5. A shepherd uses 15 sections of fence, each 2 meters long, to enclose a rectangular or square sheep pen with one side against a wall. The maximum area of the sheep pen is ( ) square meters.
(A) 100
(B) 108
(C) 112
(D) 122
Note: Be sure to pay attention that each fence is placed horizontally. At this point, assume that the two sides adjacent to the wall each have $\mathrm{n}$ fences, and the side opposite the wall has $15-2 \mathrm{n}$ fences. Each fence is 2 units long, so the perimeter of the rectangle, $2 n \times(15-2 n) \times 2$, is maximized when $2 \mathrm{n}$ and 15-2n are as close as possible, and the product is maximized when the difference is minimized. When $2 \mathrm{n}$ and $15-2 \mathrm{n}$ are as close as possible, the area is maximized, $\mathrm{n}=4$. At this time, the rectangle has $2 \mathrm{n}=8$, and the other side length is $(15-2 \mathrm{n}) \times 2=14$, and the area is 112.