Login/Sign Up
Home » A man rides his motorcycle at the speed of     km/hour. He has to spend Rs     per km on petrol. If he rides it at a faster speed of     km/hour, the petrol cost increases to Rs     per km. He has at most Rs     to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.
A man rides his motorcycle at the speed of

    \[50\]

km/hour. He has to spend Rs

    \[2\]

per km on petrol. If he rides it at a faster speed of

    \[80\]

km/hour, the petrol cost increases to Rs

    \[3\]

per km. He has at most Rs

    \[120\]

to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.
CBSE | Class 12 | Exercise 1 | Linear Programming | Maths | NCERT Exemplar
A man rides his motorcycle at the speed of

    \[50\]

km/hour. He has to spend Rs

    \[2\]

per km on petrol. If he rides it at a faster speed of

    \[80\]

km/hour, the petrol cost increases to Rs

    \[3\]

per km. He has at most Rs

    \[120\]

to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.

Let’s assume the man covers x km on his motorcycle at the speed of

    \[50\]

km/hr and covers y km at the speed of

    \[50\]

km/hr and covers y km at the speed of

    \[80\]

km/hr.

So, cost of petrol =

    \[2x\text{ }+\text{ }3y\]

The man has to spend Rs

    \[120\]

atmost on petrol

⇒

    \[2x\text{ }+\text{ }3y\le 120\]

…. (i)

Now, the man has only

    \[1\]

hr time

So,

    \[x/50\text{ }+\text{ }y/80\le 1\Rightarrow 8x\text{ }+\text{ }5y\le 400\]

… (ii)

And,

    \[x\ge 0,\text{ }y\ge 0\]

To have maximum distance Z = x + y.

Therefore, the required LPP to travel maximum distance is maximize Z = x + y, subject to the constraints

    \[2x\text{ }+\text{ }3y\le 120,\text{ }8x\text{ }+\text{ }5y\le 400,\text{ }x\ge 0,\text{ }y\ge 0\]

.

i

More Material