what are the vertices of the feasible region

To find the vertices of the feasible region, we need to determine the points of intersection of the constraint lines. The constraints are:

1. \( 3x + 4y \leq 460 \)
2. \( 3x + 4y \geq 360 \)
3. \( 2x + y \leq 190 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)

First, we need to find the equations of the lines:

1. \( 3x + 4y = 460 \)
2. \( 3x + 4y = 360 \)
3. \( 2x + y = 190 \)

Next, we find the intersection points of these lines by solving the system of equations pairwise.

### Intersection of \( 3x + 4y = 460 \) and \( 3x + 4y = 360 \)
These lines are parallel and do not intersect, so we ignore this pair for intersection points.

### Intersection of \( 3x + 4y = 460 \) and \( 2x + y = 190 \)
Solve the system:
\[
\begin{cases}
3x + 4y = 460 \\
2x + y = 190
\end{cases}
\]

From the second equation: \( y = 190 - 2x \).

Substitute \( y \) into the first equation:
\[
3x + 4(190 - 2x) = 460 \\
3x + 760 - 8x = 460 \\
-5x = -300 \\
x = 60
\]

Substitute \( x = 60 \) back into \( y = 190 - 2x \):
\[
y = 190 - 2(60) = 70
\]

So, the intersection point is \( (60, 70) \).

### Intersection of \( 3x + 4y = 360 \) and \( 2x + y = 190 \)
Solve the system:
\[
\begin{cases}
3x + 4y = 360 \\
2x + y = 190
\end{cases}
\]

From the second equation: \( y = 190 - 2x \).

Substitute \( y \) into the first equation:
\[
3x + 4(190 - 2x) = 360 \\
3x + 760 - 8x = 360 \\
-5x = -400 \\
x = 80
\]

Substitute \( x = 80 \) back into \( y = 190 - 2x \):
\[
y = 190 - 2(80) = 30
\]

So, the intersection point is \( (80, 30) \).

### Intersection with the axes
For the constraints \( 2x + y = 190 \), \( 3x + 4y = 460 \), and \( 3x + 4y = 360 \):

#### Intersection of \( 2x + y = 190 \) with \( x = 0 \):
\[
2(0) + y = 190 \\
y = 190
\]

So, the point is \( (0, 190) \).

#### Intersection of \( 3x + 4y = 460 \) with \( x = 0 \):
\[
3(0) + 4y = 460 \\
4y = 460 \\
y = 115
\]

So, the point is \( (0, 115) \).

#### Intersection of \( 3x + 4y = 360 \) with \( x = 0 \):
\[
3(0) + 4y = 360 \\
4y = 360 \\
y = 90
\]

So, the point is \( (0, 90) \).

#### Intersection of \( 2x + y = 190 \) with \( y = 0 \):
\[
2x + 0 = 190 \\
2x = 190 \\
x = 95
\]

So, the point is \( (95, 0) \).

Combining these points, we have the vertices of the feasible region:

1. \( (60, 70) \)
2. \( (80, 30) \)
3. \( (0, 190) \)
4. \( (0, 115) \)
5. \( (0, 90) \)
6. \( (95, 0) \)

These vertices define the feasible region.

Would you like to know more about how to find the optimal solution or have any questions on this topic?

Here are 8 related questions to expand on this problem:

1. How can we verify that these vertices indeed form the feasible region?
2. What is the next step to find the maximum revenue from these vertices?
3. How do we check if all constraints are satisfied at each vertex?
4. How would we interpret the constraints if they were different, e.g., \(3x + 4y \leq 500\)?
5. How can we use these vertices to plot the feasible region graphically?
6. What if the constraints were changed, how would the feasible region change?
7. Can you explain the concept of the objective function in this context?
8. How does changing the coefficients in the revenue function \( R(x, y) = 62x + 68y \) affect the optimal solution?

**Tip:** Always double-check your intersection points to ensure they lie within the feasible region by substituting them back into the original inequalities.

Explore how to find the vertices of the feasible region in linear programming with detailed step-by-step solutions. This problem involves solving constraint equations and identifying intersection points, crucial in optimization problems. Suitable for advanced high school students studying linear programming.

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