Surface area of a cuboid

Relevant wiki: Surface Area - Problem Solving

No calculations need to be done, as we can prove that this has the most surface area. To maximise the surface area, we want to minimise the number of joined faces, as each joining loses a surface area of $2 \times (1 \times 1)$ . By joining up each unit cube in a long line, this is meaning that the end cubes have 1 face joined and the rest have 2 faces joined. As this is the smallest possible way in this respect of connecting all the cubes together, it has the maximum surface area.

Relevant wiki: Surface Area of a Cuboid

Relevant wiki: Surface Area of a Cuboid

Imagine any rectangular cuboid constructed $12$ of these unit cubes floating space.
The surface area of the cuboid can now be measured by how many faces of these unit cubes are visible.
This means that if we want to maximize the surface area, we want to maximize the number of the faces being visible.
However, some of the faces are hidden inside the cuboid and since we have a fixed number of cubes, and thus faces, it is equivalent to say that we want to minimize the number of faces hidden inside the cuboid.
However, to construct a rectangular cuboid out of these unit cubes, we must connect unit cubes by their faces, thus hiding some of the faces.
To construct a cuboid, we must start with a single unit cube and then connect the other cubes to it by faces.
This means that since we have $12$ cubes, we have to hide at least $2 \times 11$ faces, because we have at least $11$ face-to-face connections between the cubes.

However, in the example cuboid, we make exactly $11$ face-to-face connections and thus hide $22$ faces.
Hence, the example cuboid has the maximum possible area meaning that the answer is $\boxed{\text{No}}$ .

For a given volume, the surface area of sphere is minimum. So, one could imagine that the arrangement in a straight line would have maximum surface area.

Set the solid in the picture as a fundamental cube. If we want the surface of this solid to be the maximum, we only choose to save the most 1×1 square surfaces. Every time we move a little cube somewhere else to make a new three dimensional solid, the fewest surfaces we delete must be 2 or 0. So, bsed on the solid in the picture 2, whatever move we make, we will only make the number of surfaces fewer or the same. As the consequence, the solid in the picture 2 has the maximum of surface area.