Cost Minimization for Longwall Face

The total cost $C$ (in lakh rupees) of a longwall face of length $L$ (in m) is given by the function:

$ C(L) = 0.1L + \frac{1562.5}{L} + 300 $

To find the length $L$ that minimizes the total cost $C$, we need to use calculus. This involves finding the first derivative of the cost function with respect to $L$, setting it to zero, and solving for $L$.

Step 1: Calculate the First Derivative

Find the derivative of $C$ with respect to $L$ ($ \frac{dC}{dL} $):

$ \frac{dC}{dL} = \frac{d}{dL} \left( 0.1L + 1562.5L^{-1} + 300 \right) $

$ \frac{dC}{dL} = 0.1 - 1562.5L^{-2} $

$ \frac{dC}{dL} = 0.1 - \frac{1562.5}{L^2} $

Step 2: Set Derivative to Zero and Solve for L

To find the critical points (where the cost might be minimum or maximum), set the first derivative equal to zero:

$ 0.1 - \frac{1562.5}{L^2} = 0 $

Rearrange the equation to solve for $ L^2 $:

$ 0.1 = \frac{1562.5}{L^2} $

$ L^2 = \frac{1562.5}{0.1} $

$ L^2 = 15625 $

Now, take the square root to find $L$:

$ L = \sqrt{15625} $

$ L = 125 $

Step 3: Conclusion

The calculation shows that the critical value for $L$ is 125 m. To confirm this is a minimum, we can check the second derivative ($ \frac{d^2C}{dL^2} = \frac{3125}{L^3} $), which is positive for positive $L$, indicating a minimum cost. Therefore, the length of the face for the minimum total cost is 125 m.