To solve the problem of maximizing and minimizing the objective function \( Z = 4x + 3y - 7 \) subject to the given constraints, we will follow these steps:
### Step 1: Identify the Constraints
The constraints provided are:
1. \( x + y \leq 10 \)
2. \( x + y \geq 3 \)
3. \( x \leq 8 \)
4. \( y \leq 9 \)
5. \( x \geq 0 \)
6. \( y \geq 0 \)
### Step 2: Convert Constraints to Equations
We will convert the inequalities into equations to find the boundary lines:
1. \( x + y = 10 \)
2. \( x + y = 3 \)
3. \( x = 8 \)
4. \( y = 9 \)
### Step 3: Find Intercepts
For each equation, we will find the intercepts:
1. **For \( x + y = 10 \)**:
- \( x \)-intercept: Set \( y = 0 \) → \( x = 10 \) → Point (10, 0)
- \( y \)-intercept: Set \( x = 0 \) → \( y = 10 \) → Point (0, 10)
2. **For \( x + y = 3 \)**:
- \( x \)-intercept: Set \( y = 0 \) → \( x = 3 \) → Point (3, 0)
- \( y \)-intercept: Set \( x = 0 \) → \( y = 3 \) → Point (0, 3)
3. **For \( x = 8 \)**:
- Vertical line at \( x = 8 \).
4. **For \( y = 9 \)**:
- Horizontal line at \( y = 9 \).
### Step 4: Graph the Constraints
We will graph the lines on the coordinate plane:
- The line \( x + y = 10 \) will intersect the axes at (10, 0) and (0, 10).
- The line \( x + y = 3 \) will intersect the axes at (3, 0) and (0, 3).
- The line \( x = 8 \) is a vertical line.
- The line \( y = 9 \) is a horizontal line.
### Step 5: Identify the Feasible Region
The feasible region is the area that satisfies all constraints. It is bounded by the lines we graphed. The region will be where:
- \( x + y \leq 10 \) (below the line)
- \( x + y \geq 3 \) (above the line)
- \( x \leq 8 \) (to the left of the line)
- \( y \leq 9 \) (below the line)
- \( x \geq 0 \) and \( y \geq 0 \) (in the first quadrant)
### Step 6: Find Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( x + y = 10 \) and \( y = 9 \):
- \( x + 9 = 10 \) → \( x = 1 \) → Point (1, 9)
2. Intersection of \( x + y = 10 \) and \( x = 8 \):
- \( 8 + y = 10 \) → \( y = 2 \) → Point (8, 2)
3. Intersection of \( x + y = 3 \) and \( y = 9 \):
- Not applicable since \( y = 9 \) is above \( y = 3 \).
4. Intersection of \( x + y = 3 \) and \( x = 8 \):
- Not applicable since \( x = 8 \) is above \( x + y = 3 \).
5. Intersection of \( x + y = 3 \) and \( y = 0 \):
- \( x + 0 = 3 \) → \( x = 3 \) → Point (3, 0)
6. Intersection of \( x + y = 3 \) and \( x = 0 \):
- \( 0 + y = 3 \) → \( y = 3 \) → Point (0, 3)
### Step 7: List of Corner Points
The corner points of the feasible region are:
- (1, 9)
- (8, 2)
- (3, 0)
- (0, 3)
### Step 8: Evaluate the Objective Function at Each Corner Point
Now we will evaluate \( Z = 4x + 3y - 7 \) at each corner point:
1. At (1, 9):
\[
Z = 4(1) + 3(9) - 7 = 4 + 27 - 7 = 24
\]
2. At (8, 2):
\[
Z = 4(8) + 3(2) - 7 = 32 + 6 - 7 = 31
\]
3. At (3, 0):
\[
Z = 4(3) + 3(0) - 7 = 12 + 0 - 7 = 5
\]
4. At (0, 3):
\[
Z = 4(0) + 3(3) - 7 = 0 + 9 - 7 = 2
\]
### Step 9: Determine Maximum and Minimum Values
From the evaluations:
- Maximum value of \( Z = 31 \) at point (8, 2).
- Minimum value of \( Z = 2 \) at point (0, 3).
### Final Solution
- **Maximum**: \( Z = 31 \) at \( (8, 2) \)
- **Minimum**: \( Z = 2 \) at \( (0, 3) \)
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