To solve the problem step by step, we will express it as a linear programming problem and find the maximum distance the young man can travel within 1 hour while adhering to the constraints of petrol cost and time.
### Step 1: Define Variables
Let:
- \( x \) = distance traveled at 25 km/hr (in km)
- \( y \) = distance traveled at 40 km/hr (in km)
### Step 2: Set Up the Objective Function
We want to maximize the total distance traveled, which can be expressed as:
\[
Z = x + y
\]
### Step 3: Establish Constraints
1. **Cost Constraint**: The cost of petrol for traveling \( x \) km at 25 km/hr is Rs 2 per km, and for \( y \) km at 40 km/hr, it is Rs 5 per km. The total cost must not exceed Rs 100:
\[
2x + 5y \leq 100
\]
2. **Time Constraint**: The total time spent traveling must not exceed 1 hour. The time taken to travel \( x \) km at 25 km/hr is \( \frac{x}{25} \) hours, and the time taken to travel \( y \) km at 40 km/hr is \( \frac{y}{40} \) hours. Therefore:
\[
\frac{x}{25} + \frac{y}{40} \leq 1
\]
### Step 4: Convert Time Constraint to Standard Form
To eliminate the fractions in the time constraint, we can multiply through by the least common multiple of 25 and 40, which is 200:
\[
8x + 5y \leq 200
\]
### Step 5: Identify the Feasible Region
We now have the following constraints:
1. \( 2x + 5y \leq 100 \)
2. \( 8x + 5y \leq 200 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)
### Step 6: Find Intersection Points
To find the feasible region, we need to find the intersection points of the constraints.
1. **For \( 2x + 5y = 100 \)**:
- When \( x = 0 \): \( 5y = 100 \) → \( y = 20 \) → Point (0, 20)
- When \( y = 0 \): \( 2x = 100 \) → \( x = 50 \) → Point (50, 0)
2. **For \( 8x + 5y = 200 \)**:
- When \( x = 0 \): \( 5y = 200 \) → \( y = 40 \) → Point (0, 40)
- When \( y = 0 \): \( 8x = 200 \) → \( x = 25 \) → Point (25, 0)
3. **Finding the intersection of the two lines**:
- Solve the equations:
\[
2x + 5y = 100 \quad (1)
\]
\[
8x + 5y = 200 \quad (2)
\]
- Subtract (1) from (2):
\[
(8x + 5y) - (2x + 5y) = 200 - 100
\]
\[
6x = 100 \quad \Rightarrow \quad x = \frac{100}{6} = \frac{50}{3}
\]
- Substitute \( x = \frac{50}{3} \) into (1):
\[
2\left(\frac{50}{3}\right) + 5y = 100
\]
\[
\frac{100}{3} + 5y = 100 \quad \Rightarrow \quad 5y = 100 - \frac{100}{3} = \frac{200}{3}
\]
\[
y = \frac{200}{15} = \frac{40}{3}
\]
- So the intersection point is \( \left(\frac{50}{3}, \frac{40}{3}\right) \).
### Step 7: Evaluate the Objective Function at Corner Points
Now we evaluate \( Z = x + y \) at the corner points:
1. \( (0, 0) \): \( Z = 0 + 0 = 0 \)
2. \( (0, 20) \): \( Z = 0 + 20 = 20 \)
3. \( (50, 0) \): \( Z = 50 + 0 = 50 \)
4. \( \left(\frac{50}{3}, \frac{40}{3}\right) \):
\[
Z = \frac{50}{3} + \frac{40}{3} = \frac{90}{3} = 30
\]
5. \( (25, 0) \): \( Z = 25 + 0 = 25 \)
### Step 8: Determine Maximum Value
The maximum value of \( Z \) occurs at the point \( (50, 0) \) where \( Z = 50 \).
### Conclusion
The maximum distance the young man can travel within 1 hour, given the constraints, is **50 km** at a speed of 25 km/hr.
