Given that $x\ge0$, $y\ge0$, $x+2y\le6$, and $2x+y\le6$, compute the maximum possible value of $x+y$.
To find the maximum possible value of \(x + y\) given the constraints \(x \ge 0\), \(y \ge 0\), \(x + 2y \le 6\), and \(2x + y \le 6\), we can follow these steps:

1. **Graph the inequalities**:
   - The inequality \(x \ge 0\) represents the region to the right of the y-axis.
   - The inequality \(y \ge 0\) represents the region above the x-axis.
   - The inequality \(x + 2y \le 6\) can be rewritten as \(y \le \frac{6 - x}{2}\). This represents the region below the line \(y = \frac{6 - x}{2}\).
   - The inequality \(2x + y \le 6\) can be rewritten as \(y \le 6 - 2x\). This represents the region below the line \(y = 6 - 2x\).

2. **Find the intersection points of the lines**:
   - To find the intersection of \(x + 2y = 6\) and \(2x + y = 6\), solve the system of equations:
     \[
     \begin{cases}
     x + 2y = 6 \\
     2x + y = 6
     \end{cases}
     \]
     Multiply the first equation by 2:
     \[
     2x + 4y = 12
     \]
     Subtract the second equation from this result:
     \[
     2x + 4y - (2x + y) = 12 - 6 \\
     3y = 6 \\
     y = 2
     \]
     Substitute \(y = 2\) back into the first equation:
     \[
     x + 2(2) = 6 \\
     x + 4 = 6 \\
     x = 2
     \]
     So, the intersection point is \((2, 2)\).

3. **Check the vertices of the feasible region**:
   - The vertices of the feasible region are where the lines intersect each other and the axes. These points are:
     \[
     (0, 0), (0, 3), (3, 0), (2, 2)
     \]
   - Evaluate \(x + y\) at each vertex:
     \[
     (0, 0): x + y = 0 + 0 = 0 \\
     (0, 3): x + y = 0 + 3 = 3 \\
     (3, 0): x + y = 3 + 0 = 3 \\
     (2, 2): x + y = 2 + 2 = 4
     \]

4. **Determine the maximum value**:
   - The maximum value of \(x + y\) among the vertices is 4, which occurs at the point \((2, 2)\).

The final answer is \(\boxed{4}\).