Which of the following is not a vertex of the feasible region bounded by the inequalities `2x+3yle6,5x+3yle15 and x,y,ge0`

Options

A. `(0,2)`
B. (0,0)
C. (3,0)
D. (0,5)

Answer

To determine which of the given points is not a vertex of the feasible region bounded by the inequalities \(2x + 3y \leq 6\), \(5x + 3y \leq 15\), and \(x, y \geq 0\), we will follow these steps:

### Step 1: Identify the inequalities and their intercepts
1. **First Inequality**: \(2x + 3y \leq 6\)
- To find the intercepts:
- Set \(x = 0\): \(3y = 6 \Rightarrow y = 2\) (y-intercept)
- Set \(y = 0\): \(2x = 6 \Rightarrow x = 3\) (x-intercept)
- Intercepts: (3, 0) and (0, 2)

2. **Second Inequality**: \(5x + 3y \leq 15\)
- To find the intercepts:
- Set \(x = 0\): \(3y = 15 \Rightarrow y = 5\) (y-intercept)
- Set \(y = 0\): \(5x = 15 \Rightarrow x = 3\) (x-intercept)
- Intercepts: (3, 0) and (0, 5)

### Step 2: Plot the lines on a graph
- Draw the lines corresponding to the equations \(2x + 3y = 6\) and \(5x + 3y = 15\) on the coordinate plane.
- The region satisfying \(x, y \geq 0\) is the first quadrant.

### Step 3: Determine the feasible region
- The feasible region is where the shaded areas from both inequalities overlap, considering the constraints \(x, y \geq 0\).

### Step 4: Identify the vertices of the feasible region
- The vertices of the feasible region can be found at the intersection points of the lines and the axes.
- The intersection points can be calculated by solving the equations:
1. \(2x + 3y = 6\)
2. \(5x + 3y = 15\)

#### Solving the equations:
- From the first equation, express \(y\):
\[ y = \frac{6 - 2x}{3} \]
- Substitute into the second equation:
\[ 5x + 3\left(\frac{6 - 2x}{3}\right) = 15 \]
\[ 5x + 6 - 2x = 15 \]
\[ 3x = 9 \Rightarrow x = 3 \]
- Substitute \(x = 3\) back into the first equation to find \(y\):
\[ 2(3) + 3y = 6 \Rightarrow 6 + 3y = 6 \Rightarrow 3y = 0 \Rightarrow y = 0 \]
- Thus, one vertex is (3, 0).

- Now, find the intersection of the two lines:
- Set \(2x + 3y = 6\) and \(5x + 3y = 15\):
\[ 5x + 3y - (2x + 3y) = 15 - 6 \Rightarrow 3x = 9 \Rightarrow x = 3 \]
- Substitute \(x = 3\) into either equation to find \(y\):
\[ 2(3) + 3y = 6 \Rightarrow 6 + 3y = 6 \Rightarrow 3y = 0 \Rightarrow y = 0 \]

- The vertices of the feasible region are:
- (0, 0)
- (3, 0)
- (0, 2)
- (0, 5)

### Step 5: Identify which point is not a vertex
- Given the options, we check each point:
- (0, 0) is a vertex.
- (3, 0) is a vertex.
- (0, 2) is a vertex.
- (0, 5) is **not** a vertex of the feasible region since it lies outside the feasible area defined by the inequalities.

### Conclusion
The point that is **not** a vertex of the feasible region is **(0, 5)**.

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