A shopkeeper has Rs. 1200 to buy pens and notebooks. A pen costs Rs. 20 and a notebook costs Rs. 40. He makes a profit of Rs. 4 on a pen and Rs. 10 on a notebook, and he sells everything he buys. The maximum profit he can make is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Rs. 300
Let him buy x pens and y notebooks. Then the problem is
Maximise \(Z = 4x + 10y\)
subject to \(20x + 40y \le 1200\), that is, \(x + 2y \le 60\), with \(x \ge 0\), \(y \ge 0\)
The feasible region is the triangle with corner points (0, 0), (60, 0) and (0, 30).
\(Z(0, 0) = 0\)
\(Z(60, 0) = 240\)
\(Z(0, 30) = 300\)
The maximum profit is Rs. 300, by buying 30 notebooks and no pens.
Why the other options are incorrect
Option (2) Rs. 240 — Incorrect. This is the profit from 60 pens only, at the corner (60, 0); notebooks give more.
Option (3) Rs. 540 — Incorrect. This adds 240 and 300, as if he could buy 60 pens and 30 notebooks; that costs Rs. 2400, more than Rs. 1200.
Option (4) Rs. 1200 — Incorrect. This is the money he spends, not his profit.