5. Ana has 16 small cuboids with edge lengths of $2 \mathrm{~cm}, 2 \mathrm{~cm}$, and $1 \mathrm{~cm}$. She can form a larger cuboid in several ways, using all 16 small cuboids without any gaps. What are the edge lengths of the cuboid that Ana formed, if it must have the smallest surface area?
Solution. The volume of the resulting cuboid is always the same, regardless of how Ana shapes and carries it, $V=16 \cdot 2 \cdot 2 \cdot 1=64 \mathrm{~cm}^{3}$. The lengths of the edges of such a resulting cuboid must be divisors of 64, that is, $1 \mathrm{~cm}, 2 \mathrm{~cm}, 4 \mathrm{~cm}, 8 \mathrm{~cm}, 16 \mathrm{~cm}, 32 \mathrm{~cm}$, or $64 \mathrm{~cm}$. The only possible cuboids that can be obtained in this way have dimensions: $32 \times 2 \times 1, 16 \times 4 \times 1, 16 \times 2 \times 2, 8 \times 8 \times 1, 8 \times 4 \times 2$, and $4 \times 4 \times 4$.

Their surface areas are:

$$
\begin{aligned}
& P_{1}=2(64+32+2)=196 \mathrm{~cm}^{2} \\
& P_{2}=2(64+16+4)=168 \mathrm{~cm}^{2} \\
& P_{3}=2(32+32+4)=136 \mathrm{~cm}^{2} \\
& P_{4}=2(64+8+8)=160 \mathrm{~cm}^{2} \\
& P_{5}=2(32+16+8)=112 \mathrm{~cm}^{2} \\
& P_{6}=2(16+16+16)=96 \mathrm{~cm}^{2}
\end{aligned}
$$

Therefore, the cuboid that Ana made has dimensions $4 \times 4 \times 4$.

## VIII Department