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

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Aicharya · Accepted Answer

$\begin{aligned} & f(x)=3 x^3-18 x^2+27 x-40 \ & \Rightarrow f^{\prime}(x)=9 x^2-36 x+27 \ & =9(x-1)(x-3)\end{aligned}$ Sign scheme for $f^{\prime}(x)$ Now, $x^2+30 \leq 11 x$ $\begin{aligned} & \Rightarrow(x-5)(x-6) \leq 0 \ & \Rightarrow x \in[5,6]\end{aligned}$ For the interval $[5,6] f(x)$ is increasing Hence $f_{\max }=f(6)$ $\forall x \in[5,6]$ i.e. on the set S $=3 	imes 6^3-18 	imes 6^2+27 	imes 6-40=122$

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

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