To solve the linear programming problem of maximizing \( Z = x + 2y \) subject to the constraints:
1. \( x - y \leq 10 \)
2. \( 2x + 3y \leq 20 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)
we will follow these steps:
### Step 1: Graph the Constraints
First, we will graph the constraints to find the feasible region.
**Constraint 1:** \( x - y \leq 10 \)
To graph this, we can rearrange it to \( y \geq x - 10 \). The line \( y = x - 10 \) has a y-intercept of -10 and an x-intercept of 10.
**Constraint 2:** \( 2x + 3y \leq 20 \)
Rearranging gives us \( 3y \leq 20 - 2x \) or \( y \leq \frac{20 - 2x}{3} \). The y-intercept is \( \frac{20}{3} \) and the x-intercept is 10.
### Step 2: Identify the Feasible Region
Next, we plot the lines on a graph and shade the appropriate regions based on the inequalities. The feasible region will be where all shaded areas overlap, considering that \( x \geq 0 \) and \( y \geq 0 \).
### Step 3: Find the Corner Points
The maximum value of \( Z \) will occur at one of the corner points of the feasible region. We need to find the intersection points of the lines.
1. **Intersection of \( x - y = 10 \) and \( 2x + 3y = 20 \)**:
- From \( x - y = 10 \), we have \( y = x - 10 \).
- Substitute into \( 2x + 3(x - 10) = 20 \):
\[
2x + 3x - 30 = 20 \implies 5x = 50 \implies x = 10
\]
Then, \( y = 10 - 10 = 0 \). So, one corner point is \( (10, 0) \).
2. **Intersection of \( x - y = 10 \) and \( y = 0 \)**:
- Setting \( y = 0 \) in \( x - y = 10 \) gives \( x = 10 \). So, we have \( (10, 0) \).
3. **Intersection of \( 2x + 3y = 20 \) and \( y = 0 \)**:
- Setting \( y = 0 \) in \( 2x + 3y = 20 \) gives \( 2x = 20 \implies x = 10 \). So, we have \( (10, 0) \).
4. **Intersection of \( 2x + 3y = 20 \) and \( x = 0 \)**:
- Setting \( x = 0 \) gives \( 3y = 20 \implies y = \frac{20}{3} \). So, we have \( (0, \frac{20}{3}) \).
5. **Intersection of \( x - y = 10 \) and \( y = 0 \)**:
- This gives \( (10, 0) \) again.
### Step 4: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( Z = x + 2y \) at each corner point:
1. At \( (10, 0) \):
\[
Z = 10 + 2(0) = 10
\]
2. At \( (0, \frac{20}{3}) \):
\[
Z = 0 + 2 \left(\frac{20}{3}\right) = \frac{40}{3} \approx 13.33
\]
### Step 5: Determine the Maximum Value
Comparing the values of \( Z \):
- At \( (10, 0) \), \( Z = 10 \)
- At \( (0, \frac{20}{3}) \), \( Z = \frac{40}{3} \approx 13.33 \)
The maximum value of \( Z \) occurs at the point \( (0, \frac{20}{3}) \) with \( Z = \frac{40}{3} \).
### Final Answer
The maximum value of \( Z \) is \( \frac{40}{3} \).
