Consider the following L.P.P minimize \(z = x-7y+190\) subject to \(x + y \le 8, x + y \ge 4, x \le 5, y \le 5\) and \(x, y \ge 0\). Then which of the following is/are true?
(A) Its feasible region is unbounded
(B) Its feasible region is bounded
(C) Its feasible region has 5 corner points
(D) Its feasible region has 6 corner points
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (D) only
The feasible region is the part of the square \(0 \le x \le 5,\ 0 \le y \le 5\) that lies between the parallel lines \(x + y = 4\) and \(x + y = 8\). Going round its boundary:
on \(y = 0\): from \((4, 0)\) to \((5, 0)\)
on \(x = 5\): from \((5, 0)\) to \((5, 3)\), since \(x + y \le 8\) gives \(y \le 3\)
on \(x + y = 8\): from \((5, 3)\) to \((3, 5)\)
on \(y = 5\): from \((3, 5)\) to \((0, 5)\)
on \(x = 0\): from \((0, 5)\) to \((0, 4)\)
on \(x + y = 4\): from \((0, 4)\) back to \((4, 0)\)
So the feasible region is a bounded hexagon with 6 corner points: \((4, 0), (5, 0), (5, 3), (3, 5), (0, 5), (0, 4)\).
Checking the statements
(A) Its feasible region is unbounded — Incorrect. \(x \le 5\) and \(y \le 5\), so the region is bounded.
(B) Its feasible region is bounded — Correct. It is a closed hexagon.
(C) Its feasible region has 5 corner points — Incorrect. There are 6 corner points.
(D) Its feasible region has 6 corner points — Correct. The hexagon has six corner points.