Question

The feasible region of the L.P.P.
\[ \text{Maximize } z = 70x + 50y \] subject to \[ 8x + 5y \le 60,\quad 4x + 5y \le 40,\quad x \ge 0,\; y \ge 0 \] is

• A. a triangle
• B. a square
• C. a pentagon
• D. a quadrilateral

Hint:

In L.P.P., the number of bounding inequalities usually indicates the number of sides of the feasible region.

Answer 1

The Correct Option is D

Step 1: Identify all boundary lines.
The given inequalities correspond to the lines: \[ 8x + 5y = 60,\quad 4x + 5y = 40,\quad x = 0,\quad y = 0 \]
Step 2: Consider the non-negativity conditions.
The conditions \(x \ge 0\) and \(y \ge 0\) restrict the region to the first quadrant.

Step 3: Find the points of intersection.
- Intersection of \(4x + 5y = 40\) with axes gives two points.
- Intersection of \(8x + 5y = 60\) with axes gives two points.
- Intersection of the two given lines gives another point inside the first quadrant.

Step 4: Determine the shape of the feasible region.
The feasible region is bounded by four line segments in the first quadrant, forming a closed figure with four sides.

Step 5: Conclusion.
Hence, the feasible region is a quadrilateral.