To maximize the function \( Z = 8x + 9y \) subject to the given constraints, we will follow these steps:
### Step 1: Identify the Constraints
The constraints given are:
1. \( 2x + 3y \leq 6 \)
2. \( 3x - 2y \leq 6 \)
3. \( y \leq 1 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)
### Step 2: Convert Inequalities to Equations
To graph the constraints, we convert the inequalities into equations:
1. \( 2x + 3y = 6 \)
2. \( 3x - 2y = 6 \)
3. \( y = 1 \)
4. \( x = 0 \) (y-axis)
5. \( y = 0 \) (x-axis)
### Step 3: Graph the Constraints
We will plot the lines corresponding to the equations on a graph and identify the feasible region.
1. For \( 2x + 3y = 6 \):
- When \( x = 0 \), \( y = 2 \) (point (0, 2))
- When \( y = 0 \), \( x = 3 \) (point (3, 0))
2. For \( 3x - 2y = 6 \):
- When \( x = 0 \), \( y = -3 \) (not in the feasible region)
- When \( y = 0 \), \( x = 2 \) (point (2, 0))
- When \( x = 2 \), \( y = 0 \) gives us the point (2, 0).
3. For \( y = 1 \):
- This is a horizontal line at \( y = 1 \).
### Step 4: Identify the Feasible Region
The feasible region is bounded by the lines we plotted and is the area where all constraints are satisfied. The vertices of this region can be found by solving the equations of the lines.
### Step 5: Find the Intersection Points
1. **Intersection of \( 2x + 3y = 6 \) and \( y = 1 \)**:
\[
2x + 3(1) = 6 \implies 2x + 3 = 6 \implies 2x = 3 \implies x = \frac{3}{2}
\]
Point B: \( \left(\frac{3}{2}, 1\right) \)
2. **Intersection of \( 2x + 3y = 6 \) and \( 3x - 2y = 6 \)**:
Multiply the first equation by 2:
\[
4x + 6y = 12
\]
Multiply the second equation by 3:
\[
9x - 6y = 18
\]
Adding these:
\[
13x = 30 \implies x = \frac{30}{13}
\]
Substitute \( x \) back to find \( y \):
\[
2\left(\frac{30}{13}\right) + 3y = 6 \implies 3y = 6 - \frac{60}{13} = \frac{78 - 60}{13} = \frac{18}{13} \implies y = \frac{6}{13}
\]
Point C: \( \left(\frac{30}{13}, \frac{6}{13}\right) \)
3. **Intersection of \( 3x - 2y = 6 \) and \( y = 0 \)**:
\[
3x = 6 \implies x = 2
\]
Point D: \( (2, 0) \)
4. **Intersection of \( y = 0 \) and \( x = 0 \)**:
Point E: \( (0, 0) \)
### Step 6: Evaluate Z at Each Vertex
Now we evaluate \( Z = 8x + 9y \) at each vertex:
1. At \( (0, 1) \):
\[
Z = 8(0) + 9(1) = 9
\]
2. At \( \left(\frac{3}{2}, 1\right) \):
\[
Z = 8\left(\frac{3}{2}\right) + 9(1) = 12 + 9 = 21
\]
3. At \( \left(\frac{30}{13}, \frac{6}{13}\right) \):
\[
Z = 8\left(\frac{30}{13}\right) + 9\left(\frac{6}{13}\right) = \frac{240}{13} + \frac{54}{13} = \frac{294}{13} \approx 22.615
\]
4. At \( (2, 0) \):
\[
Z = 8(2) + 9(0) = 16
\]
5. At \( (0, 0) \):
\[
Z = 8(0) + 9(0) = 0
\]
### Step 7: Determine Maximum Value
The maximum value of \( Z \) occurs at the point \( \left(\frac{3}{2}, 1\right) \) with \( Z = 21 \).
### Final Answer
The maximum value of \( Z \) is \( 21 \) at the point \( \left(\frac{3}{2}, 1\right) \).
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