1. If $y=(x+1)(x+2)(x+3)(x+4)+5$, where $-3 \leqslant x \leqslant 3$, then the minimum value of $y$ is ( ).
(A) 3
(B) 4
(C) 29
(D) 5.5625
- 1. B.

Notice
$$
\begin{aligned}
y & =(x+1)(x+2)(x+3)(x+4)+5 \\
& =\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)+5 \\
& =\left(x^{2}+5 x+4\right)^{2}+2\left(x^{2}+5 x+4\right)+5 .
\end{aligned}
$$

Let $z=x^{2}+5 x+4$.
From $-3 \leqslant x \leqslant 3$, we get
$$
-2.25 \leqslant z \leqslant 28, y=(z+1)^{2}+4 \text {. }
$$

Then $y_{\text {min }}=4$.