To find the maximum total surface area of the cuboid formed by merging 5150 cubes with side length 1, we need to consider the shape of the cuboid that would give the maximum surface area.

1. The maximum surface area of a cuboid is achieved when the cuboid is a cube itself. This is because for a given volume, a cube has the smallest surface area among all cuboids.

2. Therefore, we need to find the cube root of 5150 to determine the side length of the cube that would be formed. However, 5150 is not a perfect cube. The cube root of 5150 is approximately 17.27, but we can't have a fraction of a cube, so we take the integer part, which is 17.

3. The volume of the cube formed by 17^3 is 4913, which leaves 237 unit cubes (5150 - 4913) that can't be arranged into a larger cube.

4. These remaining cubes can be arranged into a cuboid of dimensions 17 x 17 x 1, which gives a volume of 289, still leaving 52 cubes (237 - 185).

5. These 52 cubes can be arranged into a cuboid of dimensions 17 x 1 x 1, which gives a volume of 17, still leaving 35 cubes (52 - 17).

6. These 35 cubes can be arranged into a cuboid of dimensions 17 x 1 x 1, which gives a volume of 17, still leaving 18 cubes (35 - 17).

7. These 18 cubes can be arranged into a cuboid of dimensions 9 x 1 x 1, which gives a volume of 9, still leaving 9 cubes (18 - 9).

8. These 9 cubes can be arranged into a cuboid of dimensions 3 x 1 x 1, which gives a volume of 3, still leaving 6 cubes (9 - 3).

9. These 6 cubes can be arranged into a cuboid of dimensions 2 x 1 x 1, which gives a volume of 2, still leaving 4 cubes (6 - 2).

10. These 4 cubes can be arranged into a cuboid of dimensions 2 x 1 x 1, which gives a volume of 2, still leaving 2 cubes (4 - 2).

11. These 2 cubes can be arranged into a cuboid of dimensions 1 x 1 x 1, which gives a volume of 1, still leaving 1 cube (2 - 1).

12. This last cube can be arranged into a cuboid of dimensions 1 x 1 x 1, which gives a volume of 1, leaving no cubes (1 - 1).

13. Now, we calculate the total surface area of all these cuboids. The surface area of a cuboid is given by the formula 2lw + 2lh + 2wh, where l, w, and h are the length, width, and height of the cuboid, respectively.

14. The total surface area is therefore 2(17*17) + 2(17*1) + 2(17*1) + 2(17*1) + 2(17*1) + 2(9*1) + 2(3*1) + 2(2*1) + 2(2*1) + 2(1*1) + 2(1*1) = 578 + 68 + 18 + 4 + 2 = 670.

So, the maximum total surface area of the cuboid formed by merging 5150 cubes with side length 1 is 670 square units.

Question

To find the maximum total surface area of the cuboid formed by merging 5150 cubes with side length 1, we need to consider the shape of the cuboid that would give the maximum surface area.

1. The maximum surface area of a cuboid is achieved when the cuboid is a cube itself. This is because for a given volume, a cube has the smallest surface area among all cuboids.

2. Therefore, we need to find the cube root of 5150 to determine the side length of the cube that would be formed. However, 5150 is not a perfect cube. The cube root of 5150 is approximately 17.27, but we can't have a fraction of a cube, so we take the integer part, which is 17.

3. The volume of the cube formed by 17^3 is 4913, which leaves 237 unit cubes (5150 - 4913) that can't be arranged into a larger cube.

4. These remaining cubes can be arranged into a cuboid of dimensions 17 x 17 x 1, which gives a volume of 289, still leaving 52 cubes (237 - 185).

5. These 52 cubes can be arranged into a cuboid of dimensions 17 x 1 x 1, which gives a volume of 17, still leaving 35 cubes (52 - 17).

6. These 35 cubes can be arranged into a cuboid of dimensions 17 x 1 x 1, which gives a volume of 17, still leaving 18 cubes (35 - 17).

7. These 18 cubes can be arranged into a cuboid of dimensions 9 x 1 x 1, which gives a volume of 9, still leaving 9 cubes (18 - 9).

8. These 9 cubes can be arranged into a cuboid of dimensions 3 x 1 x 1, which gives a volume of 3, still leaving 6 cubes (9 - 3).

9. These 6 cubes can be arranged into a cuboid of dimensions 2 x 1 x 1, which gives a volume of 2, still leaving 4 cubes (6 - 2).

10. These 4 cubes can be arranged into a cuboid of dimensions 2 x 1 x 1, which gives a volume of 2, still leaving 2 cubes (4 - 2).

11. These 2 cubes can be arranged into a cuboid of dimensions 1 x 1 x 1, which gives a volume of 1, still leaving 1 cube (2 - 1).

12. This last cube can be arranged into a cuboid of dimensions 1 x 1 x 1, which gives a volume of 1, leaving no cubes (1 - 1).

13. Now, we calculate the total surface area of all these cuboids. The surface area of a cuboid is given by the formula 2lw + 2lh + 2wh, where l, w, and h are the length, width, and height of the cuboid, respectively.

14. The total surface area is therefore 2(17*17) + 2(17*1) + 2(17*1) + 2(17*1) + 2(17*1) + 2(9*1) + 2(3*1) + 2(2*1) + 2(2*1) + 2(1*1) + 2(1*1) = 578 + 68 + 18 + 4 + 2 = 670.

So, the maximum total surface area of the cuboid formed by merging 5150 cubes with side length 1 is 670 square units.

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