A farmer mixes two cattle feeds P and Q. Each kg of P contains 4 unit…

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

This is a diet problem:

    Minimise \(Z = 20x + 30y\)

subject to

    \(4x + 2y \ge 16\), that is \(2x + y \ge 8\) (protein)
    \(x + 3y \ge 9\) (fat)
    \(x \ge 0,\ y \ge 0\)

Both constraints are of the “at least” type, so the feasible region lies away from the origin and is unbounded. The two lines meet where, putting \(y = 8 - 2x\),

    \(x + 24 - 6x = 9\)
    \(x = 3,\ y = 2\)

The corner points are (0, 8), (3, 2) and (9, 0). Values of Z:

    \(Z(0, 8) = 240\)
    \(Z(3, 2) = 60 + 60 = 120\)
    \(Z(9, 0) = 180\)

The half-plane \(20x + 30y \lt 120\), that is \(2x + 3y \lt 12\), has no point in common with the feasible region, so 120 is the minimum.

So the region is unbounded with corners (0, 8), (3, 2), (9, 0), the fat constraint is \(x + 3y \ge 9\), and the minimum cost is Rs. 120 at (3, 2).

Checking the statements

(A) The feasible region is unbounded. — Correct. Both constraints are “at least” constraints, so x and y can be as large as we like.

(B) The minimum cost is Rs. 120, using 3 kg of P and 2 kg of Q. — Correct. \(Z(3, 2) = 120\) is the least value.

(C) The fat constraint is \(3x + y \ge 9\). — Incorrect. P has 1 unit and Q has 3 units of fat, so it is \(x + 3y \ge 9\).

(D) (8, 0) is a corner point of the feasible region. — Incorrect. At (8, 0), \(x + 3y = 8 \lt 9\), so the point is not even feasible.