Question

The maximum and minimum values of $x^3–18x^2+96x$ in the interval $(0,9)$ are

• A. $160,0$
• B. $60,0$
• C. $160,128$
• D. $120,28$

Answer 1

The correct option is C

$160,128$

Explanation for the correct option

Given function is $x^3–18x^2+96x$

Let, $y=x^3–18x^2+96x$

$\frac{dy}{dx}=3x^2–36x+96$

We know that, for critical point, $\frac{dy}{dx}=0$

$$\begin{gathered} \Rightarrow x^2–12x+32=0 \\ \Rightarrow (x–4)(x–8)=0 \\ \Rightarrow x=4,8 \end{gathered}$$

We know that for maximum $\frac{d^{2}y}{d{x}^2}<0$ and for minimum $\frac{d^{2}y}{d{x}^2}>0$

$\frac{d^{2}y}{d{x}^2}=6x–36$

At, $x=4,\frac{d^{2}y}{d{x}^2}=24-36=-12<0$

Thus, $x=4$, is the point of maxima of the given function.

${\left[f\right(4\left)\right]}_{max}=64-288+384=160$

Then, $x=8$, is the point of minima of the given function.

${\left[f\right(8\left)\right]}_{min}={\left(8\right)}^3-18{\left(8\right)}^2+96\left(8\right)=128$

Hence, $\mathrm{Option}\left(\mathrm{C}\right)$ is the correct answer.