Vasilev N.B.

A cherry, a sphere of radius $r$, is dropped into a round glass, the axial cross-section of which is the graph of the function $y=x^{4}$. For what largest $r$ will the sphere touch the lowest point of the bottom? (In other words, what is the maximum radius $r$ of a circle lying in the region $y \geq x^{4}$ and containing the origin?)
Let's first solve another problem: construct a circle with its center on the $y$-axis, which touches the $x$-axis, and determine the smallest radius $r$ for which it has a common point with the curve $y=x^{4}$, different from the origin (see the figure). In other words, for what smallest $r$ does the system of equations

$$
y=x^{4}, \quad x^{2}+(y-r)^{2}=r^{2}
$$

have a non-zero solution. Intuitively, it is clear that these problems are equivalent, and we will later prove this rigorously.

Substitute $y=x^{4}$ into the equation of the circle. Combine like terms and divide by $x^{2} (x>0)$:

$$
x^{6}-2 r x^{2}+1=0
$$

Express $r$ in terms of $x$:

$$
r(x)=\frac{1}{2}\left(x^{4}+\frac{1}{x^{2}}\right).
$$

The desired number $r_{0}$ is the minimum of this function. The derivative of the function is

$$
r^{\prime}(x)=2 x^{3}-\frac{1}{x^{3}}
$$

For $x>0$, this derivative behaves as follows: it is negative for $x<x_{0}$ and positive for $x>x_{0}$. This means that $r(x)$ decreases for $0<x<x_{0}$ and increases for $x>x_{0}$.

Thus, the smallest $r$ for which the circle has a common point with the curve $y=x^{4}$ is

$$
r_{0}=r\left(x_{0}\right)=\frac{3 \sqrt[3]{2}}{4}
$$

It remains to show that this $r_{0}$ gives the answer to the original problem. First, let's show that the corresponding cherry is entirely contained within the glass. Indeed, for any $x \neq 0$, we have $r_{0} \leq r(x)$. Substituting this inequality into the expression for $r(x)$, we get that for any $x$,

$$
x^{6}-2 r_{0} x^{2}+1 \geq 0
$$

Multiplying both sides by $x^{2}$ and substituting $y=x^{4}$, we obtain

$$
x^{2}+\left(y-r_{0}\right)^{2} \geq r_{0}^{2}
$$

for all $x, y=x^{4}$. But this means that the cherry is inside the glass.

It remains to show that if $r>r_{0}$, the cherry will not touch the origin. Indeed, in this case,

$$
x_{0}^{6}-2 r x_{0}^{2}+1>0
$$

This means that for $r>r_{0}$, there is a non-zero solution, and the situation is the same as for $a=4$;

b) for $a=2$ we get $x=0$ and $r=1 / 2$ - the radius of curvature of the graph at the point $(0 ; 0)$, the maximum cherry touches the parabola at a single point

c) for $0<a<2$ no circle can touch the "bottom" of the graph.