Question

If a young man rides his motorcycle at 25 km/hr, he had to spend Rs.2 per km on petrol. If he rides at a faster speed of 40 km/hr, the petrol cost increases at Rs.5 per km. He has Rs.100 to spend on petrol and wishes to find what is the maximum distance he can travel within one hour. Express this as LPP and solve it graphically.

Answer 1

Answer:

The maximum distance a man can travel is 30 km.

Step-by-step explanation:

Let x denote the number of km he traveled when the speed is: 25 km/hr.

and y denote the number of km he travels when the speed is 40 km/hr.

so, the total distance travelled is given by:

z=(x+y) km

Also,  He has Rs.100 to spend on petrol

This means that:

2x+5y≤100

Now, time taken to travel x  km is:

$\dfrac{x}{25}hr$

since, Time is the ratio of distance and speed.

similarly,

Now, time taken to travel y  km is:

$\dfrac{y}{40}hr$

$\dfrac{x}{25}+\dfrac{y}{40}\leq 1$

on solving we have:

$8x+5y\leq 200$

Also x≥0, y≥0.

so, we have to maximize the objective function:

Maximize z=x+y

subject to the constraint:

2x+5y≤100

8x+5y≤200

x≥0 , y≥0.

so, on solving graphically,

we have the Feasible region as ABCD vertex points as:

A(0,0) , B(0,20) , C(25,0) and D(50/3,40/3)

on checking the objective function at the end points we have:

Points             z

(0,0)                0

(0,20)              20

(25,0)               25

(50/3,40/3)      90/3=30

Hence, the maximum distance a man can travel is 30 km.

And at this distance the distance traveled by man at the speed of 25km/hr is 50/3 km.

and the  distance the distance traveled by man at the speed of 40km/hr is 40/3 km.

Answer 2

$\textbf{Answer is in Attachment !}$