13. Let two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ in a plane be perpendicular to each other, and $|\boldsymbol{a}|=2,|\boldsymbol{b}|=1$. Also, $k$ and $t(t \geqslant 0)$ are two real numbers that are not both zero. If the vectors
$$
\boldsymbol{x}=\boldsymbol{a}+(3-t) \boldsymbol{b} \text { and } \boldsymbol{y}=-k \boldsymbol{a}+t^{2} \boldsymbol{b}
$$

are perpendicular to each other, then the maximum value of $k$ is $\qquad$
13.1.

Given $\boldsymbol{a} \cdot \boldsymbol{b}=0$.
$$
\begin{array}{l}
\boldsymbol{x} \cdot \boldsymbol{y}=[\boldsymbol{a}+(3-t) \boldsymbol{b}] \cdot\left(-k \boldsymbol{a}+t^{2} \boldsymbol{b}\right) \\
=-k \boldsymbol{a}^{2}+\left[-k(3-t)+t^{2}\right] \boldsymbol{a} \cdot \boldsymbol{b}+t^{2}(3-t) \boldsymbol{b}^{2} \\
=-k|\boldsymbol{a}|^{2}+t^{2}(3-t)|\boldsymbol{b}|^{2} \\
=-4 k+t^{2}(3-t) .
\end{array}
$$

According to the problem, $-4 k+t^{2}(3-t)=0$.
Therefore, $k=-\frac{1}{4} t^{3}+\frac{3}{4} t^{2}$.
Let $k=f(t)$, then
$$
f^{\prime}(t)=-\frac{3}{4} t^{2}+\frac{3}{2} t=-\frac{3}{4} t(t-2) .
$$

Since $t \neq 0$ (otherwise $k=0$), when $t=2$, $f^{\prime}(t)=0$, and $00, t>2$ when $f^{\prime}(t)<0$. Therefore, when $t=2$, $k$ reaches its maximum value of 1.