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(C) The total number of batches is $\frac{N}{x}$.The time taken per batch is $(\alpha + \beta x^2)$ seconds.Therefore,the total processing time $T$ is

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$\sqrt{\frac{\alpha}{\beta}}$

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(C) The total number of batches is $\frac{N}{x}$.The time taken per batch is $(\alpha + \beta x^2)$ seconds.Therefore,the total processing time $T$ is given by $T = \frac{N}{x}(\alpha + \beta x^2) = N(\frac{\alpha}{x} + \beta x)$.For fast processing,$T$ must be minimized.Taking the derivative with respect to $x$: $\frac{dT}{dx} = N(-\frac{\alpha}{x^2} + \beta)$.Setting $\frac{dT}{dx} = 0$ for critical points: $-\frac{\alpha}{x^2} + \beta = 0 \implies x^2 = \frac{\alpha}{\beta} \implies x = \sqrt{\frac{\alpha}{\beta}}$.Checking the second derivative: $\frac{d^2T}{dx^2} = N(\frac{2\alpha}{x^3})$. Since $x, \alpha, \beta > 0$,$\frac{d^2T}{dx^2} > 0$,which confirms that $T$ is minimum at $x = \sqrt{\frac{\alpha}{\beta}}$.

$N$ characters of information are held on magnetic tape,in batches of $x$ characters each; the batch processing time is $\alpha + \beta x^2$ seconds; $\alpha$ and $\beta$ are constants. The optimal value of $x$ for fast processing is

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