Answer: \(z_{\max} = 37\) at \(x = 4\), \(y = 5\).
Boundary lines and their intercepts:
- \(x + 4y = 24\): \((24, 0)\), \((0, 6)\)
- \(3x + y = 21\): \((7, 0)\), \((0, 21)\)
- \(x + y = 9\): \((9, 0)\), \((0, 9)\)
All three inequalities are of the \(\le\) type, so the origin side of each line is feasible. With \(x, y \ge 0\) the feasible region is the polygon OABCD (rough sketch: region in the first quadrant below all three lines).
Corner points:
- O \((0, 0)\)
- A \((7, 0)\) on \(3x + y = 21\)
- B: \(3x + y = 21\) and \(x + y = 9\) give \(2x = 12\), \(x = 6\), \(y = 3\): B \((6, 3)\)
- C: \(x + y = 9\) and \(x + 4y = 24\) give \(3y = 15\), \(y = 5\), \(x = 4\): C \((4, 5)\)
- D \((0, 6)\) on \(x + 4y = 24\)
| Corner point | \(z = 3x + 5y\) |
|---|
| O (0, 0) | 0 |
| A (7, 0) | 21 |
| B (6, 3) | 33 |
| C (4, 5) | 37 |
| D (0, 6) | 30 |
The maximum value of \(z\) is 37, at \(x = 4\), \(y = 5\).
