Solve the following linear programming problem graphically:
minimize : `z=x+2y`
Subject to: `2x+yge3`
`x+2yge6`
`xge0`
`yge0`
Show that `z` is minimum at two points.

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 minimize the objective function:
\[ z = x + 2y \]

Subject to the constraints:
1. \( 2x + y \geq 3 \) (Constraint 1)
2. \( x + 2y \geq 6 \) (Constraint 2)
3. \( x \geq 0 \) (Non-negativity constraint)
4. \( y \geq 0 \) (Non-negativity constraint)

### Step 2: Convert inequalities to equations
To graph the constraints, we first convert the inequalities into equations:
1. From \( 2x + y = 3 \)
2. From \( x + 2y = 6 \)

### Step 3: Find intercepts for each equation
**For the first equation \( 2x + y = 3 \):**
- When \( x = 0 \):
\[ y = 3 \] (Point: \( (0, 3) \))
- When \( y = 0 \):
\[ 2x = 3 \Rightarrow x = \frac{3}{2} = 1.5 \] (Point: \( (1.5, 0) \))

**For the second equation \( x + 2y = 6 \):**
- When \( x = 0 \):
\[ 2y = 6 \Rightarrow y = 3 \] (Point: \( (0, 3) \))
- When \( y = 0 \):
\[ x = 6 \] (Point: \( (6, 0) \))

### Step 4: Plot the lines on a graph
Now we will plot the points obtained:
- For \( 2x + y = 3 \): Points \( (0, 3) \) and \( (1.5, 0) \)
- For \( x + 2y = 6 \): Points \( (0, 3) \) and \( (6, 0) \)

### Step 5: Determine the feasible region
Next, we shade the feasible region based on the inequalities:
- For \( 2x + y \geq 3 \), shade above the line.
- For \( x + 2y \geq 6 \), shade above the line.
- The feasible region is where the shaded areas overlap, considering \( x \geq 0 \) and \( y \geq 0 \).

### Step 6: Identify corner points of the feasible region
The corner points of the feasible region are:
1. \( (0, 3) \)
2. \( (1.5, 0) \)
3. \( (6, 0) \)

### Step 7: Evaluate the objective function at each corner point
Now we calculate the value of \( z \) at each corner point:
1. At \( (0, 3) \):
\[ z = 0 + 2(3) = 6 \]
2. At \( (1.5, 0) \):
\[ z = 1.5 + 2(0) = 1.5 \]
3. At \( (6, 0) \):
\[ z = 6 + 2(0) = 6 \]

### Step 8: Determine the minimum value of \( z \)
From the evaluations:
- \( z = 6 \) at points \( (0, 3) \) and \( (6, 0) \)
- \( z = 1.5 \) at point \( (1.5, 0) \)

The minimum value of \( z \) is \( 6 \), which occurs at two points: \( (0, 3) \) and \( (6, 0) \).

### Conclusion
Thus, the minimum value of \( z \) is \( 6 \) at two points: \( (0, 3) \) and \( (6, 0) \).

---