11. (15 points) In the rectangular cardboard shown in the figure, each corner has been cut off with a small rectangle (including squares). If the lengths of the 8 pairs of opposite sides of the cut-off small rectangles are one 1, four 2s, two 3s, and one 4, what is the maximum area of the remaining part of the cardboard?
【Analysis】The original area of the rectangular cardboard is: $12 \times 11=132$, which is a constant. To maximize the area of the remaining part of the cardboard, the sum of the areas of the four small rectangles cut off must be minimized. Clearly, 1 and 4 should be paired, then 2 and 3 should be paired twice, and finally, the two 2s should be paired. The areas of the four small rectangles cut off are: 4, 6, 6, 4. At this point, the sum of the areas of the four small rectangles cut off is minimized, and thus the maximum area of the remaining part of the cardboard can be calculated.

【Solution】According to the analysis: The areas of the four small rectangles cut off are: $1 \times 4=4, 2 \times 2=4, 2 \times 3=6, 2 \times 3=6$,

The original area of the cardboard: $12 \times 11=132$,
The sum of the areas of the four small rectangles cut off: $4+4+6+6=20$,
The maximum area of the remaining part of the cardboard: $132-20=112$;
Answer: The maximum area of the remaining part of the cardboard is 112.