Find the minimum and maximum values of $z=4 x+6 y$, if possible, for the following set of constraints.
\[
\begin{array}{c}
5 x+4 y \geq 20 \\
x+4 y \geq 8 \\
x \geq 0, y \geq 0
\end{array}
\]
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The minimum value is (Round to the nearest tenth as needed.)
B. There is no minimum value.

Step 1 ：First, we need to graph the constraints to find the feasible region. The constraints are $5x+4y \geq 20$, $x+4y \geq 8$, and $x \geq 0, y \geq 0$. The feasible region is the area that satisfies all these inequalities.

Step 2 ：The feasible region is a polygon bounded by the lines $5x+4y = 20$, $x+4y = 8$, and the axes $x=0$ and $y=0$.

Step 3 ：The vertices of this polygon are the points where these lines intersect. We can find these points by solving the equations simultaneously.

Step 4 ：Solving $5x+4y = 20$ and $x=0$ gives $y=5$. So one vertex is $(0,5)$.

Step 5 ：Solving $5x+4y = 20$ and $x+4y = 8$ gives $x=2$ and $y=1.5$. So another vertex is $(2,1.5)$.

Step 6 ：Solving $x+4y = 8$ and $x=0$ gives $y=2$. So another vertex is $(0,2)$.

Step 7 ：Solving $x+4y = 8$ and $y=0$ gives $x=8$. So the last vertex is $(8,0)$.

Step 8 ：Now we substitute these vertices into the objective function $z=4x+6y$ to find the minimum and maximum values.

Step 9 ：Substituting $(0,5)$ gives $z=4(0)+6(5)=30$.

Step 10 ：Substituting $(2,1.5)$ gives $z=4(2)+6(1.5)=14$.

Step 11 ：Substituting $(0,2)$ gives $z=4(0)+6(2)=12$.

Step 12 ：Substituting $(8,0)$ gives $z=4(8)+6(0)=32$.

Step 13 ：So the minimum value of $z$ is 12 and the maximum value is 32.

Step 14 ：Therefore, the minimum value of $z=4x+6y$ subject to the constraints is 12 and the maximum value is 32.