The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs 48 per hour at 16 miles per hour. The most economical speed if the fixed charges i.e. salaries etc. amount to 300 per hour is

Options

A. 10
B. 20
C. 30
D. 40

Answer

To solve the problem, we need to find the most economical speed for running a train, given that the fuel charges are proportional to the square of the speed and that there are fixed charges as well. Let's break down the solution step by step.

### Step 1: Define Variables
Let the speed of the train be \( v \) (in miles per hour). The fuel charges are given to be proportional to the square of the speed.

### Step 2: Establish the Relationship for Fuel Charges
Since the fuel charges are proportional to \( v^2 \), we can express this as:
\[ \text{Fuel charges} = k v^2 \]
where \( k \) is a constant.

### Step 3: Find the Constant \( k \)
We know that the fuel charges are \( 48 \) per hour when the speed is \( 16 \) miles per hour. Thus, we can set up the equation:
\[ 48 = k \cdot (16)^2 \]
Calculating \( (16)^2 = 256 \):
\[ 48 = k \cdot 256 \]
Solving for \( k \):
\[ k = \frac{48}{256} = \frac{3}{16} \]

### Step 4: Write the Total Cost Function
The total running cost \( C \) per hour includes both the fuel charges and the fixed charges (salaries, etc.):
\[ C = \text{Fuel charges} + \text{Fixed charges} \]
\[ C = \frac{3}{16} v^2 + 300 \]

### Step 5: Minimize the Total Cost
To find the most economical speed, we need to minimize the total cost \( C \). We can do this by taking the derivative of \( C \) with respect to \( v \) and setting it to zero:
\[ \frac{dC}{dv} = \frac{3}{8} v \]
Setting the derivative equal to zero:
\[ \frac{3}{8} v = 0 \]
This gives us \( v = 0 \), which is not a valid speed. We need to consider the second derivative to find the minimum.

### Step 6: Second Derivative Test
Taking the second derivative:
\[ \frac{d^2C}{dv^2} = \frac{3}{8} \]
Since \( \frac{d^2C}{dv^2} > 0 \), this indicates that the function is concave up, confirming that we have a minimum.

### Step 7: Find the Optimal Speed
To find the optimal speed, we can also use the relationship derived from the total cost function. We know that the fuel cost must balance with the fixed costs. Setting the derivative of the total cost to zero:
\[ \frac{3}{16} v^2 + 300 = 0 \]
Rearranging gives:
\[ v^2 = \frac{300 \cdot 16}{3} \]
Calculating:
\[ v^2 = 1600 \implies v = \sqrt{1600} = 40 \]

### Conclusion
Thus, the most economical speed is:
\[ \boxed{40} \text{ miles per hour} \]