Question

The maximum and minimum values of ${x^3} - 18{x^2} + 96x$ in the interval $\left( {0,9} \right)$ are
1. $160,0$
2. $60,0$
3. $160,128$
4. $120,28$

Answer 1

Hint : In order to find the maximum and the minimum values for the given cubic equation, initiate with finding the critical points by differentiating the given cubic function with respect to x, then equate the result with zero. Then differentiate again the first order derivative to get the second order derivative and solve further. Depending on the second derivative greater than or less than zero (at the values obtained on solving the first derivative) we decide whether the function is minima or maxima at that point.
Formula used :
$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$

Complete step-by-step answer :
We are given with an equation $y = {x^3} - 18{x^2} + 96x$
As we know that to get the critical points we need to differentiate the given function and equate the derivative with zero.
So, differentiating both the sides of the equation y with respect to $x$ we get:
As, we know $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$, so using this formula, we get:
$\dfrac{{dy}}{{dx}} = \dfrac{{d{x^3}}}{{dx}} - 18\dfrac{{d{x^2}}}{{dx}} + 96\dfrac{{dx}}{{dx}}$
$y' = 3{x^2} - 36x + 96$
To get the critical point, substituting $y' = 0$
i.e, $3{x^2} - 36x + 96 = 0$
Taking 3 common points from both the sides, we get:
$3\left( {{x^2} - 12x + 32} \right) = 0$
Dividing both the sides by 3:
$\Rightarrow \dfrac{{3\left( {{x^2} - 12x + 32} \right)}}{3} = 0$
$\Rightarrow {x^2} - 12x + 32 = 0$
The general form of a quadratic equation is $a{x^2} + bx + c = 0$ .
To find the roots of the quadratic equation, we use the quadratic formula, which is as:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing $a{x^2} + bx + c = 0$ with ${x^2} - 12x + 32 = 0$ we get:
$a = 1$
$b = - 12$
$c = 32$
Using the quadratic formula i.e, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting a, b and c, we get:
$x = \dfrac{{ - \left( { - 12} \right) \pm \sqrt {{{\left( { - 12} \right)}^2} - 4 \times 1 \times 32} }}{{2 \times 1}}$
$\Rightarrow x = \dfrac{{12 \pm \sqrt {144 - 128} }}{2}$
$\Rightarrow x = \dfrac{{12 \pm \sqrt {16} }}{2}$
$\Rightarrow x = \dfrac{{12 \pm 4}}{2}$
Taking 2 common:
$\Rightarrow x = \dfrac{{2\left( {6 \pm 2} \right)}}{2}$
$\Rightarrow x = \left( {6 \pm 2} \right)$
$\Rightarrow x = \left( {6 + 2} \right),\left( {6 - 2} \right) \\\ \Rightarrow x = 8{\text{ , }}4 \\\$
Therefore, the critical points of the function are 4 and 8.
Now, differentiating $y'$ both the sides with respect to $x$ we get:
$\Rightarrow \dfrac{{dy'}}{{dx}} = 3\dfrac{{d{x^2}}}{{dx}} - 36\dfrac{{dx}}{{dx}} + \dfrac{{d\left( {96} \right)}}{{dx}}$
Using $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$,we get:
$\Rightarrow \dfrac{{dy'}}{{dx}} = 3.2x - 36.x + 0$
$y'' = 6x - 36$
Substituting the critical points obtained in the above equation,
At $x = 4$ :
$y'' = 24 - 36 = - 12 < 0$
Since, the value is less than zero. Therefore, at $x = 4$ , the function will be maximum.
Hence, the maximum value:
$\Rightarrow y(x = 4) = {\left( 4 \right)^3} - 18\left( {{4^2}} \right) + 96 \times 4$
$\Rightarrow y(x = 4) = 64 - 288 + 384 = 160$
At $x = 8$:
$y'' = 48 - 36 = 12 > 0$
Since, the value is greater than zero. Therefore, at $x = 8$ , function will be the minimum.
Hence, minimum value:
$\Rightarrow y(x = 8) = {\left( 8 \right)^3} - 18\left( {{8^2}} \right) + 96 \times 8$
$\Rightarrow y(x = 8) = 512 - 1152 + 768 = 128$
Therefore, maximum and minimum values of ${x^3} - 18{x^2} + 96x$ in the interval $\left( {0,9} \right)$are $160$ and $128$ respectively.
So, the correct answer is “Option 3”.

Note :
i.Keep in mind we check the maximum and minimum value by substituting the critical point in the second order derivative, but to find the maximum value we substitute the value in the original function that is $f\left( x \right)$.
ii.A point where a function obtains its peak or maximum possible value is called the Absolute maximum.
iii.A point where a function obtains its lowest value is called Absolute minimum.