A nursery raises x trays of flower seedlings and y trays of vegetable seedlings in a greenhouse that can hold at most 20 trays. A flower tray needs 2 litres of water a day and a vegetable tray needs 1 litre, and at most 30 litres of water are available a day. At least 4 trays of vegetable seedlings must be raised. The profit is Rs. 50 on a flower tray and Rs. 40 on a vegetable tray. The maximum profit is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Rs. 900
Let x and y be the numbers of flower and vegetable trays. The problem is:
Maximise \(Z = 50x + 40y\)
\(x + y \le 20\), \(2x + y \le 30\), \(y \ge 4\), \(x \ge 0\)
The lines \(x + y = 20\) and \(2x + y = 30\) meet where
\(x = 10, \; y = 10\)
The line \(y = 4\) meets \(2x + y = 30\) at (13, 4). So the region is bounded, with corner points (0, 4), (13, 4), (10, 10) and (0, 20).
\(Z(0, 4) = 160\)
\(Z(13, 4) = 650 + 160 = 810\)
\(Z(10, 10) = 500 + 400 = 900\)
\(Z(0, 20) = 800\)
The maximum profit is Rs. 900, with 10 trays of each kind.
Why the other options are incorrect
Option (1) Rs. 810 — Incorrect. This is the profit at the corner (13, 4), less than Rs. 900.
Option (3) Rs. 800 — Incorrect. This is the profit at the corner (0, 20), less than Rs. 900.
Option (4) Rs. 750 — Incorrect. This is the profit at (15, 0), but that plan has no vegetable trays, so it breaks \(y \ge 4\).