Question

Maximum value of $x{\left(1-x\right)}^2$ when $0\le x\le 2$, is

• A. $\frac{2}{27}$
• B. $\frac{4}{27}$
• C. $5$
• D. $0$

Answer 1

The correct option is B

$\frac{4}{27}$

Explanation for the correct option:

Finding the minimum value:

Given that $f\left(x\right)=x{\left(1-x\right)}^2$, which is

$f\left(x\right)=x^3-2x^2+x$

Now, $f\text{'}\left(x\right)=3x^2-4x+1$

Substituting $f\text{'}\left(x\right)=0$, in the above equation we get,

$\Rightarrow 3x^2-4x+1=0$

$\Rightarrow 3x^2-3x-x+1=0$

$\Rightarrow x=1,\frac{1}{3}$

Now, $f\text{'}\text{'}\left(x\right)=6x-4$

$f\text{'}\text{'}\left(1\right)=2$, which is positive and $f\text{'}\text{'}\left(\frac{1}{3}\right)=-2$, which is negative.

The maximum value is $x=\frac{1}{3}$.

$$\begin{gathered} f\left(x\right)=x{\left(1-x\right)}^2 \\ f\left(\frac{1}{3}\right)=\left(\frac{1}{3}\right){\left(1-\frac{1}{3}\right)}^2 \\ =\left(\frac{1}{3}\right){\left(\frac{2}{3}\right)}^2 \\ =\left(\frac{1}{3}\right)\left(\frac{4}{9}\right) \end{gathered}$$

$\therefore f\left(\frac{1}{3}\right)=\frac{4}{27}$

Hence, the maximum value of $f\left(x\right)$ when $x=\frac{1}{3}$ is $\frac{4}{27}$.

Therefore, the correct answer is option (B).