11. A box contains balls of red, yellow, and white. If the number of white balls is at most $\frac{1}{2}$ of the number of yellow balls, and at least $\frac{1}{3}$ of the number of red balls, and the total number of yellow and white balls is no more than 55, then the box can contain at most $\qquad$ red balls.
11.54.

Suppose there are $x$ white balls, $y$ yellow balls, and $z$ red balls. According to the problem, we can set up the following system of equations:
$$
\left\{\begin{array}{l}
x \leqslant \frac{1}{2} y, \\
x \geqslant \frac{1}{3} z, \\
y + x \leqslant 55 .
\end{array}\right.
$$

From equation (1), we get $y \geqslant 2 x$.
From equations (3) and (4), we get $3 x \leqslant 55$, which means $x \leqslant \frac{55}{3}$. Therefore, the integer $x \leqslant 18$. From equation (2), we get $z \leqslant 3 x$, which means $z \leqslant 54$.
Hence, there are at most 54 red balls in the box.