A vertex of bounded region of inequalities `xge0,x+2yge0` and `2x+yle4,` is

Options

A. `(1,1)`
B. `(0,1)`
C. `(3,0)`
D. `(0,0)`

Answer

To find the vertices of the bounded region defined by the inequalities \( x \geq 0 \), \( x + 2y \geq 0 \), and \( 2x + y \leq 4 \), we will follow these steps:

### Step 1: Convert inequalities to equations
We start by converting the inequalities into equations to find the boundary lines:
1. \( x = 0 \)
2. \( x + 2y = 0 \) (or \( y = -\frac{x}{2} \))
3. \( 2x + y = 4 \) (or \( y = 4 - 2x \))

### Step 2: Find intersection points of the lines
Next, we will find the intersection points of these lines.

#### Intersection of \( x + 2y = 0 \) and \( 2x + y = 4 \):
1. From \( x + 2y = 0 \), we can express \( y \) in terms of \( x \):
\[ y = -\frac{x}{2} \]
2. Substitute \( y \) into \( 2x + y = 4 \):
\[ 2x - \frac{x}{2} = 4 \]
\[ \frac{4x - x}{2} = 4 \implies \frac{3x}{2} = 4 \]
\[ 3x = 8 \implies x = \frac{8}{3} \approx 2.67 \]
3. Now substitute \( x \) back to find \( y \):
\[ y = -\frac{8/3}{2} = -\frac{4}{3} \approx -1.33 \]
So, the intersection point is \( \left(\frac{8}{3}, -\frac{4}{3}\right) \).

#### Intersection of \( x + 2y = 0 \) and \( x = 0 \):
1. Substitute \( x = 0 \) into \( x + 2y = 0 \):
\[ 0 + 2y = 0 \implies y = 0 \]
So, the intersection point is \( (0, 0) \).

#### Intersection of \( 2x + y = 4 \) and \( x = 0 \):
1. Substitute \( x = 0 \) into \( 2x + y = 4 \):
\[ 0 + y = 4 \implies y = 4 \]
So, the intersection point is \( (0, 4) \).

### Step 3: Identify the vertices
The vertices of the bounded region formed by the inequalities are:
1. \( A(0, 4) \)
2. \( B\left(\frac{8}{3}, -\frac{4}{3}\right) \)
3. \( O(0, 0) \)

### Conclusion
The vertices of the bounded region defined by the inequalities \( x \geq 0 \), \( x + 2y \geq 0 \), and \( 2x + y \leq 4 \) are \( (0, 4) \), \( \left(\frac{8}{3}, -\frac{4}{3}\right) \), and \( (0, 0) \).

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