$A$ man rides his motorcycle at the speed of $50 \, km/h$. He has to spend $Rs. \, 2$ per $km$ on petrol. If he rides it at a faster speed of $80 \, km/h$,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.
(A) Let the man ride his motorcycle for a distance $x \, km$ at the speed of $50 \, km/h$ and for a distance $y \, km$ at the speed of $80 \, km/h$.
The cost of petrol for distance $x$ is $2x$ and for distance $y$ is $3y$. Since he has at most $Rs. \, 120$ to spend,the cost constraint is $2x + 3y \leq 120$.
The time taken to travel distance $x$ is $\frac{x}{50}$ hours and for distance $y$ is $\frac{y}{80}$ hours. Since he has at most $1$ hour,the time constraint is $\frac{x}{50} + \frac{y}{80} \leq 1$,which simplifies to $8x + 5y \leq 400$.
Since distance cannot be negative,$x \geq 0$ and $y \geq 0$.
The objective is to maximize the total distance $Z = x + y$.
Thus,the linear programming problem is:
Maximize $Z = x + y$
Subject to:
$2x + 3y \leq 120$
$8x + 5y \leq 400$
$x, y \geq 0$
The cost of petrol for distance $x$ is $2x$ and for distance $y$ is $3y$. Since he has at most $Rs. \, 120$ to spend,the cost constraint is $2x + 3y \leq 120$.
The time taken to travel distance $x$ is $\frac{x}{50}$ hours and for distance $y$ is $\frac{y}{80}$ hours. Since he has at most $1$ hour,the time constraint is $\frac{x}{50} + \frac{y}{80} \leq 1$,which simplifies to $8x + 5y \leq 400$.
Since distance cannot be negative,$x \geq 0$ and $y \geq 0$.
The objective is to maximize the total distance $Z = x + y$.
Thus,the linear programming problem is:
Maximize $Z = x + y$
Subject to:
$2x + 3y \leq 120$
$8x + 5y \leq 400$
$x, y \geq 0$
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