Q.A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has at most Rs 120 to spend on petrol and one hour's time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.

Split the journey into a part at 50 km/h and a part at 80 km/h, and turn the budget and time limit into linear constraints.

Decision variables. Let $x$ = distance (km) at 50 km/h, $y$ = distance (km) at 80 km/h.

Time constraint. Time = distance / speed:

Petrol-cost constraint. Cost is Rs 2/km and Rs 3/km:

$2x + 3y \le 120.$ …

Setting up the model

The man can ride part of the way at 50 km/h and part at 80 km/h. Each speed uses two limited resources — money (petrol) and time — and he wants the greatest total distance. That is a linear programming problem: two decision variables, two linear constraints, and a linear objective.

1. Decision variables.

Let $x$ = distance (km) ridden at 50 km/h and $y$ = distance (km) ridden at 80 km/h, with $x \ge 0$, $y \ge 0$.

2. Objective function.

Total distance is $x + y$, to be maximised: Maximise $Z = x + y$.

3. Petrol-cost constraint.

At 50 km/h petrol costs Rs 2 per km, so $x$ km costs $2x$; at 80 km/h it costs Rs 3 per km, so $y$ km costs $3y$. Total cost is at most Rs 120:

4. Time constraint.

Time = distance divided by speed, so the two legs take $\dfrac{x}{50}$ and $\dfrac{y}{80}$ hours. Total time is at most 1 hour:

$\dfrac{x}{50} + \dfrac{y}{80} \le 1.$ …

Method: Modelling a "rates and budgets" problem as an LPP

Reach for this when a problem offers two options (two speeds, two machines, two routes), each using up limited resources at its own rate, and asks you to express the situation as a linear programming problem.

Steps

Step 1: Choose the right decision variables.

Let the variables be the amounts you accumulate (distance covered at each speed, units made on each machine) -- not the rates themselves. A speed or a price is a fixed constant, never a variable.

Step 2: Write the objective.

Express the quantity to optimise as $Z = ax + by$. When the goal is simply "the greatest total amount," the objective is often just $Z = x + y$.

Step 3: Turn each resource limit into a linear inequality, checking units.

For a cost limit, multiply each amount by its per-unit price and cap the sum. For a time limit, remember …

Common Mistakes

Mistake 1: Writing the time limit as $50x + 80y \le 1$.

Why it's wrong: that multiplies speed by distance, which is dimensionally meaningless. Time equals distance divided by speed. Correct approach: $\dfrac{x}{50} + \dfrac{y}{80} \le 1$ (hours), since $\dfrac{\text{km}}{\text{km/h}} = \text{hours}$.

Mistake 2: Pairing the wrong cost with the wrong speed. …