Solve the linear programming problem. Maximize and minimize \[ z=3 x+4 y \] Subject to \[ \begin{aligned} 2 x+y & \geq 20 \\ x+2 y & \geq 16 \\ x, y & \geq 0 \end{aligned} \]

Step 1 ：First, we need to find the feasible region defined by the constraints. The constraints are $2x + y \geq 20$, $x + 2y \geq 16$, and $x, y \geq 0$. These inequalities define a region in the $xy$-plane.

Step 2 ：We graph these inequalities to find the feasible region. The feasible region is the area that satisfies all the inequalities simultaneously.

Step 3 ：We find the vertices of the feasible region by solving the system of equations formed by the intersection of the lines $2x + y = 20$, $x + 2y = 16$, and $x = 0$, $y = 0$.

Step 4 ：Solving $2x + y = 20$ and $x + 2y = 16$ gives $x = 8$ and $y = 4$. Solving $2x + y = 20$ and $x = 0$ gives $x = 0$ and $y = 20$. Solving $x + 2y = 16$ and $x = 0$ gives $x = 0$ and $y = 8$.

Step 5 ：So the vertices of the feasible region are $(8, 4)$, $(0, 20)$, and $(0, 8)$.

Step 6 ：We substitute these vertices into the objective function $z = 3x + 4y$ to find the maximum and minimum values.

Step 7 ：Substituting $(8, 4)$ into the objective function gives $z = 3(8) + 4(4) = 40$.

Step 8 ：Substituting $(0, 20)$ into the objective function gives $z = 3(0) + 4(20) = 80$.

Step 9 ：Substituting $(0, 8)$ into the objective function gives $z = 3(0) + 4(8) = 32$.

Step 10 ：So the maximum value of $z$ is $80$ at $(0, 20)$ and the minimum value of $z$ is $32$ at $(0, 8)$.

Step 11 ：Thus, the solution to the linear programming problem is \(\boxed{\text{Maximize } z = 80 \text{ at } (0, 20) \text{ and Minimize } z = 32 \text{ at } (0, 8)}\).