A cylindrical cup weighs 200 pounds. Its center of gravity is $4.8 \mathrm{~cm}$ high, and its base area is $20 \mathrm{~cm}^{2}$. If we slowly pour water into the cup, at what water height will the center of gravity be the lowest?
I. solution. Let $x$ be the amount of water in the glass, and $h$ be the numerical value of the center of gravity's height in cm. Assuming the specific weight of water is 1 pond $/ \mathrm{cm}^{3}$, we can express $h$ as follows based on what we learned in physics:

$$
h=\frac{20 x \cdot \frac{x}{2}+200 \cdot 4.8}{20 x+200} \mathrm{~cm}=\frac{x^{2}+96}{2(x+10)} \mathrm{~cm}
$$

Our task is to determine for which value of $x \geq 0$ the value of $h$ will be the smallest. We solve the problem using differential calculus. The first derivative of the function $h=h(x)$ with respect to $x$ is:

$$
h'=\frac{2 x(2 x+20)-2\left(x^{2}+96\right)}{4(x+10)^{2}}=\frac{(x-4)(x+24)}{2(x+10)^{2}}
$$

We can observe that the derivative function is negative for $0 \leq x < 4$ and positive for $x > 4$. The numerator is a parabola opening upwards with roots at -24 and 4, and the denominator is always positive for $x \geq 0$.

This means that the function $h$ is strictly decreasing in the interval $0 \leq x < 4$ and strictly increasing in the interval $4 < x < \infty$. Therefore, the function $h=h(x)$ takes its smallest value in the interval $0 \leq x < \infty$ at $x=4$. Thus, the center of gravity will be the lowest when the water level is $4 \mathrm{~cm}$.

The minimum value of $h(x)$ is also 4, meaning the center of gravity coincides with the water surface when it is at its lowest.

Hajnal Péter (Szeged, Radnóti M. Gymn., IV. grade)

II. solution. The minimum point of the function $h=h(x)$ can be determined in several ways without using differential calculus: for example, by using the inequality between the arithmetic and geometric means, or by reasoning with the discriminant of a quadratic equation. However, in the context of this problem, we can also provide a solution that closely relates to the physical content of the problem.

From the result of the first solution, it is noteworthy that the center of gravity coincides with the water surface when it is at its lowest. This is not a coincidence. As long as we fill the water below the center of gravity, the height of the center of gravity continuously decreases. However, once the center of gravity meets the water surface, any additional amount of water will raise the center of gravity, but it will remain below the water surface. Based on this, the desired water height can be calculated from the equation

$$
\frac{x^{2}+96}{2(x+10)}=x
$$

under the condition $x \geq 0$.

Tóth Viktor (Budapest, Fazekas M. Gymn., II. grade)

Note. We neglected the thickness of the glass walls at the bottom. Our result is correct for a thick-walled glass as well, provided that the height of the center of gravity is measured from the inner surface of the bottom of the glass.