Question

Find the maximum value of the function $\dfrac{{\log x}}{x}$ in the interval $(2,\infty )$.
A. $1$
B. $\dfrac{2}{e}$
C. $e$
D. $\dfrac{1}{e}$

Answer 1

Hint: We are given the function and we have to find the maximum value in the given interval. We will use the concept of application of derivatives in the given question. So firstly we will differentiate the value and put it equal to zero and by double differentiation we will check and find the maxima and minima of the function.

Formula Used:
The following derivative formula are useful for this question
 $\dfrac{d}{{dx}}(\dfrac{u}{v}) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}} \\ \dfrac{{d(\log x)}}{{dx}} = \dfrac{1}{x} \\ \dfrac{{dx}}{{dx}} = 1$

Complete step by step solution:
Let the given function be equal to $y$ and differentiate the function with respect to $x$
$y{\rm{ }} = {\rm{ }}\dfrac{{log x}}{x}$
Firstly we will find the derivative of function with respect to $x$
$\dfrac{{dy}}{{dx}}{\rm{ }} = {\rm{ }}\dfrac{{\left[ {x\dfrac{{d(\log x)}}{{dx}}{\rm{ }}-{\rm{ }}\dfrac{{dx}}{{dx}}logx} \right]}}{{{x^2}}}{\rm{ }} \\ \dfrac{{dy}}{{dx}} = \dfrac{{[\dfrac{x}{x} - (1)\log x]}}{{{x^2}}} \\ \dfrac{{dy}}{{dx}} = \dfrac{{1 - \log x}}{{{x^2}}}$
Now we will simplify the equation by putting $\dfrac{{dy}}{{dx}}$ equal to zero
$\dfrac{{1 - \log x}}{{{x^2}}} = 0\\1 - \log x = 0 \\ \log x = 1\\x = e$
Now, we will double differentiate to check whether the function is maximum or minimum.
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{{x^2}\dfrac{{d[1 - \log x]}}{{dx}} - \left( {1 - \log x\left[ {\dfrac{{d{x^2}}}{{dx}}} \right]} \right)}}{{{{[{x^2}]}^2}}} \\ \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\dfrac{{ - {x^2}}}{x} - \left( {1 - \log x[2x]} \right)}}{{{x^4}}} \\ \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - x - 2x + 2x(\log x)}}{{{x^4}}} \\ \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 3x + 2x(\log x)}}{{{x^4}}}$
Now, we will check at $x = e$ the double derivative is greater or smaller than zero to find the function is maximum or minimum
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 3x + 2x(\log x)}}{{{x^4}}} \\ \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 3e + 2e(\log e)}}{{{e^4}}} \\ \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - e}}{{{e^4}}} \\ \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 1}}{{{e^3}}}$
We can see that at $x = e$ the double derivative is smaller than zero. Therefore the value of the function is maximum at $x = e$
Now, we will put $x = e$ in function to get the maximum value of the function
$y = \dfrac{{\log x}}{x} \\ y = \dfrac{{\log e}}{e} \\ y = \dfrac{1}{e}$
Hence, the maximum value of the function $\dfrac{{\log x}}{x}$ in the interval $(2,\infty )$ is $\dfrac{1}{e}$.

Option ‘D’ is correct

Note: Here, finding the maxima and minima of the function is easy. The thing to keep in mind is that if the double derivative is greater than zero, then the function is minimum at that point and if it is less than zero, then it is maximum at that point.