If a person rides his motorbike \(x\) km at 30 km per hour, he has to spend Rs. 3 per kilometer on petrol; if he rides \(y\) km at a faster speed of 40 km per hour, the petrol cost increases to Rs. 4 per kilometer. If he has Rs. 100 to spend on petrol and wishes to find the maximum distance he can travel within one hour, then linear programming problem (LPP) formulation is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Maximize Distance (D) = \(x + y\) subject to the constraints \(3x + 4y \le 100, 4x + 3y \le 120, x \ge 0, y \ge 0\).
He rides x km at 30 km/hr and y km at 40 km/hr, and the distance to be maximised is D = x + y. There are two limits: money for petrol and time.
Petrol costs Rs. 3 per km for x km and Rs. 4 per km for y km, and he has Rs. 100:
\(3x + 4y \le 100\)
Time = distance ÷ speed, and he has one hour:
\( \displaystyle \frac{x}{30} + \frac{y}{40} \le 1 \)
Multiplying by 120:
\(4x + 3y \le 120\)
Distances cannot be negative, so \(x \ge 0,\ y \ge 0\). This is Option (3).
Why the other options are incorrect
Option (1) Maximize Distance (D) = \(x + y\) subject to the constraints \(3x+4y \le 120, 4x+3y \ge 100, x \ge 0, y \ge 0\). — Incorrect. The right-hand sides are swapped: the petrol budget (100) belongs with \(3x + 4y\) and the time limit (120) with \(4x + 3y\); the time constraint must also be \(\le\).
Option (2) Maximize Distance (D) = \(x + y\) subject to the constraints \(3x + 4y \le 70, 4x + 3y \le 1, x \ge 0, y \ge 0\). — Incorrect. Neither 70 nor 1 is the limit on \(3x + 4y\) or \(4x + 3y\); the limits are 100 and 120.
Option (4) Maximize Distance (D) = \(x + y\) subject to the constraints \(3x + 4y \le 100, 4x + 3y \le 120, x \le 0, y \le 0\). — Incorrect. The two main constraints are right, but distances must satisfy \(x \ge 0,\ y \ge 0\), not \(x \le 0,\ y \le 0\).