Step-by-Step Solution

1. Graph the inequalities:

   • \(4x + y \ge 80\) can be rewritten as \(y \ge -4x + 80\)
   • \(3x + 2y \le 150\) can be rewritten as \(y \le -\frac{3}{2}x + 75\)
   • \(x + 5y \ge 115\) can be rewritten as \(y \ge -\frac{1}{5}x + 23\)
   • \(x \ge 0\) and \(y \ge 0\)

   Plot these inequalities on a graph. Find the intersection points of the lines.

2. Identify the feasible region:

   The feasible region is the area on the graph that satisfies all the inequalities. It is the intersection of all the shaded regions defined by the inequalities.

3. Find the corner points of the feasible region:

   The corner points are the vertices of the feasible region. These are the points where the boundary lines intersect. Solve the equations of the lines to find these points.

   • Intersection of \(4x + y = 80\) and \(3x + 2y = 150\): Solving these equations, we get \(x = 10\) and \(y = 40\). So, the point is (10, 40).
   • Intersection of \(3x + 2y = 150\) and \(x + 5y = 115\): Solving these equations, we get \(x = 40\) and \(y = 15\). So, the point is (40, 15).
   • Intersection of \(4x + y = 80\) and \(x + 5y = 115\): Solving these equations, we get \(x = 15\) and \(y = 20\). So, the point is (15, 20).

4. Evaluate the objective function at each corner point:

   Substitute the coordinates of each corner point into the objective function \(z = 6x + 3y\) to find the value of \(z\) at each point.

   • At (10, 40): \(z = 6(10) + 3(40) = 60 + 120 = 180\)
   • At (40, 15): \(z = 6(40) + 3(15) = 240 + 45 = 285\)
   • At (15, 20): \(z = 6(15) + 3(20) = 90 + 60 = 150\)

5. Determine the optimal solution:

   The maximum value of \(z\) is 285, which occurs at the point (40, 15).