To solve the problem of maximizing \( Z = 9x + 10y \) subject to the given constraints, we will follow these steps:
### Step 1: Identify the Constraints
The constraints given are:
1. \( 9x + 2y \geq 20 \)
2. \( x - 2y \geq 0 \) (or \( x \geq 2y \))
3. \( x + y \leq 9 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)
### Step 2: Convert Inequalities to Equations
To graph the constraints, we first convert the inequalities into equations:
1. \( 9x + 2y = 20 \)
2. \( x - 2y = 0 \) (or \( x = 2y \))
3. \( x + y = 9 \)
### Step 3: Find Intercepts for Each Constraint
- For \( 9x + 2y = 20 \):
- When \( x = 0 \): \( 2y = 20 \) → \( y = 10 \) (point: \( (0, 10) \))
- When \( y = 0 \): \( 9x = 20 \) → \( x = \frac{20}{9} \) (point: \( (\frac{20}{9}, 0) \))
- For \( x - 2y = 0 \) (or \( x = 2y \)):
- When \( y = 0 \): \( x = 0 \) (point: \( (0, 0) \))
- When \( x = 0 \): \( y = 0 \) (point: \( (0, 0) \))
- For \( x + y = 9 \):
- When \( x = 0 \): \( y = 9 \) (point: \( (0, 9) \))
- When \( y = 0 \): \( x = 9 \) (point: \( (9, 0) \))
### Step 4: Graph the Constraints
Plot the lines on a graph and shade the feasible region:
- The line \( 9x + 2y = 20 \) will be shaded above.
- The line \( x = 2y \) will be shaded to the right.
- The line \( x + y = 9 \) will be shaded below.
- The feasible region is where all shaded areas overlap, considering \( x \geq 0 \) and \( y \geq 0 \).
### Step 5: Identify the Corner Points of the Feasible Region
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( 9x + 2y = 20 \) and \( x = 2y \):
- Substitute \( x = 2y \) into \( 9(2y) + 2y = 20 \) → \( 18y + 2y = 20 \) → \( 20y = 20 \) → \( y = 1 \), \( x = 2 \) (point: \( (2, 1) \))
2. Intersection of \( x + y = 9 \) and \( x = 2y \):
- Substitute \( x = 2y \) into \( 2y + y = 9 \) → \( 3y = 9 \) → \( y = 3 \), \( x = 6 \) (point: \( (6, 3) \))
3. Intersection of \( 9x + 2y = 20 \) and \( x + y = 9 \):
- Solve the system:
- From \( x + y = 9 \), \( y = 9 - x \)
- Substitute into \( 9x + 2(9 - x) = 20 \) → \( 9x + 18 - 2x = 20 \) → \( 7x = 2 \) → \( x = \frac{2}{7} \), \( y = \frac{61}{7} \) (not feasible)
4. The corner points are \( (0, 10) \), \( (9, 0) \), \( (2, 1) \), and \( (6, 3) \).
### Step 6: Evaluate the Objective Function at Each Corner Point
Calculate \( Z \) at each corner point:
1. At \( (2, 1) \): \( Z = 9(2) + 10(1) = 18 + 10 = 28 \)
2. At \( (6, 3) \): \( Z = 9(6) + 10(3) = 54 + 30 = 84 \)
3. At \( (9, 0) \): \( Z = 9(9) + 10(0) = 81 + 0 = 81 \)
4. At \( (0, 10) \): \( Z = 9(0) + 10(10) = 0 + 100 = 100 \)
### Step 7: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point \( (6, 3) \) with \( Z = 84 \).
### Conclusion
The maximum value of \( Z \) is \( 84 \) at the point \( (6, 3) \).
