To solve the given linear programming problem, we need to maximize the objective function \( Z = 3x + 5y \) subject to the constraints:
1. \( x + 2y \geq 10 \)
2. \( x + y \geq 6 \)
3. \( 3x + y \geq 8 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)
### Step-by-Step Solution
**Step 1: Convert inequalities to equations**
We will first convert the inequalities into equations to find the boundary lines.
1. \( x + 2y = 10 \)
2. \( x + y = 6 \)
3. \( 3x + y = 8 \)
**Step 2: Find intercepts for each equation**
- For \( x + 2y = 10 \):
- When \( x = 0 \): \( 2y = 10 \) → \( y = 5 \) (Point: \( (0, 5) \))
- When \( y = 0 \): \( x = 10 \) (Point: \( (10, 0) \))
- For \( x + y = 6 \):
- When \( x = 0 \): \( y = 6 \) (Point: \( (0, 6) \))
- When \( y = 0 \): \( x = 6 \) (Point: \( (6, 0) \))
- For \( 3x + y = 8 \):
- When \( x = 0 \): \( y = 8 \) (Point: \( (0, 8) \))
- When \( y = 0 \): \( 3x = 8 \) → \( x = \frac{8}{3} \approx 2.67 \) (Point: \( \left(\frac{8}{3}, 0\right) \))
**Step 3: Graph the constraints**
Now, we will graph the lines based on the intercepts we calculated.
1. The line \( x + 2y = 10 \) intersects at \( (0, 5) \) and \( (10, 0) \).
2. The line \( x + y = 6 \) intersects at \( (0, 6) \) and \( (6, 0) \).
3. The line \( 3x + y = 8 \) intersects at \( (0, 8) \) and \( \left(\frac{8}{3}, 0\right) \).
**Step 4: Identify the feasible region**
The feasible region is the area where all the constraints overlap. Since all inequalities are of the form "greater than or equal to," the feasible region will be above the lines.
**Step 5: Find the corner points of the feasible region**
To find the corner points, we need to solve the equations pairwise:
1. Intersection of \( x + 2y = 10 \) and \( x + y = 6 \):
- From \( x + y = 6 \), we can express \( x = 6 - y \).
- Substitute into \( x + 2y = 10 \):
\[
(6 - y) + 2y = 10 \Rightarrow 6 + y = 10 \Rightarrow y = 4
\]
Then, \( x = 6 - 4 = 2 \) (Point: \( (2, 4) \))
2. Intersection of \( x + 2y = 10 \) and \( 3x + y = 8 \):
- From \( x + 2y = 10 \), express \( y = \frac{10 - x}{2} \).
- Substitute into \( 3x + y = 8 \):
\[
3x + \frac{10 - x}{2} = 8 \Rightarrow 6x + 10 - x = 16 \Rightarrow 5x = 6 \Rightarrow x = \frac{6}{5}
\]
Then, \( y = \frac{10 - \frac{6}{5}}{2} = \frac{10 - 1.2}{2} = \frac{8.8}{2} = 4.4 \) (Point: \( \left(\frac{6}{5}, 4.4\right) \))
3. Intersection of \( x + y = 6 \) and \( 3x + y = 8 \):
- From \( x + y = 6 \), express \( y = 6 - x \).
- Substitute into \( 3x + y = 8 \):
\[
3x + (6 - x) = 8 \Rightarrow 2x + 6 = 8 \Rightarrow 2x = 2 \Rightarrow x = 1
\]
Then, \( y = 6 - 1 = 5 \) (Point: \( (1, 5) \))
**Step 6: Evaluate the objective function at each corner point**
Now we evaluate \( Z = 3x + 5y \) at each corner point:
1. At \( (2, 4) \):
\[
Z = 3(2) + 5(4) = 6 + 20 = 26
\]
2. At \( \left(\frac{6}{5}, 4.4\right) \):
\[
Z = 3\left(\frac{6}{5}\right) + 5(4.4) = \frac{18}{5} + 22 = \frac{18 + 110}{5} = \frac{128}{5} = 25.6
\]
3. At \( (1, 5) \):
\[
Z = 3(1) + 5(5) = 3 + 25 = 28
\]
**Step 7: Determine the maximum value**
The maximum value of \( Z \) occurs at the point \( (1, 5) \) with \( Z = 28 \).
### Final Answer
The maximum value of \( Z \) is **28** at the point **(1, 5)**.
