`Z = 2x + 3y`
`x ge 0, y ge 0, 1 le x + 2y le10`

Answer

To solve the linear programming problem given by the objective function \( Z = 2x + 3y \) with the constraints \( x \geq 0 \), \( y \geq 0 \), and \( 1 \leq x + 2y \leq 10 \), we will follow these steps:

### Step 1: Identify the Constraints
The constraints can be expressed as:
1. \( x + 2y \geq 1 \)
2. \( x + 2y \leq 10 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Constraints to Equations
To find the boundary lines, we convert the inequalities into equations:
1. \( x + 2y = 1 \)
2. \( x + 2y = 10 \)

### Step 3: Find Intercepts
For the equation \( x + 2y = 1 \):
- When \( x = 0 \): \( 2y = 1 \) → \( y = \frac{1}{2} \) (y-intercept)
- When \( y = 0 \): \( x = 1 \) (x-intercept)

For the equation \( x + 2y = 10 \):
- When \( x = 0 \): \( 2y = 10 \) → \( y = 5 \) (y-intercept)
- When \( y = 0 \): \( x = 10 \) (x-intercept)

### Step 4: Plot the Lines
Now we plot the lines on a graph:
- The line \( x + 2y = 1 \) intersects the axes at \( (1, 0) \) and \( (0, \frac{1}{2}) \).
- The line \( x + 2y = 10 \) intersects the axes at \( (10, 0) \) and \( (0, 5) \).

### Step 5: Identify the Feasible Region
The feasible region is bounded by the lines and the axes. It is the area where all constraints are satisfied.

### Step 6: Determine Corner Points
The corner points of the feasible region are:
1. \( (0, \frac{1}{2}) \)
2. \( (1, 0) \)
3. \( (10, 0) \)
4. \( (0, 5) \)

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we will substitute these points into the objective function \( Z = 2x + 3y \):

1. For \( (0, \frac{1}{2}) \):
\[ Z = 2(0) + 3\left(\frac{1}{2}\right) = 0 + \frac{3}{2} = 1.5 \]

2. For \( (1, 0) \):
\[ Z = 2(1) + 3(0) = 2 + 0 = 2 \]

3. For \( (10, 0) \):
\[ Z = 2(10) + 3(0) = 20 + 0 = 20 \]

4. For \( (0, 5) \):
\[ Z = 2(0) + 3(5) = 0 + 15 = 15 \]

### Step 8: Identify Maximum and Minimum Values
From the calculations:
- Minimum value of \( Z \) is \( 1.5 \) at point \( (0, \frac{1}{2}) \).
- Maximum value of \( Z \) is \( 20 \) at point \( (10, 0) \).

### Final Solution
- Minimum value of \( Z \) is \( 1.5 \) at \( (0, \frac{1}{2}) \).
- Maximum value of \( Z \) is \( 20 \) at \( (10, 0) \).