3. (2004 National College Entrance Examination, Zhejiang Paper) Let the tangent line $l$ at point $M\left(t, \mathrm{e}^{-t}\right)$ on the curve $y=\mathrm{e}^{-x}(x \geqslant 0)$ intersect the $x$-axis and $y$-axis to form a triangle with area $S(t)$.
(I) Find the equation of the line $l$;
(II) Find the maximum value of $S(t)$.
3. ( I ) Since $f^{\prime}(x)=-\mathrm{e}^{-x}$, the slope of the tangent line $l$ is $-\mathrm{e}^{-1}$. Therefore, $l: y-\mathrm{e}^{-t}=-\mathrm{e}^{-t}(x-t)$, which simplifies to $\mathrm{e}^{-t} x+y-\mathrm{e}^{-t}(t+1)=0$.
（II ）Setting $y=0$, we get $x=t+1$; setting $x=0$, we get $y=\mathrm{e}^{-t}(t+1)$. Thus, $S(t)=\frac{1}{3}(t+1) \cdot \mathrm{e}^{t}(t+1)=\frac{1}{2}(t+1)^{2} \mathrm{e}^{-t}$. Therefore, $S^{\prime}(t)=\frac{1}{2} \mathrm{e}^{-t}(1-t)(1+t)$.
Furthermore, when $t \in(0,1)$, $S^{\prime}(t)>0$; when $t \in(1,+\infty)$, $S^{\prime}(t)<0$, hence the maximum value of $S(t)$ is $S(1)=\frac{2}{\mathrm{e}}$.