The feasible region represented by the constraints 4x + y ≥ 80 , x +…

The correct answer is Option (3) → Region C

The three boundary lines and their intercepts:

    \(4x + y = 80\): through (0, 80) and (20, 0)
    \(x + 5y = 115\): through (0, 23) and (115, 0)
    \(3x + 2y = 150\): through (0, 75) and (50, 0)

Testing the origin (0, 0) in each inequality:

    \(4(0) + 0 = 0 \ge 80\): false, so the region lies on the side of \(4x + y = 80\) away from O
    \(0 + 5(0) = 0 \ge 115\): false, so the region lies on the side of \(x + 5y = 115\) away from O
    \(3(0) + 2(0) = 0 \le 150\): true, so the region lies on the side of \(3x + 2y = 150\) towards O

So the feasible region lies above both \(4x + y = 80\) and \(x + 5y = 115\), and below \(3x + 2y = 150\). This is the triangle with corners (2, 72), (15, 20) and (40, 15), which is Region C. A check of the corners:

    (2, 72): \(4(2) + 72 = 80\), \(3(2) + 2(72) = 150\)
    (15, 20): \(4(15) + 20 = 80\), \(15 + 5(20) = 115\)
    (40, 15): \(40 + 5(15) = 115\), \(3(40) + 2(15) = 150\)

Regions A and D lie below \(4x + y = 80\), and Region B lies below \(x + 5y = 115\), so none of them satisfies all the constraints.