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Question Number 110669 by Aina Samuel Temidayo last updated on 30/Aug/20

Given that f (x) = {(3 + 2 x)}^{3} {(4 - x)}^{4} on

the interval - \frac{3}{2} < x < 4 . Find the

(a) Maximum value of f (x)

(b) The value of x that gives the

maximum in (a)

Commented by Her_Majesty last updated on 30/Aug/20

{(3 + 2 x)}^{3} (4 - x^{4}) o r r a t h e r {(3 + 2 x)}^{3} {(4 - x)}^{4} ?

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

You are correct . I have changed it .

Thanks .

Answered by Her_Majesty last updated on 30/Aug/20

f^{'} (x) = 6 {(3 + 2 x)}^{2} {(4 - x)}^{4} - 4 {(4 - x)}^{3} {(3 + 2 x)}^{3} =

= 2 {(3 + 2 x)}^{2} {(4 - x)}^{3} (3 (4 - x) - 2 (3 + 2 x)) =

= 2 {(3 + 2 x)}^{2} {(4 - x)}^{3} (6 - 7 x)

f^{'} (x) = 0 \Rightarrow x_{1, 2} = - \frac{3}{2} \land x_{3, 4, 5} = 4 \land x_{6} = \frac{6}{7}

n o w o b v i o u s l y f (- \frac{3}{2}) = f (4) = 0 b u t

f (\frac{6}{7}) = \frac{2^{4} 3^{3} 11^{7}}{7^{7}} = \frac{8418457872}{823543} \approx 10222.2

Answered by 1549442205PVT last updated on 30/Aug/20

Apply {Cauchy}^{'} s inequality we have

{(3 + 2 x)}^{3} {(4 - x)}^{4} = \frac{{(3 + 2 x)}^{3} {(6 - 1 . 5 x)}^{4}}{1 . 5^{4}} (1)

Apply {Cauchy}^{'} s inequality we have

7^{7} \sqrt{(3 + 2 x) (3 + 2 x) (3 + 2 x) (6 - 1 . 5 x) (6 - 1 . 5 x) (7 - 1 . 5 x) (6 - 1 . 5 x)}

⩽ (3 + 2 x) + (3 + 2 x) + (3 + 2 x) + (6 - 1 . 5 x) + (6 - 1 . 5 x) + (7 - 1 . 5 x) + (6 - 1 . 5 x)

= 33 (2)

\Rightarrow {(3 + 2 x)}^{3} {(6 - 1 . 5 x)}^{4} ⩽ {(\frac{33}{7})}^{7}

\frac{{(3 + 2 x)}^{3} {(6 - 1 . 5 x)}^{4}}{1 . 5^{4}} ⩽ \frac{33^{7}}{7^{7} . 1 . 5^{4}}

The equality ocurrs if and only if

3 + 2 x = 6 - 1 . 5 x \Leftrightarrow 3 . 5 x = 3 \Leftrightarrow x = \frac{6}{7}

Thus f (x) has greatest value equal to

\frac{33^{7}}{7^{7} . 1 . 5^{4}} when x = \frac{6}{7}

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

Yes . Both are correct .