A parking lot has 900 m2 of space. A car needs 10 m2 and a bus needs 30 m2, and at most 60 vehicles can be handled in a day. The parking fee is Rs. 50 for a car and Rs. 120 for a bus per day. The maximum fee that can be collected in a day is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Rs. 4050
Let x cars and y buses be parked. Then
Maximise \(Z = 50x + 120y\)
subject to
\(10x + 30y \le 900\), that is \(x + 3y \le 90\) (space)
\(x + y \le 60\) (vehicles)
\(x \ge 0,\ y \ge 0\)
The two lines meet where, subtracting,
\(2y = 30\)
\(y = 15,\ x = 45\)
The corner points are (0, 0), (60, 0), (45, 15) and (0, 30). Values of Z:
\(Z(0, 0) = 0\)
\(Z(60, 0) = 3000\)
\(Z(45, 15) = 2250 + 1800 = 4050\)
\(Z(0, 30) = 3600\)
The maximum fee is Rs. 4050, with 45 cars and 15 buses.