Consider the set of all rectangles with a given area $S$. 
Find the largest value o  $ M = \frac{16-p}{p^2+2p}$ where $p$ is the perimeter of the rectangle.
1. **Define the problem and variables:**
   We are given a set of all rectangles with a fixed area \( S \). We need to find the largest value of \( M = \frac{16 - p}{p^2 + 2p} \), where \( p \) is the perimeter of the rectangle.

2. **Express the perimeter in terms of side lengths:**
   Let \( a \) and \( b \) be the side lengths of the rectangle. The perimeter \( p \) is given by:
   \[
   p = 2(a + b)
   \]
   The area \( S \) is given by:
   \[
   S = ab
   \]

3. **Use the inequality involving the perimeter and area:**
   Using the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality), we have:
   \[
   \frac{a + b}{2} \geq \sqrt{ab}
   \]
   Multiplying both sides by 2, we get:
   \[
   a + b \geq 2\sqrt{ab}
   \]
   Therefore, the perimeter \( p \) satisfies:
   \[
   p = 2(a + b) \geq 4\sqrt{ab} = 4\sqrt{S}
   \]

4. **Substitute the inequality into the expression for \( M \):**
   Since \( p \geq 4\sqrt{S} \), we substitute this into the expression for \( M \):
   \[
   M = \frac{16 - p}{p^2 + 2p}
   \]
   To find the maximum value of \( M \), we need to consider the boundary condition when \( p = 4\sqrt{S} \):
   \[
   M \leq \frac{16 - 4\sqrt{S}}{(4\sqrt{S})^2 + 2 \cdot 4\sqrt{S}} = \frac{16 - 4\sqrt{S}}{16S + 8\sqrt{S}}
   \]

5. **Simplify the expression:**
   Simplifying the expression, we get:
   \[
   M \leq \frac{4 - \sqrt{S}}{4S + 2\sqrt{S}}
   \]

6. **Consider the equality condition:**
   The equality holds if and only if \( a = b \), i.e., the rectangle is a square. In this case, the perimeter \( p \) is exactly \( 4\sqrt{S} \).

7. **Analyze the behavior for \( p > 16 \):**
   If \( p > 16 \), then:
   \[
   M = \frac{16 - p}{p^2 + 2p} < 0
   \]
   For any \( M_0 < 0 \), there always exists a \( p \) such that \( M(p) = \frac{16 - p}{p^2 + 2p} > M_0 \).

The final answer is \( \boxed{ \frac{4 - \sqrt{S}}{4S + 2\sqrt{S}} } \)