$6 \cdot 112$ The axial cross-section of a large wine glass is the graph of the function $y=x^{4}$. A small ball with a radius of $r$ is placed inside the wine glass. What is the maximum radius $r$ for the ball to touch the lowest point of the bottom of the glass (in other words, what is the maximum $r$ for a circle with radius $r$ located in the region $y \geqslant x^{4}$ to pass through the origin)?
[Solution] The original problem is equivalent to: the circle passing through the origin
$$
(y-r)^{2}+x^{2}=r^{2}
$$

is completely within the region $y \geqslant x^{4}$, find the maximum value of $r$.
Therefore, when $0 \leqslant x \leqslant r$, we have
$$
r-\sqrt{r^{2}-x^{2}} \geqslant x^{4} \text {. }
$$

Let $x=r \sin \theta$. (1) becomes: when $0 \leqslant \theta \leqslant \frac{\pi}{2}$, we have
$$
1-\cos \theta \geqslant r^{3} \sin ^{4} \theta \text {. }
$$

When $\theta=0$, (2) is obviously true. When $0<\theta \leqslant \frac{\pi}{2}$, (2) becomes
$$
r^{3} \leqslant \frac{1}{\sin ^{2} \theta(1+\cos \theta)}
$$

Since
$$
\begin{array}{l}
\sin ^{2} \theta(1+\cos \theta)=8 \sin ^{2} \frac{\theta}{2} \cos ^{4} \frac{\theta}{2} \\
\leqslant 4\left(\frac{2 \sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}}{3}\right)^{3} \\
=\frac{32}{27} \text {. } \\
\end{array}
$$

Therefore
$$
\begin{array}{l}
r^{3} \leqslant \frac{27}{32}, \\
r \leqslant \sqrt[3]{\frac{27}{32}}=\frac{3}{4} \sqrt[3]{2} .
\end{array}
$$

The maximum value of $r$ is $\frac{3}{4} \sqrt[3]{2}$.