To solve the given linear programming problem graphically, we will follow these steps:
### Step 1: Identify the Objective Function and Constraints
We need to maximize the objective function:
\[ z = 4x + 3y \]
Subject to the constraints:
1. \( 2x + y \geq 40 \)
2. \( x + 2y \geq 50 \)
3. \( x + y \leq 35 \)
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. \( 2x + y = 40 \)
2. \( x + 2y = 50 \)
3. \( x + y = 35 \)
### Step 3: Find Intercepts of Each Line
For each equation, we will find the x-intercept and y-intercept.
1. **For \( 2x + y = 40 \)**:
- When \( x = 0 \): \( y = 40 \) (Point: \( (0, 40) \))
- When \( y = 0 \): \( 2x = 40 \) → \( x = 20 \) (Point: \( (20, 0) \))
2. **For \( x + 2y = 50 \)**:
- When \( x = 0 \): \( 2y = 50 \) → \( y = 25 \) (Point: \( (0, 25) \))
- When \( y = 0 \): \( x = 50 \) (Point: \( (50, 0) \))
3. **For \( x + y = 35 \)**:
- When \( x = 0 \): \( y = 35 \) (Point: \( (0, 35) \))
- When \( y = 0 \): \( x = 35 \) (Point: \( (35, 0) \))
### Step 4: Graph the Lines
Plot the lines on a graph using the intercepts found:
- Line 1: Connect points \( (0, 40) \) and \( (20, 0) \)
- Line 2: Connect points \( (0, 25) \) and \( (50, 0) \)
- Line 3: Connect points \( (0, 35) \) and \( (35, 0) \)
### Step 5: Determine Feasible Region
Identify the feasible region by checking which side of each line satisfies the inequalities:
- For \( 2x + y \geq 40 \): Shade above the line.
- For \( x + 2y \geq 50 \): Shade above the line.
- For \( x + y \leq 35 \): Shade below the line.
- The feasible region will be where all shaded areas overlap, and it must also be in the first quadrant (where \( x \geq 0 \) and \( y \geq 0 \)).
### Step 6: Identify Corner Points of the Feasible Region
The corner points (vertices) of the feasible region can be found by solving the equations of the lines:
1. **Intersection of \( 2x + y = 40 \) and \( x + 2y = 50 \)**:
- Solve the equations:
- Multiply the first equation by 2: \( 4x + 2y = 80 \)
- Subtract the second: \( 4x + 2y - (x + 2y) = 80 - 50 \)
- This simplifies to \( 3x = 30 \) → \( x = 10 \)
- Substitute \( x = 10 \) into \( x + 2y = 50 \): \( 10 + 2y = 50 \) → \( 2y = 40 \) → \( y = 20 \)
- Point: \( (10, 20) \)
2. **Intersection of \( 2x + y = 40 \) and \( x + y = 35 \)**:
- Solve the equations:
- Substitute \( y = 35 - x \) into \( 2x + (35 - x) = 40 \)
- This simplifies to \( x + 35 = 40 \) → \( x = 5 \)
- Substitute \( x = 5 \) into \( x + y = 35 \): \( 5 + y = 35 \) → \( y = 30 \)
- Point: \( (5, 30) \)
3. **Intersection of \( x + 2y = 50 \) and \( x + y = 35 \)**:
- Solve the equations:
- Substitute \( y = 35 - x \) into \( x + 2(35 - x) = 50 \)
- This simplifies to \( x + 70 - 2x = 50 \) → \( -x + 70 = 50 \) → \( x = 20 \)
- Substitute \( x = 20 \) into \( x + y = 35 \): \( 20 + y = 35 \) → \( y = 15 \)
- Point: \( (20, 15) \)
### Step 7: Evaluate the Objective Function at Each Corner Point
Now, we will evaluate \( z = 4x + 3y \) at each corner point:
1. At \( (10, 20) \):
\[ z = 4(10) + 3(20) = 40 + 60 = 100 \]
2. At \( (5, 30) \):
\[ z = 4(5) + 3(30) = 20 + 90 = 110 \]
3. At \( (20, 15) \):
\[ z = 4(20) + 3(15) = 80 + 45 = 125 \]
### Step 8: Identify the Maximum Value
The maximum value of \( z \) occurs at the point \( (20, 15) \):
- Maximum \( z = 125 \)
### Conclusion
The optimal solution to the linear programming problem is:
- \( x = 20 \)
- \( y = 15 \)
- Maximum value of \( z = 125 \)
