Question

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

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

Answer 1

Answer: (B)

$\frac{4}{27}$

Let $y=x(1-x{)}^2$, then for maximum/minimum
$\frac{dy}{dx}=1.(1-x{)}^2-2x(1-x)=0$
$\Rightarrow (1-x)[1-x-2x)=0\Rightarrow (1-x)(1-3x)=0$
$\Rightarrow x=1,x=\frac{1}{3}$
Now $\frac{d^{2}y}{d^{2}x}=(-1)(1-3x)+(1-x)(-3)$
$=(-1)(1-3)+(1-1)(-3)=2atx=1$
$\frac{d^{2}y}{d{x}^2}=2>0$
$\therefore$ At x =1, y is minimum
$\frac{d^{2}y}{d{x}^2}=(-1)[1-3\left(\frac{1}{3}\right)]+(1-\frac{1}{3})(-3)$
$=(-1)\left(0\right)+\left(\frac{2}{3}\right)(-3)=-2$
Maximum at $x=1/3,\frac{d^{2}y}{d{x}^2}<0$
$\therefore$ y is maximum
$\therefore y=\frac{1}{3}{(1-\frac{1}{3})}^2=\frac{1}{3}{\left(\frac{2}{3}\right)}^2=\frac{4}{27}$