7. The axial cross-section of a large wine glass is the graph of the function $y=x^{4}$. A small ball with radius $r$ is placed inside the wine glass. When $r$ is maximized to
$\qquad$ , the ball can touch the lowest point of the bottom of the glass.
Solve: $\frac{3 \sqrt[3]{2}}{4}$.
In other words, when $r$ is at most how much, a circle with radius $r$ located in the region $y \geq x^{4}$ can pass through the origin.
Thus, the original problem is equivalent to: a circle $(y-r)^{2}+x^{2}=r^{2}$ passing through the origin is completely within the region $y \geq x^{4}$. Find the maximum value of $r$.
Therefore, when $0 \leq x \leq r$, we have
$$
r-\sqrt{r^{2}-x^{2}} \geq x^{4}
$$

Let $x=r \sin \alpha\left(0 \leq \alpha \leq \frac{\pi}{2}\right)$. Then equation (1) becomes $1-\cos \alpha \geq r^{3} \sin ^{4} \alpha \cdots \cdots$ (2).
When $\alpha=0$, equation (2) clearly holds.
When $0<\alpha \leq \frac{\pi}{2}$, equation (2) becomes
$$
r^{3} \leq \frac{1}{\sin ^{2} \alpha(1+\cos \alpha)}
$$

By
$$
\sin ^{2} \alpha(1+\cos \alpha)=8 \sin ^{2} \frac{\alpha}{2} \cdot \cos ^{4} \frac{\alpha}{2} \leq 4\left(\frac{2 \sin ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}}{3}\right)^{3}=\frac{32}{27}
$$

Then $r^{3} \leq \frac{27}{32} \Rightarrow r \leq \frac{3 \sqrt[3]{2}}{4}$