smc

The value of $5$-pound rocks is $\text{textdollar}14÷5=\text{textdollar}2.80$ per pound, and the value of $4$-pound rocks is $\text{textdollar}11÷4=\text{textdollar}2.75$ per pound. Clearly, Carl should not carry more than three $1$-pound rocks. Otherwise, he can replace some $1$-pound rocks with some heavier rocks, preserving the weight but increasing the total value. We perform casework on the number of $1$-pound rocks Carl can carry: \begin{array}{c|c|c||c} & & & \\ [-2.5ex] \boldsymbol{1}\textbf{-Pound Rocks} & \boldsymbol{4}\textbf{-Pound Rocks} & \boldsymbol{5}\textbf{-Pound Rocks} & \textbf{Total Value} \\ \textbf{(}\boldsymbol{\textdollar2}\textbf{ Each)} & \textbf{(}\boldsymbol{\textdollar11}\textbf{ Each)} & \textbf{(}\boldsymbol{\textdollar14}\textbf{ Each)} & \\ [0.5ex] \hline & & & \\ [-2ex] 0 & 2 & 2 & \textdollar50 \\ & & & \\ [-2.25ex] 1 & 3 & 1 & \textdollar49 \\ & & & \\ [-2.25ex] 2 & 4 & 0 & \textdollar48 \\ & & & \\ [-2.25ex] 3 & 0 & 3 & \textdollar48 \end{array} Clearly, the maximum value of the rocks Carl can carry is $\boxed{50}$ dollars. Remark Note that an upper bound of the total value is $\text{textdollar}2.80\cdot 18=\text{textdollar}50.40,$ from which we can eliminate choices $\text{(D)}$ and $\text{(E)}.$