Use graphical methods to solve the following. \[ \begin{array}{ll} \text { Maximize: } & z=9 x+6 y \\ \text { subject to: } & 4 x+6 y \leq 24 \\ & 9 x+y \leq 24 \\ & x \geq 0, y \geq 0 \end{array} \]

Step 1 ：Plot the constraints on a graph. The first constraint is \(4x + 6y \leq 24\). This can be rearranged to \(y \leq -\frac{2}{3}x + 4\). The area below the line (including the line itself) represents the solutions to this inequality.

Step 2 ：The second constraint is \(9x + y \leq 24\). This can be rearranged to \(y \leq -9x + 24\). The area below the line (including the line itself) represents the solutions to this inequality.

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 (the area where all constraints are satisfied) is the area that is below both lines and in the first quadrant.

Step 5 ：Find the vertices of the feasible region. These are the points where the lines intersect. The vertices are \((0,0)\), \((0,4)\), and \(\left(\frac{8}{3},8\right)\).

Step 6 ：Substitute these points into the objective function \(z = 9x + 6y\) to find the maximum value. For \((0,0)\), \(z = 9(0) + 6(0) = 0\). For \((0,4)\), \(z = 9(0) + 6(4) = 24\). For \(\left(\frac{8}{3},8\right)\), \(z = 9\left(\frac{8}{3}\right) + 6(8) = 72\).

Step 7 ：\(\boxed{72}\) is the maximum value of z, which occurs at the point \(\left(\frac{8}{3},8\right)\).