Question

The feasible region for the inequalities $ x + 2y \geq 4, \quad 2x + y \leq 6, \quad x \geq 0, \quad y \geq 0 $

• A.
• B.
• C.
• D.

Hint:

The feasible region will be the area where all four inequalities intersect.

Answer 1

The Correct Option is A

To determine the feasible region defined by the inequalities:

\[ \begin{cases} x + 2y \geq 4 \\ 2x + y \leq 6 \\ x \geq 0 \\ y \geq 0 \end{cases} \]

1. Non-Negativity Constraints:
The inequalities \( x \geq 0 \) and \( y \geq 0 \) restrict our solution to:
- The first quadrant (including both axes)

2. First Inequality Analysis:
For \( x + 2y \geq 4 \):
- The boundary line is \( x + 2y = 4 \) (passes through (4,0) and (0,2))
- The feasible region is above this line

3. Second Inequality Analysis:
For \( 2x + y \leq 6 \):
- The boundary line is \( 2x + y = 6 \) (passes through (3,0) and (0,6))
- The feasible region is below this line

4. Intersection Points:
The vertices of the feasible region are:
- (0,2) - Intersection of \( x=0 \) and \( x+2y=4 \)
- (2,2) - Intersection of \( x+2y=4 \) and \( 2x+y=6 \)
- (3,0) - Intersection of \( y=0 \) and \( 2x+y=6 \)
- (0,0) is not included as it violates \( x+2y \geq 4 \)

5. Graphical Representation:
The feasible region is a closed polygon bounded by:
1. The line segment from (0,2) to (2,2)
2. The line segment from (2,2) to (3,0)
3. The x-axis from (3,0) to infinity (but limited by other constraints)
4. The y-axis from (0,2) to infinity (but limited by other constraints)

Final Conclusion:
The feasible region satisfying all given inequalities is a quadrilateral in the first quadrant bounded by the identified constraints. 
Therefore, the correct option is \((a)\).

Answer 2

The feasible region for the inequalities:

\[ x + 2y \geq 4, \quad 2x + y \leq 6, \quad x \geq 0, \quad y \geq 0 \]To find the feasible region, we need to solve the system of inequalities:

1. Inequality 1: \( x + 2y \geq 4 \)Rearranged: \( y \geq \frac{4 - x}{2} \)This represents the region above the line \( y = \frac{4 - x}{2} \).

2. Inequality 2: \( 2x + y \leq 6 \)Rearranged: \( y \leq 6 - 2x \)This represents the region below the line \( y = 6 - 2x \).

3. Inequality 3: \( x \geq 0 \)This represents the region to the right of the y-axis.

4. Inequality 4: \( y \geq 0 \)This represents the region above the x-axis.
The feasible region is where all four inequalities intersect.
The lines \( x + 2y = 4 \) and \( 2x + y = 6 \) create boundaries.The restrictions \( x \geq 0 \) and \( y \geq 0 \) limit the region to the first quadrant.

From the graphical representation, the correct region is shown in option A