Question

The maximum value of the function $f\left(x\right)=3x^3-18x^2+27x-40$ on the set $S=\{x\in R:x^2+30\le 11x\}$ is :

• A. $122$
• B. $-122$
• C. $222$
• D. $-222$

Answer 1

The correct option is A $122$

$S=\{x\in R:x^2+30\le 11x\}$
$x^2+30\le 11x$
$\Rightarrow x^2-11x+30\le 0$
$\Rightarrow x^2-6x-5x+30\le 0$
$\Rightarrow (x-5)(x-6)\le 0$
$\Rightarrow x\in [5,6]$

Now, $f\left(x\right)=3x^3-18x^2+27x-40$
$f^{\prime }\left(x\right)=9x^2-36x+27$
$\Rightarrow f^{\prime }\left(x\right)=9(x^2-4x+3)$
$f^{\prime }\left(5\right)=72>0\cdots \left(1\right)$

$f^{\prime \prime }\left(x\right)=9(2x-4)$
$f^{\prime \prime }\left(x\right)>0\forall x\in [5,6]$
$\Rightarrow f^{\prime }\left(x\right)$ is strictly increasing in the interval $[5,6]\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we conclude
$f^{\prime }\left(x\right)>0\forall x\in [5,6]$
$\Rightarrow f\left(x\right)$ is strictly increasing in the interval $[5,6]$
$\therefore$ Maximum value of $f\left(x\right)$ when $x\in [5,6]$ is $f\left(6\right)=122$