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This is a linear programming (LP) problem where the goal is to **maximize** the objective function \( Z = 20X_1 + 25X_2 \), subject to the following constraints:

1. \( 2X_1 + 3X_2 \leq 150 \)
2. \( 4X_1 + 5X_2 \leq 200 \)
3. \( X_1, X_2 \geq 0 \) (non-negativity constraint)

We can solve this using the graphical method step by step.

### Step-by-Step Solution:

#### Step 1: Plot the constraints
We will first solve each constraint as if it were an equality and find the intercepts for \( X_1 \) and \( X_2 \).

**For the first constraint \( 2X_1 + 3X_2 = 150 \):**
- When \( X_1 = 0 \): \( 3X_2 = 150 \Rightarrow X_2 = 50 \)
- When \( X_2 = 0 \): \( 2X_1 = 150 \Rightarrow X_1 = 75 \)

Thus, the line \( 2X_1 + 3X_2 = 150 \) passes through the points \( (75, 0) \) and \( (0, 50) \).

**For the second constraint \( 4X_1 + 5X_2 = 200 \):**
- When \( X_1 = 0 \): \( 5X_2 = 200 \Rightarrow X_2 = 40 \)
- When \( X_2 = 0 \): \( 4X_1 = 200 \Rightarrow X_1 = 50 \)

Thus, the line \( 4X_1 + 5X_2 = 200 \) passes through the points \( (50, 0) \) and \( (0, 40) \).

#### Step 2: Find the feasible region
The feasible region is the area where both constraints are satisfied. We can find the intersection of the two lines by solving the system of equations:

\[
2X_1 + 3X_2 = 150
\]
\[
4X_1 + 5X_2 = 200
\]

We can solve this system algebraically to get the intersection point.

#### Step 3: Calculate intersection point
Let me solve the system of equations.The solution to the system of equations results in the point \( (X_1, X_2) = (-75, 100) \), which is not valid since \( X_1 \) cannot be negative (based on the non-negativity constraints \( X_1, X_2 \geq 0 \)).

Thus, the lines do not intersect in the feasible region. We will need to focus on the boundary points of the feasible region, which are the intercepts calculated earlier:

- \( (75, 0) \)
- \( (0, 50) \)
- \( (50, 0) \)
- \( (0, 40) \)

These points form the corners of the feasible region, and we will now evaluate the objective function \( Z = 20X_1 + 25X_2 \) at each of these points to find the maximum value.

Let me calculate that next.The values of the objective function \( Z = 20X_1 + 25X_2 \) at the feasible points are:

- At \( (75, 0) \): \( Z = 1500 \)
- At \( (0, 50) \): \( Z = 1250 \)
- At \( (50, 0) \): \( Z = 1000 \)
- At \( (0, 40) \): \( Z = 1000 \)

The maximum value of \( Z \) is **1500**, which occurs at \( (75, 0) \).

### Conclusion:
The optimal solution is \( X_1 = 75 \) and \( X_2 = 0 \), giving a maximum value for \( Z \) of 1500.

---

Would you like further details on any step or have any questions? Here are some related questions:

1. How do graphical methods handle more than two variables?
2. How do you handle non-linear constraints in optimization problems?
3. What are other methods to solve LP problems besides graphical?
4. How do we verify if the constraints are active at the optimal point?
5. How would a maximization problem change if one of the constraints was non-binding?

**Tip:** In linear programming, if the feasible region is unbounded, it's essential to check if the objective function increases indefinitely, as this can lead to no optimal solution.

Maximize Z = 20X1 + 25X2, subject to 2X1 + 3X2 <= 150 and 4X1 + 5X2 <= 200.

Step-by-step solution to maximize Z = 20X1 + 25X2 subject to constraints 2X1 + 3X2 <= 150 and 4X1 + 5X2 <= 200. Learn how to solve this using the graphical method in linear programming.

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