A dealer wishes to buy ceiling fans and table lamps. A fan costs Rs. 1500 and a lamp costs Rs. 600. He has Rs. 30000 to invest and space to store at most 40 items. He makes a profit of Rs. 200 on a fan and Rs. 90 on a lamp. If he buys x fans and y lamps, then the linear programming problem for maximising his profit Z is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Maximise \(Z = 200x + 90y\) subject to \(5x + 2y \le 100\), \(x + y \le 40\), \(x \ge 0\), \(y \ge 0\)
Let x be the number of fans and y the number of lamps. The profit is the objective function:
\(Z = 200x + 90y\)
The money spent cannot exceed Rs. 30000:
\(1500x + 600y \le 30000\)
Dividing by 300:
\(5x + 2y \le 100\)
At most 40 items can be stored:
\(x + y \le 40\)
The numbers bought cannot be negative, so \(x \ge 0\), \(y \ge 0\).
Why the other options are incorrect
Option (2) Maximise \(Z = 90x + 200y\) subject to \(5x + 2y \le 100\), \(x + y \le 40\), \(x \ge 0\), \(y \ge 0\) — Incorrect. The profits are swapped: Rs. 200 is earned on a fan, which is x.
Option (3) Maximise \(Z = 200x + 90y\) subject to \(2x + 5y \le 100\), \(x + y \le 40\), \(x \ge 0\), \(y \ge 0\) — Incorrect. The prices are swapped: the cost condition is \(5x + 2y \le 100\).
Option (4) Maximise \(Z = 200x + 90y\) subject to \(5x + 2y \le 100\), \(x + y \ge 40\), \(x \ge 0\), \(y \ge 0\) — Incorrect. “At most 40 items” means \(x + y \le 40\), not \(\ge\).