The maximum value of $y=x(\log x)^2$ is
The maximum value of $y=x(\log x)^2$ is
- $e^{-2}$
- $2 e^{-2}$
- $3 e^{-2}$
- $4 e^{-2}$
Solution
$y=x(\log x)^2$
For maximum value
$
\begin{aligned}
\frac{d y}{d x} & =x\left(2 \log x \cdot \frac{1}{x}\right)+(\log x)^2 \\
& =2 \log x+(\log x)^2
\end{aligned}
$
Now, put $\frac{d y}{d x}=0$
$
\begin{aligned}
& 2 \log x+(\log x)^2=0 \\
& (\log x)[\log x+2]=0 \\
& \log x=0 \mid \log x+2=0 \\
& x=1 \\
& \log x=-2 \\
& x=e^{-2} \\
&
\end{aligned}
$
Using second order derivative test. Again differentiating $y$ w.r.t. $x$
$
\begin{aligned}
& \frac{d^2 y}{d x^2}=\frac{2}{x}+2 \log x \cdot \frac{1}{x} \\
&\left.\frac{d^2 y}{d x^2}\right|_{x=1}=2>0
\end{aligned}
$
$\therefore x=1$ is point of minima
$\begin{aligned}\left.\frac{d^2 y}{d x^2}\right|_{x=e^{-2}} & =\frac{2}{e^{-2}}+2 \cdot \log e^{-2} \cdot \frac{1}{e^{-2}} \\ & =\frac{1}{e^{-2}}(2-4)=-\frac{2}{e^{-2}} < 0 \\ \therefore \quad y_{\max } & =e^{-2}\left(\log e^{-2}\right)^2 \\ & =e^{-2}(-2)^2=4 e^{-2}\end{aligned}$
Asked in: AP EAMCET 2021 (24 Aug Shift 1)
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