5. In the Cartesian coordinate plane, if a circle passing through the origin with radius $r$ is completely contained within the region $y \geqslant x^{4}$, then the maximum value of $r$ is $\qquad$ - .
5. $\frac{3 \sqrt[3]{2}}{4}$.

By symmetry, to find the maximum value of $r$, we can set the equation of the circle as $x^{2}+(y-r)^{2}=r^{2}$.
From the problem, we know that for any $x \in[-r, r]$,
$$
r-\sqrt{r^{2}-x^{2}} \geqslant x^{4} \text {. }
$$

Let $x=r \cos \theta$.
We only need to discuss for $0 \leqslant \theta<\frac{\pi}{2}$, at this time,
$$
\begin{array}{l}
r-\sqrt{r^{2}-r^{2} \cos ^{2} \theta} \geqslant r^{4} \cos ^{4} \theta \\
\Rightarrow r-r \sin \theta \geqslant r^{4} \cos ^{4} \theta \\
\Rightarrow r^{3} \leqslant \frac{1-\sin \theta}{\cos ^{4} \theta}=\frac{1}{(1-\sin \theta)(1+\sin \theta)^{2}} .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
(1-\sin \theta)(1+\sin \theta)^{2} \\
=\frac{1}{2}(2-2 \sin \theta)(1+\sin \theta)(1+\sin \theta) \\
\leqslant \frac{1}{2} \times\left(\frac{4}{3}\right)^{3},
\end{array}
$$

the equality holds when $\sin \theta=\frac{1}{3}$.
Thus, $r^{3} \leqslant \frac{1}{\frac{1}{2} \times\left(\frac{4}{3}\right)^{3}} \Rightarrow r \leqslant \frac{3 \sqrt[3]{2}}{4}$.
Conversely, when $r \leqslant \frac{3 \sqrt[3]{2}}{4}$, the circle $x^{2}+(y-r)^{2}=r^{2}$ is completely inside $y \geqslant x^{4}$.
In summary, the maximum value of $r$ is $\frac{3 \sqrt[3]{2}}{4}$.