Solve the following linear programming problem graphically:
Maximize :`z=x+2y`
Subject to: `x-yle2`
`x+yle4`
`xge0`
`yge0`

Explanation

To solve the given linear programming problem graphically, we will follow these steps:

### Step 1: Define the Objective Function and Constraints
We need to maximize the objective function:
\[ z = x + 2y \]
Subject to the constraints:
1. \( x - y \leq 2 \)
2. \( x + y \leq 4 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To graph the constraints, we first convert the inequalities into equations:
1. \( x - y = 2 \)
2. \( x + y = 4 \)

### Step 3: Find Intercepts of the Constraints
**For the first constraint \( x - y = 2 \):**
- When \( x = 0 \):
\[ 0 - y = 2 \Rightarrow y = -2 \] (not in the feasible region)
- When \( y = 0 \):
\[ x - 0 = 2 \Rightarrow x = 2 \]
- Thus, the intercepts are \( (2, 0) \).

**For the second constraint \( x + y = 4 \):**
- When \( x = 0 \):
\[ 0 + y = 4 \Rightarrow y = 4 \]
- When \( y = 0 \):
\[ x + 0 = 4 \Rightarrow x = 4 \]
- Thus, the intercepts are \( (0, 4) \) and \( (4, 0) \).

### Step 4: Graph the Constraints
Now we will graph the lines:
1. The line \( x - y = 2 \) passes through \( (2, 0) \) and has a slope of 1, so it goes up to the left.
2. The line \( x + y = 4 \) passes through \( (0, 4) \) and \( (4, 0) \) and has a slope of -1.

### Step 5: Identify the Feasible Region
The feasible region is determined by the inequalities:
- For \( x - y \leq 2 \), we shade below the line.
- For \( x + y \leq 4 \), we shade below this line as well.
- Since \( x \geq 0 \) and \( y \geq 0 \), we only consider the first quadrant.

### Step 6: Find the Corner Points of the Feasible Region
The corner points of the feasible region are:
1. \( (0, 0) \)
2. \( (2, 0) \)
3. \( (0, 4) \)
4. The intersection of the lines \( x - y = 2 \) and \( x + y = 4 \).

To find the intersection:
- Set \( x - y = 2 \) and \( x + y = 4 \).
- From \( x + y = 4 \), we can express \( y = 4 - x \).
- Substitute into \( x - (4 - x) = 2 \):
\[ x - 4 + x = 2 \]
\[ 2x - 4 = 2 \]
\[ 2x = 6 \]
\[ x = 3 \]
- Substitute \( x = 3 \) back into \( y = 4 - x \):
\[ y = 4 - 3 = 1 \]
- Thus, the intersection point is \( (3, 1) \).

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( z = x + 2y \) at each corner point:
1. At \( (0, 0) \):
\[ z = 0 + 2(0) = 0 \]
2. At \( (2, 0) \):
\[ z = 2 + 2(0) = 2 \]
3. At \( (0, 4) \):
\[ z = 0 + 2(4) = 8 \]
4. At \( (3, 1) \):
\[ z = 3 + 2(1) = 5 \]

### Step 8: Determine the Maximum Value
The maximum value of \( z \) occurs at the point \( (0, 4) \) where:
\[ z = 8 \]

### Final Solution
The maximum value of \( z \) is \( 8 \) at the point \( (0, 4) \).

---