Posted by Sheenybeany on Tuesday, July 22, 2014 at 6:02pm.
So, consider just one face of the cubes, they have to be glued. If the smaller is put smack dab, consider that the standard surface area. The exposed surface area on that glued face only is
exposed=9-4=5
But what if the don't overlap exactly, one hangs over the side of the other.
let x be the overhang.
surface area on the joined surface of the smaller cube is 2x
and the surface area on the joined surface of the larger cube is 9-(2-x)2
total= 2x+9-4+2x=5+4x so the minumum surface area on that surface will be 5
total surface of this configuration:
5+5*9+5*4=70sqr units
check my thinking
area of 3x3x3 cube is 6*9=54
area of 2x2x2 cube is 6*4=24
If not joined, total area is 78
The maximum concealed surface area is an entire face of the 2-cube, on both solids: 2*4 = 8
So, the minimum exposed area would be 78-8 = 70
First, we note that the cubes of volume 8 and 27 must have edges of length 2 and 3, respectively. Hence, their surface areas are 6 * 2^2 = 24 and 6 * 3^2 = 54, respectively.
The largest possible surface area that the cubes could share is the area of one face of the smaller cube, namely 2^2 = 4. Therefore, the smallest surface area of the solid resulting from gluing the two cubes together is 24 + 54 - 2 * 4 = 70.