Question

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

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

Answer 1

Answer: (C)

160, 128

Answer 2

Let $y = x^{3} - 18x^{2} + 96x \Rightarrow \frac{dy}{dx} = 3x^{2} - 36x + 96 = 0$

$\therefore$ $x^{2} - 12x + 32 = 0$ ⇒ $(x - 4)(x - 8) = 0,x = 4,8$

Now, $\frac{d^{2}y}{dx^{2}} = 6x - 36$ at $x = 4,\frac{d^{2}y}{dx^{2}} = 24 - 36 = - 12 < 0$

$\therefore$ at $x = 4$ function will be maximum and

$\lbrack f(x)\rbrack_{max.}$ at

$x = 8\frac{d^{2}y}{dx^{2}} = 48 - 36 = 12 > 0$

$\therefore$ at $x = 8$ function will be minimum and $\lbrack f(x)\rbrack_{min.}$.