Four cubes of volumes and are glued together at their faces. in is value of total surface area of the resulting figure. Find minimum possible value of .
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The total of surface area of each block = 6 × ( 1 + 4 + 9 + 2 5 ) = 2 3 6
The surface area eliminated = 2 × ( 3 × ( 1 ) + 2 × ( 4 ) + 1 × ( 9 ) ) = 4 0
The minimum surface area, S = 2 3 6 − 4 0 = 1 9 4