Question

If a young man drives his vehicle at 25 km/hr, he has to spend ₹2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to ₹5 per km. He has ₹100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same. [CBSE 2007]

Answer 1

Let us assume that the man travels x km when the speed is 25 km/hour and y km when the speed is 40 km/hour.

Thus, the total distance travelled is (x + y) km.

Now, it is given that the man has Rs 100 to spend on petrol.

Total cost of petrol = 2x + 5y ≤ 100

Now, time taken to travel x km = $\frac{x}{25}$ h

Time taken to travel y km = $\frac{y}{40}$ h

Now, it is given that the maximum time is 1 hour. So,

$$\begin{gathered} \frac{x}{25}+\frac{y}{40}\le 1 \\ \Rightarrow 8x+5y\le 200 \end{gathered}$$

Thus, the given linear programming problem is

Maximise Z = x + y

subject to the constraints

2x + 5y ≤ 100

8x + 5y ≤ 200

x ≥ 0, y ≥ 0

The feasible region determined by the given constraints can be diagrammatically represented as,



The coordinates of the corner points of the feasible region are O(0, 0), A(25, 0), B$\left(\frac{50}{3},\frac{40}{3}\right)$ and C(0, 20).

The value of the objective function at these points are given in the following table.

Corner Points	Z = x + y
(0, 0)	0 + 0 = 0
(25, 0)	25 + 0 = 25
$\left(\frac{50}{3},\frac{40}{3}\right)$	$\frac{50}{3}+\frac{40}{3}=30$
(0, 20)	0 + 20 = 20

So, the maximum value of Z is 30 at $x=\frac{50}{3},y=\frac{40}{3}$.

Thus, the maximum distance that the man can travel in one hour is 30 km.

Hence, the distance travelled by the man at the speed of 25 km/hour is $\frac{50}{3}$ km, and the distance travelled by him at the speed of 40 km/hour is $\frac{40}{3}$km.