Find the minimum and maximum values of $z=5 x+8 y$, if possible, for the following set of constraints. Use graphical methods to solve.
\[
\begin{array}{c}
-x+2 y \leq 14 \\
7 x+y \geq 7 \\
x \geq 0, y \geq 0
\end{array}
\]

Use the graphing tool to graph the system. Graph the region that represents the correct solution only once.

Step 1 ：First, we need to graph the constraints. The first constraint is \(-x + 2y \leq 14\). We rearrange it to get \(y \leq \frac{14}{2} + \frac{x}{2}\), which simplifies to \(y \leq 7 + \frac{x}{2}\). This is a line with a slope of \(\frac{1}{2}\) and y-intercept of 7. The solution to this inequality is the region below this line, including the line itself.

Step 2 ：The second constraint is \(7x + y \geq 7\). We rearrange it to get \(y \geq 7 - 7x\). This is a line with a slope of -7 and y-intercept of 7. The solution to this inequality is the region above this line, including the line itself.

Step 3 ：The third and fourth constraints are \(x \geq 0\) and \(y \geq 0\). These simply mean that we are only considering the first quadrant (where both x and y are positive).

Step 4 ：The feasible region is the area that satisfies all these constraints. It is the area in the first quadrant that is below the line \(y = 7 + \frac{x}{2}\) and above the line \(y = 7 - 7x\).

Step 5 ：Now, we need to find the minimum and maximum values of \(z = 5x + 8y\) in this feasible region. These values will occur at the vertices of the feasible region.

Step 6 ：The vertices of the feasible region are \((0,0)\), \((0,7)\), and \((1,7)\).

Step 7 ：Substituting these points into the equation for z gives: At \((0,0)\), \(z = 5(0) + 8(0) = 0\). At \((0,7)\), \(z = 5(0) + 8(7) = 56\). At \((1,7)\), \(z = 5(1) + 8(7) = 61\).

Step 8 ：So, the minimum value of z is \(\boxed{0}\) and the maximum value of z is \(\boxed{61}\).