`Z = x + 2y`
`x + 2y ge 100, x - y le 0, 2x + y le 200, x, y ge 0`

Answer

To solve the linear programming problem, we need to maximize the objective function \( Z = x + 2y \) under the given constraints. Let's go through the solution step by step.

### Step 1: Identify the Constraints
The constraints given are:
1. \( x + 2y \geq 100 \)
2. \( x - y \leq 0 \) (which can be rewritten as \( x \leq y \))
3. \( 2x + y \leq 200 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines, we convert the inequalities into equations:
1. \( x + 2y = 100 \)
2. \( x - y = 0 \) (or \( x = y \))
3. \( 2x + y = 200 \)

### Step 3: Find Intercepts for Each Equation
1. For \( x + 2y = 100 \):
- When \( x = 0 \): \( 2y = 100 \) → \( y = 50 \) (Point: \( (0, 50) \))
- When \( y = 0 \): \( x = 100 \) (Point: \( (100, 0) \))

2. For \( x = y \):
- This line passes through the origin and has a slope of 1.

3. For \( 2x + y = 200 \):
- When \( x = 0 \): \( y = 200 \) (Point: \( (0, 200) \))
- When \( y = 0 \): \( 2x = 200 \) → \( x = 100 \) (Point: \( (100, 0) \))

### Step 4: Plot the Constraints
Now we will plot the lines on a graph:
- The line \( x + 2y = 100 \) intersects the axes at \( (100, 0) \) and \( (0, 50) \).
- The line \( x = y \) is a diagonal line through the origin.
- The line \( 2x + y = 200 \) intersects the axes at \( (100, 0) \) and \( (0, 200) \).

### Step 5: Determine the Feasible Region
The feasible region is where all the constraints overlap. We will check the area that satisfies all inequalities:
- The area above the line \( x + 2y = 100 \)
- The area below the line \( x = y \)
- The area below the line \( 2x + y = 200 \)
- The area in the first quadrant (where \( x \geq 0 \) and \( y \geq 0 \))

### Step 6: Identify Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( x + 2y = 100 \) and \( x = y \):
- Substitute \( y \) for \( x \): \( x + 2x = 100 \) → \( 3x = 100 \) → \( x = \frac{100}{3} \), \( y = \frac{100}{3} \) (Point: \( \left(\frac{100}{3}, \frac{100}{3}\right) \))

2. Intersection of \( x + 2y = 100 \) and \( 2x + y = 200 \):
- Solve the system:
- From \( x + 2y = 100 \): \( y = \frac{100 - x}{2} \)
- Substitute into \( 2x + y = 200 \):
- \( 2x + \frac{100 - x}{2} = 200 \)
- Multiply through by 2: \( 4x + 100 - x = 400 \)
- \( 3x = 300 \) → \( x = 100 \), \( y = 0 \) (Point: \( (100, 0) \))

3. Intersection of \( x = y \) and \( 2x + y = 200 \):
- Substitute \( y \) for \( x \): \( 2x + x = 200 \) → \( 3x = 200 \) → \( x = \frac{200}{3} \), \( y = \frac{200}{3} \) (Point: \( \left(\frac{200}{3}, \frac{200}{3}\right) \))

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we will evaluate \( Z = x + 2y \) at each corner point:
1. At \( (0, 50) \): \( Z = 0 + 2(50) = 100 \)
2. At \( (100, 0) \): \( Z = 100 + 2(0) = 100 \)
3. At \( \left(\frac{100}{3}, \frac{100}{3}\right) \):
- \( Z = \frac{100}{3} + 2\left(\frac{100}{3}\right) = \frac{100}{3} + \frac{200}{3} = 100 \)
4. At \( \left(\frac{200}{3}, \frac{200}{3}\right) \):
- \( Z = \frac{200}{3} + 2\left(\frac{200}{3}\right) = \frac{200}{3} + \frac{400}{3} = 200 \)

### Step 8: Determine Maximum and Minimum Values
- The maximum value of \( Z \) is \( 200 \) at the point \( \left(\frac{200}{3}, \frac{200}{3}\right) \).
- The minimum value of \( Z \) is \( 100 \) at the points \( (0, 50) \) and \( (100, 0) \).

### Final Solution
- Maximum value of \( Z \) is \( 200 \) at \( \left(\frac{200}{3}, \frac{200}{3}\right) \).
- Minimum value of \( Z \) is \( 100 \) at \( (0, 50) \) and \( (100, 0) \).