Q-Find-the-value-of-Max-x-0-pi-2-sin-x-cos-3-x-

Question Number 174267 by mnjuly1970 last updated on 28/Jul/22

Q :: Find the value of :

M {ax}_{x \in [0, \frac{π}{2}]} {\sin (x) . \cos^{3} (x) ∣} = ?

Commented by kaivan.ahmadi last updated on 28/Jul/22

f^{'} (x) = {c o s}^{4} x - 3 {s i n}^{2} {x c o s}^{2} x = 0

\Rightarrow {c o s}^{2} x ({c o s}^{2} x - 3 {s i n}^{2} x) = 0 \Rightarrow

{c o s}^{2} x = 0 \Rightarrow x = \frac{π}{2}

{c o s}^{2} x - 3 {s i n}^{2} x = 0 \Rightarrow {t a n}^{2} x = \frac{1}{3}

\Rightarrow t a n x = \frac{\sqrt{3}}{3} \Rightarrow x = \frac{π}{6}

f (0) = 0

f (\frac{π}{2}) = 0

f (\frac{π}{3}) = \frac{1}{2} \times {(\frac{\sqrt{3}}{2})}^{3} = \frac{3 \sqrt{3}}{16}

M a x f = \frac{3 \sqrt{3}}{16}

Commented by mnjuly1970 last updated on 28/Jul/22

t h a n k s a l o t s i r

a l s o \dots \frac{3 \sqrt{3}}{16} t y p o

Commented by kaivan.ahmadi last updated on 28/Jul/22

a n o t h e r w a y

a = {s i n}^{2} x, b = {c o s}^{2} x, a + b = 1

M a x (f) = M a x {({s i n}^{2} x)}^{\frac{1}{2}} . {({c o s}^{2} x)}^{\frac{3}{2}} =

a^{\frac{1}{2}} b^{\frac{3}{2}} : \frac{a}{\frac{1}{2}} = \frac{b}{\frac{3}{2}}

b = 3 a, a + b = 1 \Rightarrow a = \frac{1}{4}, b = \frac{3}{4}

M a x (f) = {(\frac{1}{4})}^{\frac{1}{2}} {(\frac{3}{4})}^{\frac{3}{2}} =

\frac{1}{2} \sqrt{{(\frac{3}{4})}^{3}} = \frac{1}{2} \times \frac{3 \sqrt{3}}{8} = \frac{3 \sqrt{3}}{16}

Commented by infinityaction last updated on 28/Jul/22

h o w

\frac{a}{\frac{1}{2}} = \frac{b}{\frac{3}{2}}

Commented by mnjuly1970 last updated on 28/Jul/22

b e c a u s e a + b = c t e

Commented by kaivan.ahmadi last updated on 28/Jul/22

p r o p o s i t i o n :

i f a, b > 0 a n d a + b = k > 0 (c t e)

\Rightarrow M a x (a^{m} . b^{n}) : \frac{a}{m} = \frac{b}{n}

Commented by Tawa11 last updated on 28/Jul/22

Great sirs

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