To solve the given Linear Programming Problem (LPP) graphically, we will follow these steps:
### Step 1: Identify the Objective Function and Constraints
We need to maximize the objective function:
\[ Z = 4x + y \]
Subject to the constraints:
1. \( x + y \leq 50 \)
2. \( 3x + y \leq 90 \)
3. \( x \geq 10 \)
4. \( y \geq 0 \)
### Step 2: Convert Inequalities to Equations
To graph the constraints, we first convert the inequalities into equations:
1. \( x + y = 50 \)
2. \( 3x + y = 90 \)
3. \( x = 10 \)
4. \( y = 0 \)
### Step 3: Find Intercepts for Each Constraint
Next, we find the intercepts for each equation:
1. For \( x + y = 50 \):
- x-intercept: Set \( y = 0 \) → \( x = 50 \) (point: (50, 0))
- y-intercept: Set \( x = 0 \) → \( y = 50 \) (point: (0, 50))
2. For \( 3x + y = 90 \):
- x-intercept: Set \( y = 0 \) → \( 3x = 90 \) → \( x = 30 \) (point: (30, 0))
- y-intercept: Set \( x = 0 \) → \( y = 90 \) (point: (0, 90))
3. For \( x = 10 \): This is a vertical line at \( x = 10 \).
4. For \( y = 0 \): This is the x-axis.
### Step 4: Plot the Constraints on a Graph
Now, we plot the points and the lines on a graph:
- Draw the line for \( x + y = 50 \) connecting (50, 0) and (0, 50).
- Draw the line for \( 3x + y = 90 \) connecting (30, 0) and (0, 90).
- Draw the vertical line for \( x = 10 \).
- Draw the horizontal line for \( y = 0 \).
### Step 5: Determine the Feasible Region
The feasible region is the area that satisfies all constraints. This region is bounded by the lines we plotted and is located in the first quadrant (since \( y \geq 0 \) and \( x \geq 10 \)).
### Step 6: Identify the Corner Points of the Feasible Region
The corner points of the feasible region are where the lines intersect:
1. Intersection of \( x + y = 50 \) and \( x = 10 \):
- \( 10 + y = 50 \) → \( y = 40 \) (point: (10, 40))
2. Intersection of \( 3x + y = 90 \) and \( x = 10 \):
- \( 3(10) + y = 90 \) → \( y = 60 \) (point: (10, 60) - not in feasible region)
3. Intersection of \( x + y = 50 \) and \( 3x + y = 90 \):
- Subtract the first from the second:
- \( 3x + y - (x + y) = 90 - 50 \)
- \( 2x = 40 \) → \( x = 20 \)
- Substitute \( x = 20 \) into \( x + y = 50 \):
- \( 20 + y = 50 \) → \( y = 30 \) (point: (20, 30))
4. Intersection of \( x + y = 50 \) and \( y = 0 \):
- \( x + 0 = 50 \) → \( x = 50 \) (point: (50, 0))
5. Intersection of \( 3x + y = 90 \) and \( y = 0 \):
- \( 3x + 0 = 90 \) → \( x = 30 \) (point: (30, 0))
### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( Z = 4x + y \) at each corner point:
1. At (10, 40): \( Z = 4(10) + 40 = 40 + 40 = 80 \)
2. At (20, 30): \( Z = 4(20) + 30 = 80 + 30 = 110 \)
3. At (30, 0): \( Z = 4(30) + 0 = 120 + 0 = 120 \)
4. At (50, 0): \( Z = 4(50) + 0 = 200 + 0 = 200 \) (not feasible)
### Step 8: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point (30, 0):
- Maximum \( Z = 120 \)
### Conclusion
The optimal solution is:
- \( x = 30 \)
- \( y = 0 \)
- Maximum value of \( Z = 120 \)
