How will you arrange 12 cubes of equal length to form a cuboid of smallest surface area?

Solution

Case I

Length (L) = 12b
Breadth (B) = b
Height (H) = b
∴ Total surface area of the cuboid = 2[LB+BH+HL]
= 2(12b x b) + (b x b) + (b x 12b)]
= 2[12b² + b² + 12b²]
= 2[25b²]
= 50b²
Case II

Length (L) = 6b
Breadth (B) = 2b
Height (L) = b
∴ Surface area of the cuboid = 2[LB + BH + HL]
= 2[(6b x 2b) + (2b x b) + (b x 6b)]
= 2[12b² + 2b² + 6b²]
= 2[20b²]
= 40b²
Case III

Length (L) = 3b
Breadth (B) = 2b
Height (H) = 2b
∴ Surface area of the cuboid = 2(LB + BH + HL)
= 2[3b x 2b) + (2b x 2b) + (2b x 3b)
= 2[6b² + 4b² + 6b²]
= 2[16b²]
= 32b²Obviously, the case III arrangement of 12 cubes, has the minimum surface area.