A company produces rectangular advertising boards such that the width is always at least 8 cm shorter than the length, and both the length and width are integers. If the area of a board cannot be more than $320\text{ cm}^2$ and the width is $x$ cm, then find the maximum possible value of $x$.

Let the width be $x$ cm. Since the length is at least 8 cm more than the width, the smallest possible length is $x+8$ cm.

To maximise the width while keeping the area at most $320\text{ cm}^2$, use the minimum allowed length:

$x(x+8)\leq320$

For $x=14$, $14\times22=308\leq320$.

For $x=15$, $15\times23=345>320$.

The maximum possible integer width is $14$ cm.