Question

For the L.P.P. Maximize \[ z=10x+6y \] subjected to: \[ 3x+y\leq12 \] \[ 2x+5y\leq34 \] \[ x,y\geq0 \] Then the feasible region represented by system of inequalities is:

• A. Unbounded in first quadrant
• B. Bounded in first quadrant
• C. Unbounded in second quadrant
• D. Not possible (Empty)

Hint:

For L.P.P., inequalities with: \[ x\geq0,\ y\geq0 \] usually restrict the feasible region to the first quadrant.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A Linear Programming Problem (LPP) involves finding the optimal value of a linear objective function subject to linear constraints. The "feasible region" is the collection of all points \((x, y)\) that satisfy all constraints simultaneously. The constraints \(x \geq 0\) and \(y \geq 0\) are the non-negativity constraints, which immediately restrict the search space to the first quadrant of the coordinate plane. To determine if a region is "bounded" or "unbounded," we examine whether it can be contained within a circle of finite radius. Generally, constraints with \(\leq\) signs and positive coefficients for variables tend to create closed, finite polygons (bounded regions) in the first quadrant.
Step 2: Key Formula or Approach:
We analyze the constraints by treating them as equalities to find the boundary lines:
Line 1: \(3x + y = 12\)
Line 2: \(2x + 5y = 34\)
We find the intercepts of these lines on the axes to visualize the region.
Step 3: Detailed Explanation:
Let us analyze each constraint individually:
1. \(x \geq 0\) and \(y \geq 0\): These represent the y-axis and x-axis respectively, with the feasible region being in the first quadrant.
2. \(3x + y \leq 12\):
- Finding x-intercept: Set \(y=0 \implies 3x = 12 \implies x = 4\). Point is \((4, 0)\).
- Finding y-intercept: Set \(x=0 \implies y = 12\). Point is \((0, 12)\).
- Test the origin \((0,0)\): \(3(0) + 0 \leq 12 \implies 0 \leq 12\) (True). The region is towards the origin.
3. \(2x + 5y \leq 34\):
- Finding x-intercept: Set \(y=0 \implies 2x = 34 \implies x = 17\). Point is \((17, 0)\).
- Finding y-intercept: Set \(x=0 \implies 5y = 34 \implies y = 6.8\). Point is \((0, 6.8)\).
- Test the origin \((0,0)\): \(2(0) + 5(0) \leq 34 \implies 0 \leq 34\) (True). The region is towards the origin.
The feasible region is the intersection of these two half-planes in the first quadrant. It is a convex quadrilateral with vertices:
- Origin \((0, 0)\)
- Point on x-axis: \((4, 0)\) (since 4 is smaller than 17)
- Point on y-axis: \((0, 6.8)\) (since 6.8 is smaller than 12)
- Intersection point of the two lines:
Solving \(3x + y = 12\) and \(2x + 5y = 34\).
From first eq, \(y = 12 - 3x\). Substitute into second:
\(2x + 5(12 - 3x) = 34 \implies 2x + 60 - 15x = 34 \implies -13x = -26 \implies x = 2\).
Then \(y = 12 - 3(2) = 6\). Intersection is \((2, 6)\).
The region is enclosed by the points \((0,0), (4,0), (2,6), (0,6.8)\). Since the region is finite and entirely in the first quadrant, it is "bounded."
Step 4: Final Answer:
The feasible region is a closed polygon in the first quadrant. Thus, it is bounded in the first quadrant. Correct option is (B).