Question

The maximum value of the function f(x) = \[3{x^3}-18{x^2} + 27x-40\] on the set \[S = \{ x\; \in \;R:{x^2} + 30\; \leqslant \;11x\} \] is:
(a) 122
(b) –222
(c) −122
(d) 222

Answer 1

Hint: Here, first find the solution set from the given set S. Then differentiate the given function and put it as greater than 0 to obtain the maximum value of the function f(x). From the solution set choose the value of x at which f(x) is greatest; at this value of x function f(x) will be maximum. Obtained the maximum value of f(x) by putting the value of x from the solution set at which f’(x) was maximum.

Complete step-by-step answer:
We are given \[3{x^3}-18{x^2} + 27x-40\] and \[S = \{ x\; \in \;R:{x^2} + 30\; \leqslant \;11x\} \]
Now, $ S = \{ x \in R:{x^2} - 11x + 30 \leqslant 0\} = \{ x \in R:x \in [5,6]\} $
And f(x) = \[3{x^3}-18{x^2} + 27x-40\]
Differentiating both sides with respect to x, we get
f’(x) = \[9{x^2}-36x + 27 = 9[{x^2} - 4x + 3]\]
For maximum value f’(x) should be greater than 0.
 $ f'(x) = 9(x - 1)(x - 3) > 0\forall x \in [5,6] $
So f(x) is increasing in [5, 6]
Hence, maximum value is f(6)
f(6) = \[3{(6)^3}-18{(6)^2} + 27 \times 6-40\]
 $ \Rightarrow $ f(6) = 3 × 216 – 18 × 36 + 27 × 6 – 40
 $ \Rightarrow $ f(6) = 648 – 648 + 162 – 40 = 810 −688 = 122

So, the correct answer is “Option A”.

Note: In these types of questions, where the solution set is given, always find and consider the value of the domain within the given set. If any value of domain is outside the solution set at which function is maximum or minimum but asked in particular solution set then choose only values in solution set and ignore the values outside. A function can have more than one maximum value and more than one minimum value. You can also draw the graph and observe the maximum and minimum values of function. But graphical methods can be used for simple functions, for complex functions or functions with higher order, always use differentiation methods.