Maximize P=8x+15y when 5x+y$$ \le10,x+5y\le15,y\ge0,x\ge0 $$

To maximize the function $ P = 8x + 15y $ subject to the constraints:

1. $ 5x + y \le 10 $
2. $ x + 5y \le 15 $
3. $ y \ge 0 $
4. $ x \ge 0 $

we'll solve this as a linear programming problem.

### Step 1: Identify the Constraints and Plot the Feasible Region
The feasible region is determined by the intersection of the inequalities in the constraints. Let’s rewrite and analyze each constraint:

1. **Constraint 1**: $ 5x + y \le 10 $
   - Rewrite as $ y \le -5x + 10 $.

2. **Constraint 2**: $ x + 5y \le 15 $
   - Rewrite as $ y \le -\frac{1}{5}x + 3 $.

3. **Constraints 3 and 4**: $ y \ge 0 $ and $ x \ge 0 $, which indicate the solution must lie in the first quadrant.

### Step 2: Find Intersection Points of the Constraints
To determine the vertices of the feasible region, we need to find the intersection points of the lines:

1. **Intersection of $ 5x + y = 10 $ and $ x + 5y = 15 $**:
   - Solve this system:
     \[
     5x + y = 10
     \]
     \[
     x + 5y = 15
     \]
   - Multiply the first equation by 5 to eliminate $ y $:
     \[
     25x + 5y = 50
     \]
     \[
     x + 5y = 15
     \]
   - Subtract the second equation from the first:
     \[
     24x = 35 \Rightarrow x = \frac{35}{24}
     \]
   - Substitute $ x = \frac{35}{24} $ back into $ 5x + y = 10 $ to find $ y $:
     \[
     5\left(\frac{35}{24}\right) + y = 10
     \]
     \[
     \frac{175}{24} + y = 10 \Rightarrow y = 10 - \frac{175}{24} = \frac{65}{24}
     \]
   - So, one vertex is $ \left(\frac{35}{24}, \frac{65}{24}\right) $.

2. **Intersection of $ 5x + y = 10 $ with $ y = 0 $**:
   - Set $ y = 0 $ in $ 5x + y = 10 $:
     \[
     5x = 10 \Rightarrow x = 2
     \]
   - So, another vertex is $ (2, 0) $.

3. **Intersection of $ x + 5y = 15 $ with $ x = 0 $**:
   - Set $ x = 0 $ in $ x + 5y = 15 $:
     \[
     5y = 15 \Rightarrow y = 3
     \]
   - So, another vertex is $ (0, 3) $.

### Step 3: Evaluate $ P = 8x + 15y $ at Each Vertex
Now, we substitute these vertices into the objective function $ P = 8x + 15y $ to find the maximum value.

1. At $ \left(\frac{35}{24}, \frac{65}{24}\right) $:
   \[
   P = 8\left(\frac{35}{24}\right) + 15\left(\frac{65}{24}\right) = \frac{280}{24} + \frac{975}{24} = \frac{1255}{24} \approx 52.29
   \]

2. At $ (2, 0) $:
   \[
   P = 8(2) + 15(0) = 16
   \]

3. At $ (0, 3) $:
   \[
   P = 8(0) + 15(3) = 45
   \]

### Conclusion
The maximum value of $ P = 8x + 15y $ occurs at the point $ \left(\frac{35}{24}, \frac{65}{24}\right) $ with $ P \approx 52.29 $.

Would you like a more detailed explanation of any step? 

---

1. What are the general steps for solving linear programming problems graphically?
2. How do you find intersection points for inequalities in linear programming?
3. Why is the feasible region only in the first quadrant in this problem?
4. How do you determine if a solution is maximized or minimized?
5. What tools can you use to graph linear inequalities for verification?

**Tip:** Always check each vertex of the feasible region when maximizing or minimizing linear functions in linear programming problems.

Maximize P=8x+15y when 5x+y≤10, x+5y≤15, y≥0, x≥0

Explore how to maximize the function P=8x+15y subject to constraints 5x+y≤10 and x+5y≤15. This detailed solution covers finding intersection points and evaluating vertices for the maximum value. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation