right. \)\(
\[T=\frac{1}{AM^{2} } +\frac{1}{AN^{2} } =\frac{1}{y^{2} } +\frac{1}{x^{2} } =\frac{5}{4} .\]
\)