If a 20 year old girl drives her car at 25 km/h, she has to spend ₹4/km on petrol. If she drives her car at 40 km/h, the petrol cost increases to ₹5/km. She has ₹200 to spend on petrol and wishes to find the maximum distance she can travel within one hour. Express the above problem as a Linear Programming Problem.

Maximize $Z = x + y$; Subject to: $4x + 5y \leq 200, \enspace 8x + 5y \leq 200, \enspace x, y \geq 0$

The correct answer is Option (1) → Maximize $Z = x + y$; Subject to: $4x + 5y \leq 200, \enspace 8x + 5y \leq 200, \enspace x, y \geq 0$ ##

Let the distance covered with speed of 25 km/h = $x$ km 

and the distance covered with speed of 40 km/h = $y$ km. 

Total distance covered = $z$ km

The LPP of the above problem, therefore, is

Maximize $z = x + y$

Subject to constraints

$4x + 5y \leq 200$

$\frac{x}{50} + \frac{y}{40} \leq 1$

and $\frac{x}{25} + \frac{y}{40} \leq 1$

and $x \geq 0, \quad y \geq 0$