The set of all solutions of the system  \[ \begin{cases}& 2x+y \le 4 \\& x+y \ge 1 \\& x \ge 0 \\& y \ge 0 \end{cases} \] is a quadrilateral region. If the number of units in the length of the longest side is $a\sqrt{b}$ (expressed in simplest radical form), find $a+b$.
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
The set of all solutions of the system  \[ \begin{cases}& 2x+y \le 4 \\& x+y \ge 1 \\& x \ge 0 \\& y \ge 0 \end{cases} \] is a quadrilateral region. If the number of units in the length of the longest side is $a\sqrt{b}$ (expressed in simplest radical form), find $a+b$.
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
7
The set of all solutions of the system  \[ \begin{cases}& 2x+y \le 4 \\& x+y \ge 1 \\& x \ge 0 \\& y \ge 0 \end{cases} \] is a quadrilateral region. If the number of units in the length of the longest side is $a\sqrt{b}$ (expressed in simplest radical form), find $a+b$.
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