jmc

By the quadratic formula, the equation $5x^2+11x+c=0$ has solutions $$x=\frac{-(11)\pm \sqrt{(11{)}^2-4(5)(c)}}{2(5)}=\frac{-11\pm \sqrt{121-20c}}{10}.$$For these solutions to be rational, the quantity under the square root (i.e., the discriminant) must be a perfect square. So, we seek the possible (positive integer) values of $c$ for which $121-20c$ is a square. The possible nonnegative values for $121-20c$ are $101$, $81$, $61$, $41$, $21$, or $1$. The only squares in this list are $81$, coming from $c=2$, and $1$, coming from $c=6$. So the product of the two possible $c$ values is $2\cdot 6=\boxed{12}$.