Consider the following L.P.P.
Minimize \(z = 30x - 30y + 1800\);
subject to \(x + y \le 30, x \le 15, y \le 20, x + y \ge 15\) and \(x, y \ge 0\).
Then it attains its optimal value at the point
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (0, 20)
The feasible region lies between the lines x + y = 15 and x + y = 30, with \(0 \le x \le 15\) and \(0 \le y \le 20\). Its corner points are
(0, 15) on x = 0 and x + y = 15
(0, 20) on x = 0 and y = 20
(10, 20) on y = 20 and x + y = 30
(15, 15) on x = 15 and x + y = 30
(15, 0) on x = 15 and x + y = 15
Values of \(z = 30x - 30y + 1800\):
\(z(0, 15) = 1350,\quad z(0, 20) = 1200,\quad z(10, 20) = 1500\)
\(z(15, 15) = 1800,\quad z(15, 0) = 2250\)
The region is bounded, so the minimum, 1200, occurs at the corner (0, 20).
The point (20, 40) of Option (2) is not even feasible (it breaks \(x \le 15\) and \(y \le 20\)).