Problem 31 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

N characters of information are held on magnetic tape, in batches of \(x\) characters each; the batch processing time is \(\alpha+\beta x^{2}\) seconds; \(\alpha, \beta\) are constants. The optimum value of \(x\) for fast processing is (A) \(\frac{\alpha}{\beta}\) (B) \(\frac{\beta}{\alpha}\) (C) \(\sqrt{\frac{\alpha}{\beta}}\) (D) \(\sqrt{\frac{\beta}{\alpha}}\)

Step-by-Step Solution

Answer

The optimum value of \(x\) is \( \sqrt{\frac{\alpha}{\beta}} \) (Option C).

Step 1: Define the Problem

We need to find the optimum value of batch size \(x\) to minimize the total processing time for \(N\) characters given the batch processing time formula \(T = \alpha + \beta x^2\).

Step 2: Determine Total Processing Time

The total number of batches is \(\frac{N}{x}\). Thus, the total processing time \(T_{total}\) can be expressed as: \[ T_{total} = \left(\frac{N}{x}\right)(\alpha + \beta x^2) = \frac{N}{x}\alpha + N\beta x. \]

Step 3: Apply Calculus for Optimization

To find the optimum value of \(x\), we need to minimize \(T_{total}\). This involves taking the derivative of \(T_{total}\) with respect to \(x\) and setting it to zero: \[ \frac{d}{dx}\left( \frac{N}{x}\alpha + N\beta x \right) = -\frac{N\alpha}{x^2} + N\beta = 0. \]

Step 4: Solve the Derivative Equation

Setting the derivative to zero gives: \[ -\frac{N\alpha}{x^2} + N\beta = 0. \] Solve for \(x\): \[ \frac{N\alpha}{x^2} = N\beta \implies \frac{\alpha}{x^2} = \beta \implies x^2 = \frac{\alpha}{\beta}. \] Taking the square root, we find: \[ x = \sqrt{\frac{\alpha}{\beta}}. \]

Step 5: Verify the Nature of the Critical Point

To ensure it is a minimum, check the second derivative: \[ \frac{d^2}{dx^2}\left( \frac{N}{x}\alpha + N\beta x \right) = \frac{2N\alpha}{x^3}. \] Since \(\frac{2N\alpha}{x^3} > 0\), the function has a minimum at \( x = \sqrt{\frac{\alpha}{\beta}} \).

Key Concepts

Calculus Optimization

Calculus Optimization involves finding the best solution to a problem using calculus principles. In our context, we are interested in minimizing the total processing time for data held on magnetic tape. This involves optimizing the batch size, or the number of characters processed together. In mathematical terms, optimization refers to the process of finding maxima or minima of a function. Determining whether a point is a maximum or minimum is accomplished by employing calculus techniques, notably derivatives. For the batch processing scenario, we need to minimize the total time function: \[ T_{total} = \frac{N}{x}\alpha + N\beta x. \] By applying calculus, particularly through differentiation, we identify the critical point of this function. A critical point is where the function's derivative equals zero. Calculus helps us both find and classify these points.

Processing Time Minimization

Processing Time Minimization is central in computing and data handling environments, where the speed of operations can determine the effectiveness of a solution. When we minimize processing time, we are essentially optimizing system performance.In our batch processing example, the need is to minimize the total detailed by: \[ T_{total} = \frac{N}{x}\alpha + N\beta x. \] This requires determine the best size for \(x\), or the batch size, which translates into faster data processing. It means balancing between too few batches that take a long time individually due to increasing \(x\), and too many small batches that each have considerable setup time with constant \(\alpha\).By correctly applying minimization techniques, systems can achieve a sweet spot where processing time is minimized. This enhances throughput and efficiency without unnecessarily utilizing resources.

Derivative Applications

Derivatives are a fundamental concept in calculus, used extensively in optimization. They measure the rate of change of a function relative to one of its variables. By understanding how a function changes, we can make informed decisions about optimization.For batch processing, we derive the total processing time \(T_{total}\) with respect to batch size \(x\). The derivative \[ \frac{d}{dx}\left( \frac{N}{x}\alpha + N\beta x \right) = -\frac{N\alpha}{x^2} + N\beta. \] Setting the derivative to zero allows us to find the critical point, indicating potential minima or maxima. Solving this provides the optimal \(x\) value:\[ x = \sqrt{\frac{\alpha}{\beta}}. \] To confirm it's a minimum, we calculate the second derivative: \[ \frac{d^2}{dx^2}\left( \frac{N}{x}\alpha + N\beta x \right) = \frac{2N\alpha}{x^3}. \] Since it is positive, it confirms that we have a minimum at this point. Understanding derivatives enables us to optimize processes like batch processing effectively.