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For a Linear Programming Problem (LPP), the given objective function \(Z=3x+2y\) is subject to constraints: \(x+2y\le10\), \(3x+y\le15\), \(x, y\ge0\). The correct feasible region is:
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Q: For a Linear Programming Problem (LPP), the given objective function \(Z=3x+2y\) is subject to constraints: \(x+2y\le10\), \(3x+y\le15\), \(x, y\ge0\). The correct feasible region is:

Question Analysis & Solution

Step-by-Step Solution

  1. Graph the constraints:

    • \(x + 2y \le 10\): The boundary line is \(x + 2y = 10\). When \(x=0\), \(y=5\). When \(y=0\), \(x=10\). Plot the points (0,5) and (10,0) and draw the line. Since it's \(\le\), shade the region below the line.
    • \(3x + y \le 15\): The boundary line is \(3x + y = 15\). When \(x=0\), \(y=15\). When \(y=0\), \(x=5\). Plot the points (0,15) and (5,0) and draw the line. Since it's \(\le\), shade the region below the line.
    • \(x \ge 0\) and \(y \ge 0\): This restricts the solution to the first quadrant.

  2. Identify the Feasible Region:

    The feasible region is the area where all shaded regions overlap, which is a polygon bounded by the x-axis, y-axis, and the two lines. The vertices of this region are the intersection points of the lines.

  3. Find the intersection point of \(x + 2y = 10\) and \(3x + y = 15\):

    Multiply the second equation by 2: \(6x + 2y = 30\). Subtract the first equation from this: \(5x = 20\), so \(x = 4\). Substitute \(x = 4\) into the first equation: \(4 + 2y = 10\), so \(2y = 6\) and \(y = 3\). The intersection point is (4, 3).

  4. Determine the vertices of the feasible region:

    The vertices are (0, 0), (5, 0), (4, 3), and (0, 5).

  5. Match the vertices to the given options:

    Based on the vertices, the correct feasible region is AOEC, where A is (0,0), O is the origin, E is (5,0) and C is (4,3).

Correct Answer: AOEC

AI Suggestion: Option B
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