Cho hình lăng trụ đứng $ABC.{A}'{B}'{C}'$ có đáy $ABC$ là tam giác...

Chọn hệ trục $Oxyz$ như hình vẽ với ${A}'\equiv 0\Rightarrow {A}'\left( 0;0;0 \right)$
$D\left( \dfrac{\sqrt{3}}{2}a;\dfrac{a}{2};2a \right)$, $I\left( 0;0;x \right)$, ${C}'\left( 0;a;4a \right)$ $\left( 0\le x\le 4a \right)$
${{S}_{\Delta DI{C}'}}=\dfrac{1}{2}\left| \left[ \overrightarrow{{C}'D},\overrightarrow{{C}'I} \right] \right|$
$\overrightarrow{{C}'I}=\left( 0;-a;x-4a \right)$
$\overrightarrow{{C}'D}=\left( \dfrac{\sqrt{3}}{2}a;-\dfrac{a}{2};-2a \right)$
Ta có $\left[ \overrightarrow{I{C}'},\overrightarrow{I{D}'} \right]=\left( -\dfrac{1}{2}ax;-\dfrac{\sqrt{3}}{2}ax+2\sqrt{3}{{a}^{2}};-\dfrac{\sqrt{3}}{2}{{a}^{2}} \right)$
Suy ra
$\begin{aligned}
& {{S}_{\Delta DI{C}'}}=\dfrac{1}{2}\left| \left[ \overrightarrow{{C}'D},\overrightarrow{{C}'I} \right] \right|=\dfrac{1}{2}\sqrt{{{\left( \dfrac{1}{2}ax \right)}^{2}}+{{\left( -\dfrac{\sqrt{3}}{2}ax+2\sqrt{3}{{a}^{2}} \right)}^{2}}+\left( -\dfrac{\sqrt{3}}{2}{{a}^{2}} \right)} \\
& =\dfrac{1}{2}\sqrt{\dfrac{1}{4}{{a}^{2}}{{x}^{2}}+\dfrac{3}{4}{{a}^{2}}{{x}^{2}}-6{{a}^{3}}x+12{{a}^{4}}+\dfrac{3}{4}{{a}^{4}}} \\
\end{aligned}$
$=\dfrac{1}{2}\sqrt{{{a}^{2}}{{x}^{2}}-6{{a}^{3}}x+\dfrac{51}{4}{{a}^{4}}}=\dfrac{1}{4}\sqrt{4{{a}^{2}}{{x}^{2}}-24{{a}^{3}}x+51{{a}^{4}}}=\dfrac{1}{4}a\sqrt{4{{x}^{2}}-24ax+51{{a}^{2}}}$
$=\dfrac{1}{4}a\sqrt{{{\left( 2x-6a \right)}^{2}}+15{{a}^{2}}}\ge \dfrac{1}{4}a\sqrt{15}a=\dfrac{\sqrt{15}}{4}{{a}^{2}}$
Suy ra $\min {{S}_{\Delta ID{C}'}}=\dfrac{\sqrt{15}}{4}{{a}^{2}}\Rightarrow m=\dfrac{\sqrt{15}}{4}{{a}^{2}}$
$\Rightarrow \max {{S}_{ID{C}'}}=\dfrac{a}{4}.\sqrt{51{{a}^{2}}}=\dfrac{\sqrt{51}{{a}^{2}}}{4}\Rightarrow M=\dfrac{\sqrt{51}{{a}^{2}}}{4}$
Vậy $\sqrt{15}m+\sqrt{51}M=\dfrac{33}{2}{{a}^{2}}$.