The problem requires us to cover a rectangular area of $3' \times 100'$ using tiles of sizes $3' \times 3'$, $3' \times 7'$, and $3' \times 11'$.

Since the width of the area is $3'$ and all the available tiles also have a width of $3'$, the tiles must be placed end-to-end along the $100'$ length. They cannot be rotated because a width of $7'$ or $11'$ would not fit inside the $3'$ wide area.

Thus, the problem simplifies to finding the number of tiles of lengths $3'$, $7'$, and $11'$ that add up exactly to $100'$. Let $a$, $b$, and $c$ be the number of $3'$, $7'$, and $11'$ tiles used, respectively. We need to find non-negative integers $a, b, c$ such that: $3a + 7b + 11c = 100$

We need to find the maximum ($x$) and minimum ($y$) total number of tiles ($a + b + c$).

Step 1: Find the maximum number of tiles ($x$) To maximize the total number of tiles, we should use as many of the smallest tiles ($3'$) as possible.

• If we use $33$ tiles of $3'$, the length covered is $33 \times 3 = 99'$, leaving $1'$, which cannot be filled by $7'$ or $11'$ tiles.
• If we use $32$ tiles of $3'$, the length covered is $32 \times 3 = 96'$, leaving $4'$, which cannot be filled.
• If we use $31$ tiles of $3'$, the length covered is $31 \times 3 = 93'$, leaving $7'$. This remaining length can be exactly filled by one $7'$ tile.

So, a valid combination is $a = 31$, $b = 1$, $c = 0$. The maximum number of tiles is $x = 31 + 1 + 0 = 32$.

Step 2: Find the minimum number of tiles ($y$) To minimize the total number of tiles, we should use as many of the largest tiles ($11'$) as possible.

• If we use $9$ tiles of $11'$, the length covered is $9 \times 11 = 99'$, leaving $1'$, which cannot be filled.
• If we use $8$ tiles of $11'$, the length covered is $8 \times 11 = 88'$, leaving $12'$. This remaining length can be exactly filled by four $3'$ tiles ($4 \times 3 = 12'$).

So, a valid combination is $a = 4$, $b = 0$, $c = 8$. The minimum number of tiles is $y = 4 + 0 + 8 = 12$.

(Note: We can mathematically verify that no combination yields fewer than 12 tiles. If $a+b+c = 11$, the maximum possible length is $11 \times 11 = 121$. To get exactly 100, we check the equation $11(a+b+c) - (3a+7b+11c) = 8a+4b$. Substituting the sums gives $121 - 100 = 21$, which means $8a+4b = 21$. Since $8a+4b$ is always even and $21$ is odd, $11$ tiles is impossible.)

Step 3: Calculate $x - y$ The difference between the maximum and minimum number of tiles is: $x - y = 32 - 12 = 20$