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:

Maximize Distance (D) = x + y subject to the constraints $3x + 4y ≤ 100, 4x + 3y ≤ 120, x≥0, y ≥0$.

The correct answer is Option (3) → Maximize Distance (D) = x + y subject to the constraints $3x + 4y ≤ 100, 4x + 3y ≤ 120, x≥0, y ≥0$.

Let $x$ be distance travelled at $30$ km per hour.

Let $y$ be distance travelled at $40$ km per hour.

Petrol cost constraint

$3x+4y\le100$

Time constraint

$\frac{x}{30}+\frac{y}{40}\le1$

Non negativity conditions

$x\ge0,\;y\ge0$

Objective function

Maximize $Z=x+y$

LPP is: Maximize $Z=x+y$ subject to $3x+4y\le100,\;\frac{x}{30}+\frac{y}{40}\le1,\;x\ge0,\;y\ge0$