Question

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

Answer 1

Let young man drives x km at a speed of 25 km/hr and y km at a speed of $40\mathrm{km}/\mathrm{hr}$.

Clearly, $x,y\ge 0$

It is given that, he spends Rs 2 per km if he drives at a speed of $25\mathrm{km}/\mathrm{hr}$ and Rs 5 per km if he drives at a speed of $40k\mathrm{m}/\mathrm{hr}$. Therefore, money spent by him when he travelled x km and y km is Rs 2x and Rs 5y respectively.

It is given that he has a maximum of Rs 100 to spend.

Thus, $2x+5y\le 100$

$$\begin{gathered} \mathrm{Time}\mathrm{spent}\mathrm{by}\mathrm{him}\mathrm{when}\mathrm{travelling}\mathrm{with}\mathrm{a}\mathrm{speed}\mathrm{of}25\mathrm{km}/\mathrm{hr}=\frac{x}{25}\mathrm{hr} \\ \mathrm{Time}\mathrm{spent}\mathrm{by}\mathrm{him}\mathrm{when}\mathrm{travelling}\mathrm{with}\mathrm{a}\mathrm{speed}\mathrm{of}40\mathrm{km}/\mathrm{hr}=\frac{x}{40}\mathrm{hr} \end{gathered}$$

Also, the available time is of 1 hour.

$$\begin{gathered} \frac{x}{25}+\frac{y}{40}\le 1 \\ \Rightarrow 40x+25y\le 1000 \end{gathered}$$


The distance covered is Z = $x+y$ which is to be maximised.

Thus, the mathematical formulat​ion of the given linear programmimg problem is

Max Z = $x+y$

subject to

$$\begin{gathered} 2x+5y\le 100 \\ 40x+25y\le 1000 \end{gathered}$$
$x,y\ge 0$

First we will convert inequations into equations as follows:
2x + 5y = 100, 40x + 25y = 1000, x = 0 and y = 0

Region represented by 2x + 5y ≤ 100:
The line 2x + 5y = 100 meets the coordinate axes at $A\left(50,0\right)$ and $B\left(0,20\right)$ respectively. By joining these points we obtain the line
2x + 5y = 100. Clearly (0,0) satisfies the 2x + 5y = 100. So,the region which contains the origin represents the solution set of the inequation 2x + 5y ≤ 100.

Region represented by 40x + 25y ≤ 1000:
The line 40x + 25y = 1000 meets the coordinate axes at $C\left(25,0\right)$ and $D\left(0,40\right)$ respectively. By joining these points we obtain the line
2x + y = 12.Clearly (0,0) satisfies the inequation 40x + 25y ≤ 1000. So,the region which contains the origin represents the solution set of the inequation 40x + 25y ≤ 1000.


Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x + 5y ≤ 100, 40x + 25y ≤ 1000, x ≥ 0, and y ≥ 0 are as follows



The corner points are O(0, 0), B(0, 20), $E\left(\frac{50}{3},\frac{40}{3}\right)$ and C(25, 0).

The values of Z at these corner points are as follows

Corner point	Z = x + y
O	0
B	20
E	30
C	25

The maximum value of Z is 30 which is attained at E.

Thus, the maximum distance travelled by the young man is 30 kms, if he drives $\frac{50}{3}\mathrm{km}$ at a speed of $25\mathrm{km}/\mathrm{hr}$ and $\frac{40}{3}\mathrm{km}$ at a speed of $40\mathrm{km}/\mathrm{hr}$ .