Three cubes of volume 1, 8, and 64 are glued together.
What is the smallest possible surface area of the resulting configuration?

-Problem taken from UNL Math Day

Arrange the cubes so that one face of the one with volume 8 is glued with the one of volume 64 and another face facing to the ground. Then the little one with volume 1 glue it in the corner touching the ground, the one with volume 8 and the one with volume 64.

This arrange would give an area of:
80+16+11+3=110

Posted by Antonio on 2003-08-22 02:17:20