Maximize P = 100x + 50y subject to these constraints:
          20x + 35y ≤ 280
          10x + 10y ≤ 86
          20x + 15y ≤ 140
          40x + 15y ≤ 196
          50x + 10y ≤ 230
          x ≥ 0, y ≥ 0
Maximum value for P = 

To solve this linear programming problem, we will use the method of **graphical analysis** or apply the **Simplex method**. The objective is to maximize \( P = 100x + 50y \), subject to the following constraints:

1. \( 20x + 35y \leq 280 \)
2. \( 10x + 10y \leq 86 \)
3. \( 20x + 15y \leq 140 \)
4. \( 40x + 15y \leq 196 \)
5. \( 50x + 10y \leq 230 \)
6. \( x \geq 0 \), \( y \geq 0 \)

### Steps:
1. **Convert the inequalities to equations** to find corner points of the feasible region.
2. **Find the intersection points** by solving pairs of equations.
3. **Evaluate the objective function** \( P = 100x + 50y \) at each of these corner points to find the maximum value.

Let’s solve this systematically.

### Step 1: Solve pairs of equations

#### 1. \( 20x + 35y = 280 \)
We can rewrite this in terms of \( y \):
\[
y = \frac{280 - 20x}{35}
\]
#### 2. \( 10x + 10y = 86 \)
Simplifying, \( x + y = 8.6 \), so:
\[
y = 8.6 - x
\]

#### 3. \( 20x + 15y = 140 \)
Simplifying:
\[
y = \frac{140 - 20x}{15}
\]

#### 4. \( 40x + 15y = 196 \)
Simplifying:
\[
y = \frac{196 - 40x}{15}
\]

#### 5. \( 50x + 10y = 230 \)
Simplifying:
\[
y = \frac{230 - 50x}{10}
\]

### Step 2: Find intersection points
We now solve these equations in pairs to get the coordinates of the feasible region.

### Step 3: Evaluate \( P = 100x + 50y \) at these points

I'll calculate these points and evaluate \( P \).The intersection points (feasible region corners) and the corresponding values of \( P = 100x + 50y \) are:

1. \( \left( \frac{7}{5}, \frac{36}{5} \right) \), \( P = 500 \)
2. \( \left( \frac{7}{4}, 7 \right) \), \( P = 525 \)
3. \( \left( \frac{133}{55}, \frac{364}{55} \right) \), \( P \approx 572.73 \)
4. \( \left( \frac{105}{31}, \frac{188}{31} \right) \), \( P \approx 641.94 \)
5. \( \left( \frac{11}{5}, \frac{32}{5} \right) \), \( P = 540 \)
6. \( \left( \frac{67}{25}, \frac{148}{25} \right) \), \( P = 564 \)
7. \( \left( \frac{18}{5}, 5 \right) \), \( P = 610 \)
8. \( \left( \frac{14}{5}, \frac{28}{5} \right) \), \( P = 560 \)
9. \( \left( \frac{41}{11}, \frac{48}{11} \right) \), \( P \approx 590.91 \)
10. \( \left( \frac{149}{35}, \frac{12}{7} \right) \), \( P \approx 511.43 \)

### Maximum value:
The maximum value of \( P \approx 641.94 \) occurs at the point \( \left( \frac{105}{31}, \frac{188}{31} \right) \).

Would you like further details on how these points were calculated?

Here are five related questions for further exploration:
1. How do you graph linear inequalities to find feasible regions?
2. What is the significance of corner points in linear programming?
3. How can the Simplex method be used to solve linear programming problems?
4. How would changing the objective function coefficients affect the solution?
5. What happens if the constraints change and no feasible solution exists?

**Tip:** In linear programming, the maximum or minimum of the objective function always occurs at one of the vertices of the feasible region.

Maximize P = 100x + 50y subject to these constraints:
20x + 35y ≤ 280
10x + 10y ≤ 86
20x + 15y ≤ 140
40x + 15y ≤ 196
50x + 10y ≤ 230
x ≥ 0, y ≥ 0
Maximum value for P = 

Learn how to solve a linear programming optimization problem where the objective is to maximize P = 100x + 50y under several inequality constraints. This detailed solution involves finding the feasible region and evaluating the objective function at corner points for maximum value. Suitable for university-level students studying optimization.

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