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

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