Find the area of the quadrilateral formed by the solution set of the inequations `2x + 3y le 12, x 0, y ge 0` and `x le 3`.

Options

A. 6
B. 8
C. 9
D. 12

Answer

To find the area of the quadrilateral formed by the solution set of the inequalities \(2x + 3y \leq 12\), \(x \geq 0\), \(y \geq 0\), and \(x \leq 3\), we will follow these steps:

### Step 1: Graph the inequalities
1. **Graph the line** \(2x + 3y = 12\):
- To find the intercepts, set \(x = 0\):
\[ 3y = 12 \implies y = 4 \quad \text{(y-intercept)} \]
- Now set \(y = 0\):
\[ 2x = 12 \implies x = 6 \quad \text{(x-intercept)} \]
- The line intersects the axes at points \( (0, 4) \) and \( (6, 0) \).

2. **Identify the bounded region**:
- Since \(x \geq 0\) and \(y \geq 0\), we are only interested in the first quadrant.
- The line \(x = 3\) is a vertical line that intersects the x-axis at \( (3, 0) \).

### Step 2: Find the vertices of the quadrilateral
1. **Vertices**:
- The vertices of the quadrilateral are:
- \(O(0, 0)\) (origin)
- \(A(0, 4)\) (y-intercept)
- \(B(3, 2)\) (intersection of the line \(2x + 3y = 12\) and \(x = 3\))
- \(C(3, 0)\) (intersection of the line \(x = 3\) and the x-axis)

### Step 3: Calculate the area of the quadrilateral
1. **Area of quadrilateral OABC**:
- The quadrilateral can be divided into a rectangle \(OACB\) and triangle \(OAB\).
- **Area of rectangle OACB**:
- Width = \(3\) (from \(O\) to \(C\))
- Height = \(4\) (from \(O\) to \(A\))
\[ \text{Area}_{rectangle} = \text{width} \times \text{height} = 3 \times 4 = 12 \text{ square units} \]

2. **Area of triangle OAB**:
- Base = \(3\) (from \(O\) to \(B\))
- Height = \(2\) (from \(B\) to the line \(y = 0\))
\[ \text{Area}_{triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 2 = 3 \text{ square units} \]

3. **Total Area**:
\[ \text{Total Area} = \text{Area}_{rectangle} + \text{Area}_{triangle} = 12 + 3 = 15 \text{ square units} \]

### Final Answer
The area of the quadrilateral formed by the solution set of the inequalities is \(15\) square units.