Solve the following Linear Programming Problems graphically:
Maximize (6-15):
OBJECTIVE FUNCTION CONSTRAINTS
6. `Z = 6x + 8y`
`x + y le 6, 3x + y ge 6, x - y ge 0, x ge 0, y ge 0`

Answer

To solve the given linear programming problem graphically, we will follow these steps:

### Step 1: Define the Objective Function and Constraints
The objective function is given as:
\[ Z = 6x + 8y \]

The constraints are:
1. \( x + y \leq 6 \)
2. \( 3x + y \geq 6 \)
3. \( x - y \geq 0 \) (which implies \( x \geq y \))
4. \( x \geq 0 \)
5. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
We will convert the inequalities into equations to find the boundary lines.

1. For \( x + y = 6 \):
- x-intercept: Set \( y = 0 \) → \( x = 6 \) → Point (6, 0)
- y-intercept: Set \( x = 0 \) → \( y = 6 \) → Point (0, 6)

2. For \( 3x + y = 6 \):
- x-intercept: Set \( y = 0 \) → \( 3x = 6 \) → \( x = 2 \) → Point (2, 0)
- y-intercept: Set \( x = 0 \) → \( y = 6 \) → Point (0, 6)

3. For \( x - y = 0 \) (or \( x = y \)):
- This line passes through the origin (0, 0) and has a slope of 1.

### Step 3: Plot the Lines on a Graph
Now we will plot the lines on a graph:

- The line \( x + y = 6 \) will connect points (6, 0) and (0, 6).
- The line \( 3x + y = 6 \) will connect points (2, 0) and (0, 6).
- The line \( x = y \) will be a diagonal line through the origin.

### Step 4: Determine the Feasible Region
Next, we will determine the feasible region by testing points in relation to the inequalities:

- For \( x + y \leq 6 \), the region below the line is feasible.
- For \( 3x + y \geq 6 \), the region above the line is feasible.
- For \( x \geq y \), the region above the line \( x = y \) is feasible.
- The non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \) restrict the feasible region to the first quadrant.

### Step 5: Identify Corner Points
The feasible region is bounded by the lines, and we need to find the corner points:

1. Intersection of \( x + y = 6 \) and \( x = y \):
- Substitute \( y = x \) into \( x + y = 6 \):
\[ 2x = 6 \Rightarrow x = 3, y = 3 \] → Point (3, 3)

2. Intersection of \( 3x + y = 6 \) and \( x = y \):
- Substitute \( y = x \) into \( 3x + y = 6 \):
\[ 3x + x = 6 \Rightarrow 4x = 6 \Rightarrow x = \frac{3}{2}, y = \frac{3}{2} \] → Point \(\left(\frac{3}{2}, \frac{3}{2}\right)\)

3. Intersection of \( x + y = 6 \) and \( 3x + y = 6 \):
- Solve the system:
\[ x + y = 6 \quad (1) \\ 3x + y = 6 \quad (2) \]
Subtract (1) from (2):
\[ 2x = 0 \Rightarrow x = 0 \]
Substitute \( x = 0 \) into (1):
\[ y = 6 \] → Point (0, 6)

4. Intersection of \( x + y = 6 \) and \( x = 0 \):
- Substitute \( x = 0 \) into \( x + y = 6 \):
\[ y = 6 \] → Point (0, 6)

5. Intersection of \( 3x + y = 6 \) and \( x = 0 \):
- Substitute \( x = 0 \) into \( 3x + y = 6 \):
\[ y = 6 \] → Point (0, 6)

### Step 6: Evaluate the Objective Function at Each Corner Point
Now we will evaluate \( Z = 6x + 8y \) at each corner point:

1. At (2, 0):
\[ Z = 6(2) + 8(0) = 12 \]

2. At (6, 0):
\[ Z = 6(6) + 8(0) = 36 \]

3. At \(\left(\frac{3}{2}, \frac{3}{2}\right)\):
\[ Z = 6\left(\frac{3}{2}\right) + 8\left(\frac{3}{2}\right) = 9 + 12 = 21 \]

4. At (3, 3):
\[ Z = 6(3) + 8(3) = 18 + 24 = 42 \]

### Step 7: Identify the Maximum Value
The maximum value of \( Z \) occurs at the point (3, 3):
\[ Z_{\text{max}} = 42 \]

### Conclusion
The maximum value of the objective function \( Z = 6x + 8y \) is 42 at the point (3, 3).

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