Find the positive integer $k$ for which the coefficient of $x^k$ in the expansion of $(x + 3)^{50}$ is maximized.
Let $c_k$ denote the coefficient of $x^k$ in the expansion of $(x + 3)^{50},$ so
\[c_k = \binom{50}{k} 3^{50 - k}.\]Then
\[c_{k + 1} = \binom{50}{k + 1} 3^{50 - k - 1} = \binom{50}{k + 1} 3^{49 - k}.\]The ratio of these coefficients is
\begin{align*}
\frac{c_{k + 1}}{c_k} &= \frac{\binom{50}{k + 1} 3^{49 - k}}{\binom{50}{k} 3^{50 - k}} \\
&= \frac{\frac{50!}{(k + 1)! (49 - k)!}}{\frac{50!}{k! (50 - k)!} \cdot 3} \\
&= \frac{k! (50 - k)!}{3 (k + 1)! (49 - k)!} \\
&= \frac{50 - k}{3(k + 1)}.
\end{align*}Consider the inequality
\[\frac{50 - k}{3(k + 1)} \ge 1.\]This is equivalent to $50 - k \ge 3(k + 1) = 3k + 3.$  Then $4k \le 47,$ or $k \le \frac{47}{4}.$  Since $k$ is an integer, this is equivalent to $k \le 11.$

This means that the sequence $c_0,$ $c_1,$ $c_2,$ $\dots,$ $c_{11},$ $c_{12}$ is increasing, but the sequence $c_{12},$ $c_{13},$ $c_{14},$ $\dots$ is decreasing.  Hence, $c_k$ is maximized for $k = \boxed{12}.$