Minimise Z = -50x + 20y subject to the constraints: 2x - y ≥ -5,\ 3x… | Maths. Section B1 CUET question

Minimise $Z = -50x + 20y$

subject to the constraints: $2x - y \ge -5,\ 3x + y \ge 3,\ 2x - 3y \le 12,\ x \ge 0,\ y \ge 0$.

Then which of the following is/are true:

(A) Feasible region is unbounded.
(B) Z has no minimum value.
(C) The minimum value of Z is 100.
(D) The minimum value of Z is −300.

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 1

The correct answer is Option (1) → (A) and (B) only

$Graph of the constraints$

Writing each constraint in terms of y:

$2x - y \ge -5 \Rightarrow y \le 2x + 5$
$3x + y \ge 3 \Rightarrow y \ge 3 - 3x$

$ \displaystyle 2x - 3y \le 12 \Rightarrow y \ge \frac{2x - 12}{3} $

In the first quadrant the region lies above 3x + y = 3, the x-axis and 2x − 3y = 12, and below 2x − y = −5. Its corner points are (0, 3), (1, 0), (6, 0) and (0, 5). The upper line has slope 2 and the lower line slope 2/3, so they move apart as x increases and the region goes on without end to the right: it is unbounded.

Values of Z at the corners:

    $Z(0, 3) = 60$
    $Z(1, 0) = -50$
    $Z(6, 0) = -300$
    $Z(0, 5) = 100$

In an unbounded region the smallest corner value need not be the minimum. Along the boundary line 2x − 3y = 12, which belongs to the region for every x ≥ 6:

$ \displaystyle Z = -50x + 20\times\frac{2x - 12}{3} $

$ \displaystyle = \frac{-150x + 40x - 240}{3} $

$ \displaystyle = \frac{-110x - 240}{3} $

This keeps decreasing as x increases. For example, the point (9, 2) is in the region and

$Z(9, 2) = -450 + 40 = -410 \lt -300$

So the feasible region is unbounded and Z has no minimum value; −300 is only the smallest corner value.

Checking the statements

(A) Feasible region is unbounded. — Correct. The region goes on without end to the right.

(B) Z has no minimum value. — Correct. On 2x − 3y = 12, $Z = \frac{-110x - 240}{3}$ decreases without limit.

(C) The minimum value of Z is 100. — Incorrect. Z has no minimum; even the corner (1, 0) gives Z = −50, less than 100.

(D) The minimum value of Z is −300. — Incorrect. Z(6, 0) = −300 is only the smallest corner value; Z(9, 2) = −410 is smaller.

Tip: In an unbounded region, the smallest corner value m is the minimum only if the half-plane Z < m has no point in common with the region. Here −50x + 20y < −300 contains (9, 2), so −300 is not the minimum.

Note: NTA dropped this question. For the benefit of students and to give them additional practice, the question has been suitably modified so that it has one clear answer.