Question

If a 20 year old girl drives her car at 25km/hr, she has to spend Rs.4/km on petrol. If she drives her car at 40km/hr the petrol cost increases to Rs.5/km. She has Rs.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.

Answer 1

Hint:
In this question, we are given a condition and statement in which a 20-year-old girl drives her car. We need to form a linear programming problem, for such a question if we need to maximize get distance traveled in 1 hour. We will find the objective function and then the constraints. Both object function and constraints will give us our required linear programming problem. We will also use the formula: $\text{Distance}=\text{Speed}\times \text{Time}$.

Complete step by step answer:
Here a 20-year-old girl is driving her car under some statement and she wishes to maximize her distance in 1 hour under the given condition. We need to form a linear programming problem for that.
Let us assume that, the distance covered with speed 25km/hr as x km and the distance covered with the speed of 40km/hr to y km.
Let us assume that z km is the total distance covered.
According to the statement, she wishes to maximize z i.e. Maximize z = x+y which is our objective function.
Now let us find the constraints.
We are given that she could spend only Rs.200 on petrol.
It costs Rs.4km/hr when the speed is 25km/hr. Therefore it costs Rs.4km/hr for x km. So it will cost Rs.4x for x km.
It costs Rs.5km/hr for y km. So it will cost Rs.5y for y km.
Her budget is for Rs.200. So we have total cost as less than or equal to 200 we get,
$4x+5y\le 200$.
Now for x distance, speed is 25km/hr so time will be given by $\dfrac{\text{distance}}{\text{Speed}}\text{ i}\text{.e }\dfrac{x}{25}$.
For y distance, speed is 40km/hr so time will be $\dfrac{y}{40}$.
She has to cover the distance within 1 hour, so total time should be less than or equal to 1 hour. Therefore we have, $\dfrac{x}{25}+\dfrac{y}{40}\le 1$.
Taking LCM as 200 we get, $\dfrac{8x+5y}{200}\le 1\Rightarrow 8x+5y\le 200$.
Also we know that distance cannot be negative, so $x\ge 0,y\ge 0$.
Combining all we get constraints as $4x+5y\le 200,8x+5y\le 200\text{ and }x\ge 0,y\ge 0$.
So our linear programming problem looks like this, Maximize z = x+y.
Subject to the constraints,
$\begin{align}
  & 4x+5y\le 200 \\
 & 8x+5y\le 200 \\
 & x\ge 0,y\ge 0 \\
\end{align}$.

Note:
 Students should carefully consider every limitation and convert it into a constraint for getting a proper answer. Students often forget about taking $x\ge 0,y\ge 0$ but it is compulsory in a linear programming problem. Try to simplify the constraints as much as possible.