If a young man drives his car at 40 km per hour, he has to spend ₹ 5 per km on petrol, if he drives it at a slower speed of 25 km per hour, the petrol cost decreases to ₹ 2 per km. He has ₹ 100 to spend on petrol and wishes to find thhe maximum distance he can travel within one hour.Express this as a linear programming problem and then solve it.

Answer

To solve the problem step by step, we will express it as a linear programming problem and then find the maximum distance the young man can travel within one hour.

### Step 1: Define the Variables
Let:
- \( x \) = distance traveled at 40 km/h (in km)
- \( y \) = distance traveled at 25 km/h (in km)

### Step 2: Set Up the Constraints
1. **Time Constraint**: The total time spent traveling should not exceed 1 hour.
\[ \frac{x}{40} + \frac{y}{25} \leq 1 \]
To eliminate the fractions, we can multiply the entire inequality by 200 (the LCM of 40 and 25):
\[ 5x + 8y \leq 200 \]

2. **Cost Constraint**: The total cost of petrol should not exceed ₹100.
\[ 5x + 2y \leq 100 \]

3. **Non-negativity Constraints**: Both distances must be non-negative.
\[ x \geq 0, \quad y \geq 0 \]

### Step 3: Formulate the Objective Function
We want to maximize the total distance traveled:
\[ Z = x + y \]

### Step 4: Graph the Constraints
1. **Graph the first constraint** \( 5x + 8y \leq 200 \):
- Find the intercepts:
- \( x \)-intercept: Set \( y = 0 \) → \( 5x = 200 \) → \( x = 40 \)
- \( y \)-intercept: Set \( x = 0 \) → \( 8y = 200 \) → \( y = 25 \)
- Plot the line connecting (40, 0) and (0, 25).

2. **Graph the second constraint** \( 5x + 2y \leq 100 \):
- Find the intercepts:
- \( x \)-intercept: Set \( y = 0 \) → \( 5x = 100 \) → \( x = 20 \)
- \( y \)-intercept: Set \( x = 0 \) → \( 2y = 100 \) → \( y = 50 \)
- Plot the line connecting (20, 0) and (0, 50).

3. **Identify the feasible region**: The area where both inequalities are satisfied, along with the non-negativity constraints.

### Step 5: Find the Corner Points of the Feasible Region
The corner points can be found by solving the equations of the lines:
1. **Intersection of \( 5x + 8y = 200 \) and \( 5x + 2y = 100 \)**:
- Set the equations equal to each other:
\[ 5x + 8y = 200 \quad \text{(1)} \]
\[ 5x + 2y = 100 \quad \text{(2)} \]
Subtract (2) from (1):
\[ 6y = 100 \quad \Rightarrow \quad y = \frac{100}{6} \approx 16.67 \]
Substitute \( y \) back into (2):
\[ 5x + 2 \left(\frac{100}{6}\right) = 100 \quad \Rightarrow \quad 5x + \frac{200}{6} = 100 \]
\[ 5x = 100 - \frac{200}{6} \quad \Rightarrow \quad 5x = \frac{600 - 200}{6} = \frac{400}{6} \quad \Rightarrow \quad x = \frac{80}{6} \approx 13.33 \]
Thus, one corner point is \( \left(\frac{80}{6}, \frac{100}{6}\right) \).

2. **Other corner points**: (0, 25), (20, 0), and (40, 0).

### Step 6: Evaluate the Objective Function at Each Corner Point
1. At \( (0, 25) \): \( Z = 0 + 25 = 25 \)
2. At \( (20, 0) \): \( Z = 20 + 0 = 20 \)
3. At \( (40, 0) \): \( Z = 40 + 0 = 40 \)
4. At \( \left(\frac{80}{6}, \frac{100}{6}\right) \):
\[ Z = \frac{80}{6} + \frac{100}{6} = \frac{180}{6} = 30 \]

### Step 7: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point \( (40, 0) \) where \( Z = 40 \).

### Conclusion
The maximum distance the young man can travel in one hour is **40 km**.