Suppose $F$ is a polygon with lattice vertices and sides parralell to x-axis and y-axis.Suppose $S(F),P(F)$ are area and perimeter of $F$.
Find the smallest k that:
$S(F) \leq k.P(F)^2$
1. **Understanding the Problem:**
   We need to find the smallest constant \( k \) such that for any polygon \( F \) with lattice vertices and sides parallel to the x-axis and y-axis, the inequality \( S(F) \leq k \cdot P(F)^2 \) holds, where \( S(F) \) is the area and \( P(F) \) is the perimeter of \( F \).

2. **Analyzing the Case of a Rectangle:**
   Consider a rectangle with side lengths \( a \) and \( b \). The area \( S \) and perimeter \( P \) of the rectangle are given by:
   \[
   S = a \cdot b
   \]
   \[
   P = 2a + 2b
   \]
   We need to express \( S \) in terms of \( P \):
   \[
   S = a \cdot b
   \]
   \[
   P = 2(a + b) \implies a + b = \frac{P}{2}
   \]
   Using the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality), we have:
   \[
   \frac{a + b}{2} \geq \sqrt{a \cdot b} \implies \frac{P}{4} \geq \sqrt{S} \implies \left(\frac{P}{4}\right)^2 \geq S \implies \frac{P^2}{16} \geq S
   \]
   Therefore, for a rectangle, we have:
   \[
   S \leq \frac{P^2}{16}
   \]

3. **Generalizing to Other Polygons:**
   For polygons with lattice vertices and sides parallel to the x-axis and y-axis, we can consider the following:
   - If the polygon is concave, we can extend its border to form a convex polygon (specifically a rectangle) without increasing the perimeter.
   - The area of the concave polygon will be less than or equal to the area of the rectangle formed by extending its borders.
   - Therefore, the ratio \( \frac{S(F)}{P(F)^2} \) for a concave polygon will be less than or equal to that of a rectangle.

4. **Conclusion:**
   Since the inequality \( S(F) \leq \frac{P(F)^2}{16} \) holds for rectangles and the area of any concave polygon is less than or equal to the area of the rectangle with the same perimeter, the smallest \( k \) that satisfies the inequality for all such polygons is \( \frac{1}{16} \).

\[
\boxed{\frac{1}{16}}
\]