A square region with perimeter 60 inches is made with square inch tiles. Bob removes one tile from the square and rearranges the remaining tiles without any overlap to make a rectangular region with minimum perimeter. How many inches are in the perimeter?
If the original perimeter was 60 inches, there must have been 15 tiles per side and 15 * 15 tiles = 225 tiles in all for an area of 225 in^2
Removing one tile will give us an area of 224 in^2 with 224 tiles
It can be shown that the perimeter will be minimized whenever the difference between the length and width is minimized
The divisors of 224 are : 1 | 2 | 4 | 7 | 8 | 14 | 16 | 28 | 32 | 56 | 112 | 224
And the possible lengths and widhts giving and area of 224 in^2 with the associated perimeters, P, is as follows :
1 , 224 P = 450
2 , 112 P = 228
4 , 56 P = 120
7, 32 P = 78
8, 28 P = 72
14, 16 P = 60
So......14 x 16 tiles minimizes the perimeter = 60 inches [ oddly..... the same perimeter before we removed a tile !!! ]
Note......this will always happen when we remove exactly one tile from any square as in this situation..... if s is the side of the square....4s is the perimeter.....if we remove one tile, the new length and width that minimizes the perimeter is
(s - 1), (s +1)....but the perimeter is also 2 [(s - 1) + (s +1)] = 2 [ 2s ] = 4s !!!!