3. If for any rectangular prism $A$, there exists a rectangular prism $B$ of the same height as $A$, such that the ratio of the lateral surface area of $B$ to that of $A$ and the ratio of the volume of $B$ to that of $A$ are both equal to $\lambda$, then the range of values for $\lambda$ is
A. $\lambda>0$
B. $0<\lambda<1$
C. $\lambda>1$
D. $\lambda \geqslant 1$
3. D. Let the length and width of the base of rectangular prism $A$ be $a_{1}$ and $a_{2}$, and the length and width of the base of rectangular prism $B$ be $b_{1}$ and $b_{2}$. From the problem, we have
$$
\lambda=\frac{b_{1}+b_{2}}{a_{1}+a_{2}}=\frac{b_{1} b_{2}}{a_{1} a_{2}},
$$

which implies $b_{1}+b_{2}=\lambda\left(a_{1}+a_{2}\right), b_{1} b_{2}=\lambda a_{1} a_{2}$.
Therefore, $b_{1}$ and $b_{2}$ are the two positive roots of the quadratic equation $x^{2}-\lambda\left(a_{1}+a_{2}\right) x+\lambda a_{1} a_{2}=0$.
Thus, $\Delta=\lambda^{2}\left(a_{1}+a_{2}\right)^{2}-4 \lambda a_{1} a_{2} \geqslant 0$. Hence, $\lambda \geqslant \frac{4 a_{1} a_{2}}{\left(a_{1}+a_{2}\right)^{2}}$.
Since $\frac{4 a_{1} a_{2}}{\left(a_{1}+a_{2}\right)^{2}} \leqslant 1$, it follows that $\lambda \geqslant 1$.