Question

The local maximum value of $x{(1-x)}^2,0\le x\le 2$ is

• A. $2$
• B. $\frac{4}{27}$
• C. $5$
• D. $2,\frac{4}{27}$

Answer 1

Answer: (B)

$\frac{4}{27}$

$\begin{array}{c}Letf\left(x\right)=x{\left(1-x\right)}^2\\ f^{\prime }\left(x\right)-1{\left(1-x\right)}^2+x\times 2\left(1-x\right)\times -1\\ ={\left(1-x\right)}^2-2x\left(1-x\right)\end{array}$

$\begin{array}{c}{f}^{\prime \prime }\left(x\right)=-2\left(1-x\right)-2\left[1-x+x\times -1\right]\\ =-2+2x-2\left[1-x-x\right]\\ =-2+2x-2+4x\\ =6x-4\end{array}$

For critical points, $F^{\prime }\left(x\right)=0$

$\begin{array}{c}\Rightarrow {\left(1-x\right)}^2-2x\left(1-x\right)=0\\ \Rightarrow \left(1-x\right)\left(1-x-2x\right)=0\\ \Rightarrow \left(1-x\right)\left(1-3x\right)=0\\ \Rightarrow \left(x-1\right)\left(3x-1\right)=0\\ x=1,\frac{1}{3}\end{array}$

$f^{\prime }\left(1\right)=6\left(1\right)-4=6-4=2>0minimum$

$f^{\prime \prime }\left(\frac{1}{3}\right)=6\times \frac{1}{3}-4=2-4=-2<0miximum$

$f\left(\frac{1}{3}\right)=\frac{1}{3}{\left(1-\frac{1}{3}\right)}^2=\frac{1}{3}\times \frac{4}{9}=\frac{4}{27}$

So, $x=\frac{1}{3}$ is the point of local Maximum and the local maximum value is $\frac{4}{27}$