Find the maximum and minimum values of the function and the values of x and y where they occur.

F=5x+36y, subject to
7x+5y<=41,
5x+7y<=35,
x>=0,
y>=0.

What is the maximum value of the function?

To find the maximum and minimum values of the function \( F = 5x + 36y \) subject to the given constraints, we can follow these steps:

### 1. Identify the constraints
The problem gives the following constraints:
\[
7x + 5y \leq 41
\]
\[
5x + 7y \leq 35
\]
\[
x \geq 0
\]
\[
y \geq 0
\]

### 2. Graph the constraints
These constraints can be represented as lines in the \( xy \)-plane. The feasible region is the area where all these inequalities overlap.

### 3. Find the corner points of the feasible region
The maximum or minimum value of a linear function subject to linear constraints occurs at one of the corner points (vertices) of the feasible region.

The corner points can be found by solving the system of equations formed by the intersection of the constraint lines.

**Intersection of \( 7x + 5y = 41 \) and \( 5x + 7y = 35 \):**

1. Solve for \( x \) and \( y \) by solving the system:
   \[
   7x + 5y = 41
   \]
   \[
   5x + 7y = 35
   \]

2. Substitute \( y \) from one equation into the other to find \( x \), and then solve for \( y \).

**Intersection of \( 7x + 5y = 41 \) with the axes:**

For \( y = 0 \):
\[
7x = 41 \quad \Rightarrow \quad x = \frac{41}{7} \approx 5.86
\]

For \( x = 0 \):
\[
5y = 41 \quad \Rightarrow \quad y = \frac{41}{5} \approx 8.2
\]

**Intersection of \( 5x + 7y = 35 \) with the axes:**

For \( y = 0 \):
\[
5x = 35 \quad \Rightarrow \quad x = 7
\]

For \( x = 0 \):
\[
7y = 35 \quad \Rightarrow \quad y = 5
\]

### 4. Evaluate the objective function \( F(x, y) = 5x + 36y \) at each corner point

Next, evaluate \( F(x, y) = 5x + 36y \) at the vertices of the feasible region.

### 5. Determine the maximum and minimum values

Let's calculate these:

- **At \( (0, 5) \):**  
  \( F(0, 5) = 5(0) + 36(5) = 180 \)

- **At \( (7, 0) \):**  
  \( F(7, 0) = 5(7) + 36(0) = 35 \)

- **At \( ( \frac{41}{7}, 0) \approx (5.86, 0) \):**  
  \( F(5.86, 0) \approx 5(5.86) + 36(0) = 29.3 \)

- **At \( (0, \frac{41}{5}) \approx (0, 8.2) \):**  
  \( F(0, 8.2) \approx 5(0) + 36(8.2) = 295.2 \)

- **At the intersection of \( 7x + 5y = 41 \) and \( 5x + 7y = 35 \):**  
  Solving the system gives us \( x = 3 \) and \( y = 4 \):
  \[
  F(3, 4) = 5(3) + 36(4) = 15 + 144 = 159
  \]

### Conclusion:
The maximum value of \( F(x, y) = 295.2 \) occurs at \( (0, 8.2) \). The minimum value within the feasible region is \( 29.3 \), which occurs at \( (5.86, 0) \).

Would you like further details or any other questions on this? Here are some related questions:

1. How do we solve the intersection of linear equations graphically?
2. Why does the maximum or minimum value of a linear function occur at a vertex of the feasible region?
3. How can the Simplex Method be used to solve linear programming problems?
4. What are some real-life applications of linear programming?
5. How can constraints change the shape and position of the feasible region?

**Tip:** When solving linear programming problems, ensure to evaluate the function at every vertex of the feasible region to find the maximum and minimum values.

Find the maximum and minimum values of the function F = 5x + 36y, subject to 7x + 5y <= 41, 5x + 7y <= 35, x >= 0, y >= 0. What is the maximum value of the function?

Discover how to find the maximum and minimum values of the linear function F = 5x + 36y under a set of constraints using linear programming techniques. This problem involves concepts like feasible regions and optimization, suitable for students in Grades 10-12.

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