jmc

We can make use of the fact that the sum of the roots of the quadratic equation $a{x}^2+bx+c=0$ is $-b/a$ and the product of the roots is $c/a$. Choosing $a$, $b$, and $c$ so that $-b/a=11/12$ and $c/a=1/6$, we find that the fractions are the solutions to $12x^2-11x+2=0$. Factoring this, we get $$12x^2-11x+2=(3x-2)(4x-1).$$ Therefore, the solutions of $12x^2-11x+2=0$ are $x=\frac{1}{4}$ and $x=\frac{2}{3}$. The smaller of these fractions is $\boxed{\frac{1}{4}}$.

An alternative way to obtain the equation $12x^2-11x+2=0$ is to begin with the given equations $x+y=\frac{11}{12}$ and $xy=\frac{1}{6}$. Solve the first equation for $y$ and substitute $y=\frac{11}{12}-x$ into the second equation. Distributing, clearing denominators, and rearranging gives $12x^2-11x+2=0$. Then we proceed as before.