To solve the given Linear Programming Problem (LPP) graphically, we will follow these steps:
### Step 1: Formulate the Objective Function and Constraints
We need to minimize and maximize the objective function:
\[ z = 3x + 9y \]
Subject to the constraints:
1. \( x + 3y \leq 60 \)
2. \( x + y \geq 10 \)
3. \( x \leq y \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)
### Step 2: Convert Inequalities to Equations
To graph the constraints, we convert the inequalities into equations:
1. \( x + 3y = 60 \)
2. \( x + y = 10 \)
3. \( x = y \)
### Step 3: Plot the Lines on the Graph
We will plot each line on the XY-plane.
1. **For \( x + 3y = 60 \)**:
- When \( x = 0 \): \( 3y = 60 \) → \( y = 20 \) (Point: \( (0, 20) \))
- When \( y = 0 \): \( x = 60 \) (Point: \( (60, 0) \))
- Plot the line through points \( (0, 20) \) and \( (60, 0) \).
2. **For \( x + y = 10 \)**:
- When \( x = 0 \): \( y = 10 \) (Point: \( (0, 10) \))
- When \( y = 0 \): \( x = 10 \) (Point: \( (10, 0) \))
- Plot the line through points \( (0, 10) \) and \( (10, 0) \).
3. **For \( x = y \)**:
- This is a diagonal line through the origin at a 45-degree angle.
### Step 4: Determine the Feasible Region
To find the feasible region, we need to test points in relation to the inequalities:
- For \( x + 3y \leq 60 \): Test point \( (0, 0) \) → \( 0 + 3(0) = 0 \leq 60 \) (satisfied).
- For \( x + y \geq 10 \): Test point \( (0, 0) \) → \( 0 + 0 = 0 < 10 \) (not satisfied, so the region is above this line).
- For \( x \leq y \): Test point \( (1, 0) \) → \( 1 \not\leq 0 \) (not satisfied, so the region is above the line \( x = y \)).
The feasible region is bounded by the lines and lies in the first quadrant.
### Step 5: Identify Corner Points of the Feasible Region
The corner points of the feasible region can be found by solving the intersections of the lines:
1. **Intersection of \( x + 3y = 60 \) and \( x = y \)**:
\[
x + 3x = 60 \implies 4x = 60 \implies x = 15 \implies y = 15 \quad \text{(Point B: \( (15, 15) \))}
\]
2. **Intersection of \( x + y = 10 \) and \( x = y \)**:
\[
x + x = 10 \implies 2x = 10 \implies x = 5 \implies y = 5 \quad \text{(Point C: \( (5, 5) \))}
\]
3. **Intersection of \( x + 3y = 60 \) and \( x + y = 10 \)**:
\[
x + 3(10 - x) = 60 \implies x + 30 - 3x = 60 \implies -2x = 30 \implies x = -15 \quad \text{(not in the feasible region)}
\]
4. **Intersection of \( x + 3y = 60 \) and \( y = 0 \)**:
\[
x + 3(0) = 60 \implies x = 60 \quad \text{(Point D: \( (60, 0) \))}
\]
5. **Intersection of \( x + y = 10 \) and \( y = 0 \)**:
\[
x + 0 = 10 \implies x = 10 \quad \text{(Point A: \( (10, 0) \))}
\]
### Step 6: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( z = 3x + 9y \) at each corner point:
- Point A \( (10, 0) \): \( z = 3(10) + 9(0) = 30 \)
- Point B \( (15, 15) \): \( z = 3(15) + 9(15) = 180 \)
- Point C \( (5, 5) \): \( z = 3(5) + 9(5) = 60 \)
- Point D \( (60, 0) \): \( z = 3(60) + 9(0) = 180 \)
### Step 7: Identify Maximum and Minimum Values
- **Minimum value of \( z \)**: \( 30 \) at point \( (10, 0) \)
- **Maximum value of \( z \)**: \( 180 \) at points \( (15, 15) \) and \( (60, 0) \)
### Conclusion
- **Minimum**: \( z = 30 \) at \( (10, 0) \)
- **Maximum**: \( z = 180 \) at \( (15, 15) \) and \( (60, 0) \)
