Question

Maximum value of $x{(1 - x)^2}$ when $0 \leqslant x \leqslant 2$, is
A. $\dfrac{2}{{27}}$
B. $\dfrac{4}{{27}}$
C. 5
D. 0

Answer 1

Hint: Here we have to find the maximum value of the function at given points. This topic will come under the application of differentiation. So here the given function is considered as $f(x)$ and then we have to find the second derivative of the function and then we have to substitute the given points then we have to determine whether it is positive or negative.

Complete step by step solution:
 As we know that the word maxima means which will be the maximum value and further it can’t be reached more. Maxima and minima is the application of derivatives. To find the maxima value and minima value we use the differentiation concept.
On considering the given question.
$f(x) = x{(1 - x)^2}$
On simplifying the above function
$f(x) = x(1 + {x^2} + 2x)$
On multiplying the x to inside the bracket
$f(x) = x + {x^3} - 2{x^2}$
On differentiating the above function with respect to x we get
$f'(x) = 1 + 3{x^2} - 4x$
Now we will equate the first differentiation of the function to zero, then we get the values of x.
$3{x^2} - 4x + 1 = 0$
On simplifying the above quadratic equation
$3{x^2} - 3x - x + 1 = 0$
$ \Rightarrow 3x(x - 1) - 1(x - 1) = 0$
$ \Rightarrow (3x - 1)(x - 1) = 0$
On further simplification, the values of x will be
$x = \dfrac{1}{3},x = 1$
Now we will find the second derivation of the given function
$f''(x) = 0 + 6x - 4$
$ \Rightarrow f''(x) = 6x - 4$
Now substituting the values of x in the above equation. So now we consider the value of x as 1.
$ \Rightarrow f''(1) = 6(1) - 4$
On simplifying
$ \Rightarrow f''(1) = 2 = + ve$
Now we will substitute the value of x as $\dfrac{1}{3}$
$ \Rightarrow f''(\dfrac{1}{3}) = 6(\dfrac{1}{3}) - 4$
On simplifying
$ \Rightarrow f''(\dfrac{1}{3}) = - 2 = - ve$
Therefore the maximum will be at the point $\dfrac{1}{3}$
On finding the value of given function at this point
$f(\dfrac{1}{3}) = \dfrac{1}{3}{(1 - \dfrac{1}{3})^2}$
$ \Rightarrow f\left( {\dfrac{1}{3}} \right) = \dfrac{1}{3}\left( {\dfrac{4}{9}} \right)$
On multiplying
$ \Rightarrow f\left( {\dfrac{1}{3}} \right) = \dfrac{4}{{27}}$
Therefore the maximum value of $x{(1 - x)^2}$when $0 \leqslant x \leqslant 2$, is $\dfrac{4}{{27}}$

Option ‘B’ is correct

Note: As the student thinks if the second derivative is positive then it will be maximum but it is not correct. When the value of the second derivative is negative then it will be maximum at that point. Suppose the second derivative is positive then the function is increasing. So the student must be careful between the maxima and increasing function.