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 $a+b$.
I want to find the vertices of the quadrilateral region, since they will determine the lengths of the sides. To find the vertices, I need to solve for the intersection points of the boundary lines of the inequalities. The boundary lines are $2x+y = 4, x+y = 1, x = 0,$ and $y = 0$. To find the intersection of two lines, I can use substitution or elimination. However, this point is not in the region, since it does not satisfy $y \ge 0$. So I need to try a different pair of lines. Another pair of lines that intersect are $2x+y = 4$ and $x = 0$. If I plug in $x = 0$ into the first equation, I get $y = 4$. This point is in the region, since it satisfies all the inequalities. So one vertex of the quadrilateral is $(0,4)$. I can repeat this process for the other pairs of lines that intersect, and check if the points are in the region or not. Another pair of lines that intersect are $x+y = 1$ and $x = 0$. If I plug in $x = 0$ into the first equation, I get $y = 1$. This point is also in the region, since it satisfies all the inequalities. So another vertex of the quadrilateral is $(0,1)$. Another pair of lines that intersect are $x+y = 1$ and $y = 0$. If I plug in $y = 0$ into the first equation, I get $x = 1$. This point is also in the region, since it satisfies all the inequalities. So another vertex of the quadrilateral is $(1,0)$. The last pair of lines that intersect are $2x+y = 4$ and $y = 0$. If I plug in $y = 0$ into the first equation, I get $2x = 4$, so $x = 2$. This point is also in the region, since it satisfies all the inequalities. So the last vertex of the quadrilateral is $(2,0)$. Now I have all the vertices of the quadrilateral: $(0,4), (0,1), (1,0),$ and $(2,0)$. To find the length of the longest side, I need to compare the distances between these vertices. I can use the distance formula to find the distances. For example, the distance between $(0,4)$ and $(0,1)$ is $\sqrt{(0-0)^2 + (1-4)^2} = \sqrt{9} = 3$. Similarly, the distance between $(0,1)$ and $(1,0)$ is $\sqrt{(1-0)^2 + (0-1)^2} = \sqrt{2}$. The distance between $(1,0)$ and $(2,0)$ is $\sqrt{(2-1)^2 + (0-0)^2} = 1$. And the distance between $(2,0)$ and $(0,4)$ is $\sqrt{(0-2)^2 + (4-0)^2} = \sqrt{20}$. Comparing these distances, I see that the longest one is $\sqrt{20}$. This is the length of the longest side of the quadrilateral. To express this in simplest radical form, I can factor out a 4 from under the square root, since 4 is a perfect square. So the length of the longest side is $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$. The number of units in this length is 2, and the coefficient of the square root is 5. So $a = 2$ and $b = 5$. Therefore, the answer is $a + b = 2 + 5 = 7$.

