By graphical method, the solutions of linear programming problem maximise `Z=3x_(1)+5x_(2)` subject to constraints `3x_(1)+2x_(2) le 18, x_(1) le 4, x_(2) le 6 x_(1) ge0,x_(2) ge 0 ` are

Options

A. `x_(1)=2,x_(2)=0,Z=6`
B. `x_(1)=2,x_(2)=6,Z=36`
C. `x_(1)=4,x_(2)=3,Z=27`
D. `x_(1)=4,x_(2)=6,Z=42`

Answer

To solve the linear programming problem using the graphical method, we will follow these steps:

### Step 1: Define the Objective Function and Constraints
We want to maximize the objective function:
\[ Z = 3x_1 + 5x_2 \]

Subject to the following constraints:
1. \( 3x_1 + 2x_2 \leq 18 \)
2. \( x_1 \leq 4 \)
3. \( x_2 \leq 6 \)
4. \( x_1 \geq 0 \)
5. \( x_2 \geq 0 \)

### Step 2: Convert Constraints to Equations
To graph the constraints, we convert the inequalities into equations:
1. \( 3x_1 + 2x_2 = 18 \)
2. \( x_1 = 4 \)
3. \( x_2 = 6 \)

### Step 3: Find Intercepts for Each Constraint
- For \( 3x_1 + 2x_2 = 18 \):
- When \( x_1 = 0 \): \( 2x_2 = 18 \) → \( x_2 = 9 \) (y-intercept)
- When \( x_2 = 0 \): \( 3x_1 = 18 \) → \( x_1 = 6 \) (x-intercept)

- For \( x_1 = 4 \):
- This is a vertical line at \( x_1 = 4 \).

- For \( x_2 = 6 \):
- This is a horizontal line at \( x_2 = 6 \).

### Step 4: Graph the Constraints
Plot the lines on a graph:
- Draw the line for \( 3x_1 + 2x_2 = 18 \) between the points (6, 0) and (0, 9).
- Draw the vertical line at \( x_1 = 4 \).
- Draw the horizontal line at \( x_2 = 6 \).

### Step 5: Identify the Feasible Region
The feasible region is where all the constraints overlap. This will be the area bounded by the lines you just drew, including the axes.

### Step 6: Determine the Corner Points
Identify the corner points of the feasible region:
1. Intersection of \( 3x_1 + 2x_2 = 18 \) and \( x_1 = 4 \):
- Substitute \( x_1 = 4 \) into \( 3(4) + 2x_2 = 18 \) → \( 12 + 2x_2 = 18 \) → \( 2x_2 = 6 \) → \( x_2 = 3 \) → Point (4, 3)

2. Intersection of \( 3x_1 + 2x_2 = 18 \) and \( x_2 = 6 \):
- Substitute \( x_2 = 6 \) into \( 3x_1 + 2(6) = 18 \) → \( 3x_1 + 12 = 18 \) → \( 3x_1 = 6 \) → \( x_1 = 2 \) → Point (2, 6)

3. Intersection of \( x_1 = 4 \) and \( x_2 = 6 \):
- Point (4, 6)

4. Intersection at the origin (0, 0).

### Step 7: Evaluate the Objective Function at Each Corner Point
Calculate \( Z \) at each corner point:
1. At (0, 0): \( Z = 3(0) + 5(0) = 0 \)
2. At (4, 0): \( Z = 3(4) + 5(0) = 12 \)
3. At (4, 3): \( Z = 3(4) + 5(3) = 12 + 15 = 27 \)
4. At (2, 6): \( Z = 3(2) + 5(6) = 6 + 30 = 36 \)
5. At (0, 6): \( Z = 3(0) + 5(6) = 30 \)

### Step 8: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point (2, 6) with \( Z = 36 \).

### Final Solution
The optimal solution is:
- \( x_1 = 2 \)
- \( x_2 = 6 \)
- Maximum value of \( Z = 36 \)

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