Cho hình lăng trụ đứng ABC A'B'C' có đáy

C

- Hướng dẫn giải

Đáp án C

 Ta có 

$\begin{array}{l}BC\perp AC,BC\perp AA\text{'}\\ \Rightarrow BC\perp \left(A\text{'}ACC\text{'}\right)\\ \Rightarrow BC\perp A\text{'}C.\end{array}$

Suy ra

$\begin{array}{l}\left(\widehat{\left(A\text{'}CB\right),\left(ABC\right)}\right)\\ =\left(\widehat{A\text{'}C,AC}\right)=\widehat{A\text{'}CA}\\ =x,\left(0<x<\frac{\pi }{2}\right).\end{array}$ 

$\Delta A\text{'}AC$ vuông tại B nên 

$\begin{array}{l}AA\text{'}=A\text{'}C.\sin \widehat{A\text{'}CA}\\ =a\sin x;AC=a\cos x.\end{array}$

Suy ra 

$\begin{array}{l}{V}_{A\text{'}.ABC}=\frac{1}{3}.AA\text{'}.S_{\Delta ABC}\\ =\frac{1}{3}.a\sin x.\frac{{\left(a\cos x\right)}^2}{2}\\ =\frac{a^3}{6}\sin x{\cos }^{2}x.\end{array}$

Xét hàm số

$\begin{array}{l}f\left(x\right)=\sin x{\cos }^{2}x\\ =\sin x\left(1-{\sin }^{2}x\right)\end{array}$ trên $\left(0;\frac{\pi }{2}\right).$ 

Đặt $t=\sin x$, do $x\in \left(0;\frac{\pi }{2}\right)\Rightarrow t\in \left(0;1\right)$. Xét hàm số  $g\left(t\right)=t\left(1-t^2\right)$ trên $\left(0;1\right).$

Ta có

$\begin{array}{l}f\text{'}\left(t\right)=1-3t^2;\\ f\text{'}\left(t\right)=0\iff t=\pm \frac{1}{\sqrt{3}}\end{array}$.

Do $t\in \left(0;1\right)$ nên $t=\frac{1}{\sqrt{3}}.$

Lập bảng biến thiên, suy ra $\underset{x\in \left(0;\frac{\pi }{2}\right)}{\max }f\left(x\right)=\underset{t\in \left(0;1\right)}{\max }g\left(t\right)=g\left(\frac{1}{\sqrt{3}}\right)=\frac{2\sqrt{3}}{9}.$ 

Vậy $$$V_{\max }=\frac{a^3}{6}.\frac{2\sqrt{3}}{9}=\frac{a^3\sqrt{3}}{27}$ (đvtt).