Maximize P = 18x+ 10y

12x+ 6y ≤ 2400

8x + 2y ≤ 1200

subject to:

x≥ 0

y ≥0

This is a linear programming problem, where the objective is to maximize \( P = 18x + 10y \), subject to the given constraints. We will solve this problem using the **graphical method** by following these steps:

### Constraints:
1. \( 12x + 6y \leq 2400 \)
2. \( 8x + 2y \leq 1200 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 1: Simplify the constraints

#### Constraint 1:
\[ 12x + 6y \leq 2400 \]
Divide by 6:
\[ 2x + y \leq 400 \]

#### Constraint 2:
\[ 8x + 2y \leq 1200 \]
Divide by 2:
\[ 4x + y \leq 600 \]

### Step 2: Graph the constraints

We will convert these inequalities into equations to find the boundary lines.

1. \( 2x + y = 400 \)
   - When \( x = 0 \), \( y = 400 \)
   - When \( y = 0 \), \( x = 200 \)

2. \( 4x + y = 600 \)
   - When \( x = 0 \), \( y = 600 \)
   - When \( y = 0 \), \( x = 150 \)

Next, graph the lines and identify the feasible region where all constraints are satisfied.

### Step 3: Identify the corner points

The corner points of the feasible region are obtained by solving for the intersections of the boundary lines:

1. \( (0, 400) \) from \( 2x + y = 400 \)
2. \( (200, 0) \) from \( 2x + y = 400 \)
3. \( (150, 0) \) from \( 4x + y = 600 \)
4. Intersection of \( 2x + y = 400 \) and \( 4x + y = 600 \)

To find the intersection of these two lines, solve the system:

\[
2x + y = 400
\]
\[
4x + y = 600
\]

Subtract the first equation from the second:

\[
(4x + y) - (2x + y) = 600 - 400
\]
\[
2x = 200 \implies x = 100
\]

Substitute \( x = 100 \) into \( 2x + y = 400 \):

\[
2(100) + y = 400 \implies y = 200
\]

So, the intersection point is \( (100, 200) \).

### Step 4: Evaluate the objective function at the corner points

Now, evaluate the objective function \( P = 18x + 10y \) at each corner point:

1. At \( (0, 400) \):
   \[
   P = 18(0) + 10(400) = 4000
   \]

2. At \( (200, 0) \):
   \[
   P = 18(200) + 10(0) = 3600
   \]

3. At \( (150, 0) \):
   \[
   P = 18(150) + 10(0) = 2700
   \]

4. At \( (100, 200) \):
   \[
   P = 18(100) + 10(200) = 1800 + 2000 = 3800
   \]

### Step 5: Conclusion

The maximum value of \( P \) occurs at \( (0, 400) \), where \( P = 4000 \).

### Final Answer:
The maximum value of \( P \) is **4000**, and it occurs when \( x = 0 \) and \( y = 400 \).

---

Would you like a detailed explanation of how I found the intersection or more details about the graphical method?

Here are 5 related questions:
1. How do you graphically represent inequalities on a coordinate plane?
2. Can linear programming problems have more than one optimal solution?
3. What happens if the feasible region is unbounded in a linear programming problem?
4. How would you use the simplex method for more complex linear programming problems?
5. Can linear programming problems have no feasible solution?

**Tip**: Always check the corner points in linear programming, as the optimal solution always lies at one of them.

Maximize P = 18x + 10y

12x + 6y ≤ 2400

8x + 2y ≤ 1200

subject to:
x ≥ 0
y ≥ 0

Explore a linear programming problem to maximize P = 18x + 10y subject to constraints 12x + 6y ≤ 2400 and 8x + 2y ≤ 1200. Learn to solve using the graphical method, including constraint simplifications and boundary point evaluations. Suitable for students in Grades 10-12.

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