jmc

Let $c_k$ denote the coefficient of $x^k$ in the expansion of $(x+3{)}^{50},$ so $$c_k=\left(\genfrac{}{}{0ex}{}{50}{k}\right)3^{50-k}.$$Then $$c_{k+1}=\left(\genfrac{}{}{0ex}{}{50}{k+1}\right)3^{50-k-1}=\left(\genfrac{}{}{0ex}{}{50}{k+1}\right)3^{49-k}.$$The ratio of these coefficients is $$\begin{array}{rl}\frac{c_{k+1}}{c_k}& =\frac{\left(\genfrac{}{}{0ex}{}{50}{k+1}\right)3^{49-k}}{\left(\genfrac{}{}{0ex}{}{50}{k}\right)3^{50-k}}\\ & =\frac{\frac{50!}{(k+1)!(49-k)!}}{\frac{50!}{k!(50-k)!}\cdot 3}\\ & =\frac{k!(50-k)!}{3(k+1)!(49-k)!}\\ & =\frac{50-k}{3(k+1)}.\end{array}$$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 $k=\boxed{12}.$