Dễ thấy tứ giác MNPQ là hình bình hành.

Do $\text{CD//}\left(\text{MNPQ}\right)$ và $\text{S}\in \text{CD}$ nên $\text{d}(\text{S},(\text{MNPQ}\left)\right)=\text{d}(\text{C},(\text{MNPQ}\left)\right)$

Suy ra ${\text{V}}_{\text{S}.\text{MNPQ}}={\text{V}}_{\text{C}\cdot \text{MNPQ}}=2 {\text{V}}_{\text{C}.\text{MNQ}}$

Đặt $\text{MC}=\text{x},\text{MB}=\text{y}$

Vì $\text{MN//PQ}$ nên $\frac{\text{CM}}{\text{CB}}=\frac{\text{CN}}{\text{CA}}=\frac{\text{x}}{\text{x}+\text{y}}$

$\Rightarrow \frac{{\text{S}}_{\Delta \text{CMN}}}{{\text{S}}_{\Delta \text{CBA}}}=\frac{\text{CM}}{\text{CB}}.\frac{\text{CN}}{\text{CA}}=\frac{{\text{x}}^2}{{(\text{x}+\text{y})}^2}\Rightarrow {\text{V}}_{\text{C}.\text{MNQ}}={\text{V}}_{\text{Q}.\text{CMN}}=\frac{1}{3} \text{d}(\text{Q},(\text{MNC}\left)\right).{\text{S}}_{\Delta \text{MNC}}$

Mà $\frac{\text{d}(\text{Q},(\text{MNC}\left)\right)}{\text{d}(\text{D},(\text{MNC}\left)\right)}=\frac{\text{QB}}{\text{DB}}=\frac{\text{BM}}{\text{BC}}=\frac{\text{x}}{\text{x}+\text{y}}$

$\Rightarrow V_{C.MNQ}=\frac{1}{3}\frac{y}{x+y}d(D,(MNC\left)\right).\frac{x^2}{{(x+y)}^2}{S}_{\Delta ABC}=\frac{x^{2}y}{{(x+y)}^3}{V}_{ABCD}$

$=\frac{27x^{2}y}{{(x+y)}^3}=\frac{27\frac{y}{x}}{{\left(1+\frac{y}{x}\right)}^3}=\frac{27t}{{(1+t)}^3}$ với $t=\frac{y}{x}>0$

Xét hàm số $f\left(t\right)=\frac{27t}{{(1+t)}^3},t>0$ có $f^{\text{'}}\left(t\right)=\frac{27(1-2t)}{{(1+t)}^4};f^{\text{'}}\left(t\right)=0\iff t=\frac{1}{2}$

Lập bảng biến thiên ta được $\max \left(\text{t}\right)=\text{f}\left(\frac{1}{2}\right)=4$