Question

The region represented by the system of linear inequalities \( 3x + y \ge 3 \), \( 2x - y \ge -5 \), and \( x, y \ge 0 \) is:

• A. unbounded in 1st quadrant
• B. bounded in 1st quadrant
• C. unbounded in 2nd quadrant
• D. bounded in 2nd quadrant

Hint:

A system of linear inequalities is bounded only if it is enclosed from all directional sides by boundary lines. Here, as \(x \to \infty\) and \(y \to \infty\), both conditions \(3x + y \ge 3\) and \(2x - y \ge -5\) continue to remain true simultaneously, confirming that the region cannot be closed or bounded.

Answer 1

The Correct Option is A

Concept: To find the geometric layout of the region represented by a system of inequalities:
• Convert each inequality into a boundary straight-line equation.
• Identify the key intercepts to accurately sketch these lines.
• Determine the valid half-plane side by substituting a test point (such as the origin \((0,0)\)) into the inequalities.
• Incorporate non-negativity constraints (\(x \ge 0, y \ge 0\)), which restrict the final intersection space entirely to the first quadrant.

Step 1: Analyzing the non-negativity constraints.
We are given: \[ x \ge 0 \quad \text{and} \quad y \ge 0 \] This explicitly restricts our complete feasible area to the first quadrant only. Therefore, options (C) and (D) are immediately eliminated.

Step 2: Testing and plotting the line \( 3x + y = 3 \).
Let's compute the boundary line's coordinate plane intercepts:
• If \( x = 0 \), then \( y = 3 \implies (0, 3) \)
• If \( y = 0 \), then \( 3x = 3 \implies x = 1 \implies (1, 0) \) Now, substitute the test point \((0,0)\) into the original inequality \( 3x + y \ge 3 \): \[ 3(0) + 0 \ge 3 \implies 0 \ge 3 \quad (\text{False}) \] Since this is false, the valid region is the half-plane that does not contain the origin (i.e., the region situated above/to the right of the line).

Step 3: Testing and plotting the line \( 2x - y = -5 \).
Let's find the coordinate intercepts for this second line:
• If \( x = 0 \), then \( -y = -5 \implies y = 5 \implies (0, 5) \)
• If \( y = 0 \), then \( 2x = -5 \implies x = -2.5 \implies (-2.5, 0) \) Now, substitute the test point \((0,0)\) into the inequality \( 2x - y \ge -5 \): \[ 2(0) - 0 \ge -5 \implies 0 \ge -5 \quad (\text{True}) \] Since this is true, the valid region is the half-plane that contains the origin (i.e., the region lying down and to the right of this boundary line).

Step 4: Finding the common intersection area.
Let's examine what happens as \(x\) becomes infinitely large along the positive x-axis (as \(x \to \infty\)):
• For any massive positive value of \(x\), say \(x = 100\), and choosing a moderately large \(y\), say \(y = 50\):
• \(3(100) + 50 = 350 \ge 3\) (Satisfied)
• \(2(100) - 50 = 150 \ge -5\) (Satisfied)
• \(x \ge 0, y \ge 0\) (Satisfied) As there is no upper bounding constraint limiting how far \(x\) or \(y\) can extend outward towards infinity in the upper-right area of the first quadrant, the feasible region stretches out infinitely. Hence, the solution space is an unbounded region in the 1st quadrant, matching option (A).