Triangles of integral sides

For $x$ , we must satisfy both $3+x>5$ and $3+5>x$ . Thus $2<x<8$ . Similarly, $y$ must satisfy both $4+y>6$ and $4+6>y$ . Thus $2<y<10$ . Since $x$ and $y$ must be integers, we can rewrite these inequalities as $3\leq x\leq 7$ and $3\leq y\leq 9$ . Finally, $|x-y|$ is maximized when $x=3$ and $y=9$ , with $|x-y|=|3-9| = \boxed{6}$ .

Applying the cosine rule of triangles, since cosine of the angle between two sides must lie in (-1,1), 2<x<8 and 2<y<10. Therefore minimum value of x is 3 and the maximum value of y is 9. Hence maximum value of their difference is 6