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) $ \dfrac{\alpha }{\beta } $
(B) $ \dfrac{\beta }{\alpha } $
(C) $ \sqrt {\dfrac{\alpha }{\beta }} $
(D) $ \sqrt {\dfrac{\beta }{\alpha }} $

Answer 1

Hint: This question can be solved by applying the solving methods of the Maxima and Minima. According to this method the maximum or minimum value of a variable is calculated by differentiating the variable. From the first differentiation the value of the variable is calculated and from the second differentiation it is determined whether the function is maximum or minimum. If the value of the second differential is negative or less than zero then the function is maximum and if the value of the second differential is positive or greater than zero then the function is minimum.

Complete step-by-step answer:
Given:
The number of characters on magnetic tape $ = N $
The number of characters on each batch $ = x $
So, the number of batches $ = \dfrac{N}{x} $
The processing time per batch $ = \left( {\alpha + \beta {x^2}} \right){\rm{ seconds}} $
Where, $ \alpha ,\beta $ are two constants.
So, the total time in processing each batch T can be calculated by –
 $
\Rightarrow T = \dfrac{N}{x} \times \left( {\alpha + \beta {x^2}} \right)\\
 = N\left( {\dfrac{\alpha }{x} + \beta x} \right){\rm{ seconds}}
 $
Now the number of characters on each batch x is optimum when the value of the processing time is least.
Now to find the least processing time we have to use the Maxima and Minima method. In this method we can find the maximum or minimum value of the function by differentiating the function with respect to the changing variable and then equate the differential with the zero.
Now differentiating the both sides of the above expression with respect to x we get,
 $\Rightarrow \dfrac{{dT}}{{dx}} = N\dfrac{d}{{dx}}\left( {\dfrac{\alpha }{x} + \beta x} \right) $
This is our first equation.
For least processing time the value of $ \dfrac{{dT}}{{dx}} $ should be zero. So, substituting $ \dfrac{{dT}}{{dx}} = 0 $ in the first equation we get,
 $\Rightarrow 0 = N\dfrac{d}{{dx}}\left( {\dfrac{\alpha }{x} + \beta x} \right) $
Now solving for the value of x we get,
 $ $ \[
\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{\alpha }{x} + \beta x} \right) = 0\\
\Rightarrow \left( {\dfrac{{ - \alpha }}{{{x^2}}} + \beta } \right) = 0\\
\Rightarrow \beta = \dfrac{\alpha }{{{x^2}}}\\
{x^2} = \dfrac{\alpha }{\beta }
\]
Taking square roots of both sides we get,
\[\Rightarrow x = \sqrt {\dfrac{\alpha }{\beta }} \]
Now to check whether the value of T is minimum or maximum, differentiating first equation with request to x again we get,
 $
\Rightarrow \dfrac{{{d^2}T}}{{d{x^2}}} = N\dfrac{{{d^2}}}{{d{x^2}}}\left( {\dfrac{\alpha }{x} + \beta x} \right)\\
 = N\dfrac{d}{{dx}}\left( {\dfrac{{ - \alpha }}{{{x^2}}} + \beta } \right)\\
 = N\left( {\dfrac{{2a}}{{{x^3}}}} \right)
 $
Since, the value of $ N\left( {\dfrac{{2a}}{{{x^3}}}} \right) $ is greater than zero, which means that “the value of T is minimum”.
Therefore, the optimum value of x for fast processing is $ \sqrt {\dfrac{\alpha }{\beta }} $ and the correct option is –
(C) $ \sqrt {\dfrac{\alpha }{\beta }} $

So, the correct answer is “Option C”.

Note: It should be noted that we do not need to find the second differential if it is given in the question that the function has a maximum or minimum value. The second differential is only performed to check whether the value of the function is maximum or minimum.