4. Five cubes with edge lengths of $15,16,20,24$ and $30 \mathrm{~cm}$ respectively, need to be melted into one sphere. Calculate the diameter of the sphere and the ratio of the sum of the areas of the cubes to the area of the sphere.
Solution. Let's denote by $V(P)$ the volume (area) of the sphere whose radius needs to be determined, and by $V_{1}, V_{2}, V_{3}, V_{4}$ and $V_{5}\left(P_{1}, P_{2}, P_{3}, P_{4}\right.$ and $\left.P_{5}\right)$ the volumes (areas) of the cubes whose edge lengths are 15, 16, 20, 24, and 30 cm, respectively.

Clearly, the sum of the volumes of the cubes is equal to the volume of the sphere, i.e.,

$$
V_{1}+V_{2}+V_{3}+V_{4}+V_{5}=V
$$

i.e.,

$$
15^{3}+16^{3}+20^{3}+24^{3}+30^{3}=\frac{4}{3} R^{3} \pi
$$

from which we get $R=\sqrt[3]{\frac{42221.25}{\pi}} \mathrm{cm}$, so the diameter of the sphere is $D=2 R=2 \sqrt[3]{\frac{42221.25}{\pi}} \mathrm{cm}$.

Now let's determine the ratio of the sum of the areas of the cubes to the area of the sphere. We have:

$$
\frac{P_{1}+P_{2}+P_{3}+P_{4}+P_{5}}{P}=\frac{6\left(15^{2}+16^{2}+20^{2}+24^{2}+30^{2}\right)}{4 \cdot 23.8^{2} \pi} \approx 1.9 .
$$

## 2nd year