Solve the following linear programming model graphically:
Maximize
Z 14X 10Y
Subject to:
8X  6Y  480 4X  2Y  200 X ,Y  0


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

### Problem Statement

Maximize:  
\[ Z = 14X + 10Y \]

Subject to:  
1. \( 8X + 6Y \leq 480 \)
2. \( 4X + 2Y \leq 200 \)
3. \( X, Y \geq 0 \)

### Step 1: Convert the inequalities into equations
Convert the inequalities into equalities to find the boundary lines:
1. \( 8X + 6Y = 480 \)
2. \( 4X + 2Y = 200 \)

### Step 2: Find the intercepts for the boundary lines

For the first equation \( 8X + 6Y = 480 \):
- Set \( X = 0 \):  
  \( 6Y = 480 \)  
  \( Y = 80 \)  
- Set \( Y = 0 \):  
  \( 8X = 480 \)  
  \( X = 60 \)

So, the intercepts for the first line are \( (60, 0) \) and \( (0, 80) \).

For the second equation \( 4X + 2Y = 200 \):
- Set \( X = 0 \):  
  \( 2Y = 200 \)  
  \( Y = 100 \)  
- Set \( Y = 0 \):  
  \( 4X = 200 \)  
  \( X = 50 \)

So, the intercepts for the second line are \( (50, 0) \) and \( (0, 100) \).

### Step 3: Plot the lines and identify the feasible region
Plot the boundary lines on the graph using the intercepts calculated. The feasible region is where all the constraints overlap.

### Step 4: Identify the corner points
The feasible region's corner points are:
- \( (0, 0) \)
- \( (0, 80) \) (from the first constraint)
- \( (50, 0) \) (from the second constraint)
- The intersection point of \( 8X + 6Y = 480 \) and \( 4X + 2Y = 200 \)

### Step 5: Calculate the intersection point
To find the intersection, solve the system of equations:
1. \( 8X + 6Y = 480 \)
2. \( 4X + 2Y = 200 \)

You can simplify the second equation by dividing by 2:
\[ 2X + Y = 100 \]

Now substitute \( Y = 100 - 2X \) into the first equation:
\[ 8X + 6(100 - 2X) = 480 \]
\[ 8X + 600 - 12X = 480 \]
\[ -4X = -120 \]
\[ X = 30 \]

Substituting \( X = 30 \) back into the simplified equation:
\[ Y = 100 - 2(30) = 40 \]

So the intersection point is \( (30, 40) \).

### Step 6: Calculate Z at each corner point
- At \( (0, 0) \):  
  \( Z = 14(0) + 10(0) = 0 \)
- At \( (0, 80) \):  
  \( Z = 14(0) + 10(80) = 800 \)
- At \( (50, 0) \):  
  \( Z = 14(50) + 10(0) = 700 \)
- At \( (30, 40) \):  
  \( Z = 14(30) + 10(40) = 420 + 400 = 820 \)

### Step 7: Determine the optimal solution
The maximum value of \( Z \) occurs at the point \( (30, 40) \) with \( Z = 820 \).

### Final Answer
The optimal solution is:
\[ X = 30, Y = 40 \]
\[ Z = 820 \]

Would you like more details on any step? 

Here are some related questions you might find useful:
1. What are the benefits of using the graphical method in linear programming?
2. How would the solution change if the objective function was to minimize instead of maximize?
3. What are the limitations of the graphical method for solving linear programming problems?
4. How would you solve this problem using the Simplex method?
5. Can this problem be solved using software like Excel or MATLAB?
6. How do you interpret the feasible region in the context of a linear programming problem?
7. What is the significance of the intersection point in linear programming problems?
8. How do changes in the constraints affect the feasible region and the optimal solution?

**Tip:** Always ensure that the feasible region is correctly identified, as the optimal solution must lie within this region.

Learn how to solve a linear programming model graphically to maximize Z = 14X + 10Y. This problem covers key concepts in linear programming and the graphical method. Suitable for undergraduate students and those interested in operations research.

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