A dairy makes x kg of paneer and y kg of butter a day. It must make at least 2 kg of paneer and at least 3 kg of butter, and it can make at most 10 kg of the two together. The profit is Rs. 120 per kg of paneer and Rs. 90 per kg of butter. The maximum daily profit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Rs. 1,110
The problem is
Maximise \(Z = 120x + 90y\)
subject to
\(x \ge 2,\ y \ge 3\)
\(x + y \le 10\)
The feasible region is the triangle with corner points
\(x = 2,\ y = 3\): (2, 3)
\(y = 3\) on \(x + y = 10\): (7, 3)
\(x = 2\) on \(x + y = 10\): (2, 8)
Values of Z:
\(Z(2, 3) = 240 + 270 = 510\)
\(Z(7, 3) = 840 + 270 = 1110\)
\(Z(2, 8) = 240 + 720 = 960\)
The maximum profit is Rs. 1,110: 7 kg of paneer and the least allowed 3 kg of butter.
Why the other options are incorrect
Option (2) Rs. 1,200 — Incorrect. This needs 10 kg of paneer and no butter, but at least 3 kg of butter is required.
Option (3) Rs. 960 — Incorrect. This is the profit at (2, 8), less than at (7, 3).
Option (4) Rs. 510 — Incorrect. This is the least profit, at (2, 3).