Let $S$ be the set of points $(x,y)$ in the coordinate plane that satisfy the inequalities
\begin{cases}
x &\ge 0 \\
y &\ge 0 \\
x + y \le 1
\end{cases}
Then the set $S$ is a polygon. The length of the longest side in $S$ is $a\sqrt{b}$ (expressed in simplest radical form). Find $a+b$.
Graphing the inequalities, we find that $S$ is the region of the coordinate plane bounded by the lines $x=0$, $y=0$, and $x+y=1$. Thus, $S$ is a right triangle with legs of length $1$ and hypotenuse $1$, so the answer is $1+1=\boxed{2}$. The only other step that should be done is to say we exclude the line $y+x=1.$ this is because a line does not have area.
To determine the shape of the set $S$, let's consider each inequality and how it restricts the values of $x$ and $y$.
The inequality $x \geq 0$ represents the condition that $x$ must be non-negative, i.e., $x$ can be any real number greater than or equal to $0$.
Similarly, the inequality $y \geq 0$ represents the condition that $y$ must be non-negative, i.e., $y$ can be any real number greater than or equal to $0$.
Finally, the inequality $x + y \leq 1$ represents the condition that the sum of $x$ and $y$ must be less than or equal to $1$. Let's consider the boundary line $x + y = 1$.
When $x = 0$, we have $0 + y = 1$, which implies $y = 1$. Therefore, the line passes through the point $(0, 1)$.
When $y = 0$, we have $x + 0 = 1$, which implies $x = 1$. Therefore, the line passes through the point $(1, 0)$.
Plotting these points and connecting them, we obtain the line segment connecting the points $(0, 1)$ and $(1, 0)$. Since the inequality is less than or equal to $1$, the points in $S$ are all the points that lie below this line segment, including the line segment itself.
Thus, the set $S$ is a triangle with vertices at $(0, 0)$, $(0, 1)$, and $(1, 0)$.
To find the length of the longest side of the triangle, we can use the distance formula. The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
In this case, we can use the distance formula to find the lengths of the three sides of the triangle.
The length of the side connecting $(0, 0)$ and $(0, 1)$ is $D_1 = \sqrt{(0 - 0)^2 + (1 - 0)^2} = 1$.
The length of the side connecting $(0, 0)$ and $(1, 0)$ is $D_2 = \sqrt{(1 - 0)^2 + (0 - 0)^2} = 1$.
The length of the side connecting $(1, 0)$ and $(0, 1)$ is $D_3 = \sqrt{(0 - 1)^2 + (1 - 0)^2} = \sqrt{2}$.
Therefore, the length of the longest side of the triangle is $\sqrt{2}$.
Finally, $a + b = 1 + 2 = 3$.
Hence, the value of $a + b$ is $3$.