If a person rides his motorbike $x$ km at 30 km per hour, he has to spend ₹ 3 per kilometer on petrol. If he rides $y$ km at a faster speed of 40 km per hour, the petrol cost increases to ₹ 4 per kilometer. If he has ₹ 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:

The person rides $x$ km at 30 km/hr with a petrol cost of ₹3 per km, and $y$ km at 40 km/hr with a petrol cost of ₹4 per km. The total money available is ₹100 and total time available is 1 hour.

The goal is to travel the maximum distance possible.

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Total Distance $= x + y$

Objective Function: Maximize $D = x + y$

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The person has only ₹100 to spend.

Cost for traveling $x$ km $= 3x$

Cost for traveling $y$ km $= 4y$

Total cost cannot exceed ₹100.

Money Constraint: $3x + 4y \leq 100$

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The person has only 1 hour.

Time $= \dfrac{\text{Distance}}{\text{Speed}}$

Time for $x$ km at 30 km/hr $= \dfrac{x}{30}$ hours

Time for $y$ km at 40 km/hr $= \dfrac{y}{40}$ hours

Total time cannot exceed 1 hour:

$\dfrac{x}{30} + \dfrac{y}{40} \leq 1$

Multiplying the entire inequality by 120:

$120 \times \dfrac{x}{30} + 120 \times \dfrac{y}{40} \leq 120 \times 1$

$4x + 3y \leq 120$

Time Constraint: $4x + 3y \leq 120$

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Distance cannot be negative.

Non-negativity Constraints: $x \geq 0, y \geq 0$

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Maximize $D = x + y$

Subject to:

$3x + 4y \leq 100$

$4x + 3y \leq 120$

$x \geq 0, y \geq 0$