5. (3 points) 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
【Analysis】According to the problem, the length of these 15 sections of fence is the sum of one length and two widths of the rectangular sheep pen formed, from which we can list the values of length and width: when the width is 1 section, the length is $15-1 \times 2=13$ sections; when the width is 2 sections, the length is $15-2 \times 2=11$ sections; when the width is 3 sections, the length is $15-3 \times 2=9$ sections; when the width is 4 sections, the length is 15 $-4 \times 2=7$ sections; when the width is 5 sections, the length is $15-5 \times 2=5$ sections; when the width is 6 sections, the length is $15-6 \times 2=3$ sections; when the width is 7 sections, the length is $15-7 \times 2=1$ section. Since the area of a rectangle = length $\times$ width, we can substitute the data to find the area of the different rectangles formed, and then compare them to solve the problem.

【Solution】Solution: When the width is 1 section, the length is $15-1 \times 2=13$ sections, i.e., the width is 2 meters, and the length is $13 \times 2=26$ (meters)

Then the area is $26 \times 2=52$ (square meters)

When the width is 2 sections, the length is $15-2 \times 2=11$ sections
i.e., the width is $2 \times 2=4$ (meters), and the length is $11 \times 2=22$ (meters)
Then the area is $22 \times 4=88$ (square meters)

When the width is 3 sections, the length is $15-3 \times 2=9$ sections
i.e., the width is $3 \times 2=6$ (meters), and the length is $9 \times 2=18$ (meters)
Then the area is: $6 \times 18=108$ (square meters)

When the width is 4 sections, the length is $15-4 \times 2=7$ sections
i.e., the width is $4 \times 2=8$ (meters), and the length is $7 \times 2=14$ (meters)
Then the area is $8 \times 14=112$ (square meters)

When the width is 5 sections, the length is $15-5 \times 2=5$ sections,
i.e., both the length and width are $5 \times 2=10$ (meters), forming a square
Then the area is: $10 \times 10=100$ (square meters)

When the width is 6 sections, the length is $15-6 \times 2=3$ sections
i.e., the width is $6 \times 2=12$ (meters), and the length is $3 \times 2=6$ (meters)
Then the area is $12 \times 6=72$ (square meters)

When the width is 7 sections, the length is $15-7 \times 2=1$ section
i.e., the width is $7 \times 2=14$ (meters), and the length is $1 \times 2=2$ (meters)
Then the area is $14 \times 2=28$ (square meters)
From the above calculations, the maximum area of the rectangular sheep pen is 112 square meters.
Therefore, the answer is: $C$.