To solve the linear programming problem of maximizing \( z = 3x + 4y \) subject to the constraints \( x + y \leq 40 \), \( x + 2y \leq 10 \), and \( x, y \geq 0 \), we will follow these steps:
### Step 1: Identify the Constraints
The constraints given are:
1. \( x + y \leq 40 \)
2. \( x + 2y \leq 10 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)
### Step 2: Graph the Constraints
We will graph the inequalities to find the feasible region.
1. **For \( x + y = 40 \)**:
- When \( x = 0 \), \( y = 40 \) (point: \( (0, 40) \))
- When \( y = 0 \), \( x = 40 \) (point: \( (40, 0) \))
2. **For \( x + 2y = 10 \)**:
- When \( x = 0 \), \( y = 5 \) (point: \( (0, 5) \))
- When \( y = 0 \), \( x = 10 \) (point: \( (10, 0) \))
### Step 3: Determine the Feasible Region
Plot the lines on a graph and shade the feasible region that satisfies all the constraints. The feasible region will be bounded by the axes and the lines \( x + y = 40 \) and \( x + 2y = 10 \).
### Step 4: Identify the Corner Points
The corner points of the feasible region can be found by determining the intersection points of the lines:
1. **Intersection of \( x + y = 40 \) and \( x + 2y = 10 \)**:
- From \( x + y = 40 \), we can express \( y = 40 - x \).
- Substitute into \( x + 2(40 - x) = 10 \):
\[
x + 80 - 2x = 10 \implies -x + 80 = 10 \implies x = 70 \text{ (not feasible)}
\]
- Thus, this intersection is outside the feasible region.
2. **Evaluate the corner points**:
- \( (0, 0) \)
- \( (10, 0) \)
- \( (0, 5) \)
### Step 5: Calculate the Objective Function at Each Corner Point
Now, we will calculate \( z = 3x + 4y \) at each of the corner points:
1. At \( (0, 0) \):
\[
z = 3(0) + 4(0) = 0
\]
2. At \( (10, 0) \):
\[
z = 3(10) + 4(0) = 30
\]
3. At \( (0, 5) \):
\[
z = 3(0) + 4(5) = 20
\]
### Step 6: Determine the Maximum Value
The maximum value of \( z \) from the calculated values is:
- \( z(0, 0) = 0 \)
- \( z(10, 0) = 30 \)
- \( z(0, 5) = 20 \)
Thus, the maximum value of \( z \) is \( 30 \) at the point \( (10, 0) \).
### Final Answer
The maximum value of \( z \) is \( \boxed{30} \).
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