Given-xy-1-3-for-x-y-R-find-min-value-of-4-x-6-9-y-6-

Question Number 115733 by bemath last updated on 28/Sep/20

G i v e n x y = \frac{1}{3} f o r x, y \in R

f i n d m i n v a l u e o f \frac{4}{x^{6}} + \frac{9}{y^{6}}

Answered by MJS_new last updated on 28/Sep/20

y = \frac{1}{3 x} \Rightarrow \frac{4}{x^{6}} + \frac{9}{y^{6}} = \frac{4}{x^{6}} + 6561 x^{6}

\frac{d}{d x} [\frac{4}{x^{6}} + 6561 x^{6}] = - \frac{24}{x^{7}} + 39366 x^{5} = 0

39366 x^{12} = 24

x^{12} = \frac{24}{39366} = \frac{4}{6561} \Rightarrow x = \pm \frac{2^{1 / 6}}{3^{2 / 3}}

\Rightarrow

\min = 324

Commented by bemath last updated on 28/Sep/20

t h a n k y o u s i r

Answered by Olaf last updated on 28/Sep/20

y = \frac{1}{3 x}

(x \neq 0 otherwise \frac{4}{x^{6}} + \frac{9}{y^{6}} {doesn}^{'} t exist)

f (x) = \frac{4}{x^{6}} + \frac{9}{{(\frac{1}{3 x})}^{6}} = \frac{4}{x^{6}} + 3^{8} x^{6}

f is a even function .

I suppose x > 0 .

f^{'} (x) = - \frac{4.6}{x^{7}} + 6 . 3^{8} x^{5}

f^{'} (x) = 0 \Leftrightarrow \frac{6}{x^{7}} (3^{8} x^{12} - 4) = 0

\Leftrightarrow x = \sqrt[12]{\frac{4}{3^{8}}} = x_{1}

f^{'} (x) < 0 in] 0, x_{1} [.

f^{'} (x) > 0 in] x_{1}, + \infty [.

\Rightarrow {it}^{'} s a minimum .

x^{6} = \sqrt{\frac{4}{3^{8}}} = \frac{2}{3^{4}} = \frac{2}{81}

y^{6} = \frac{1}{3^{6} x^{6}} = \frac{1}{3^{6}} . \frac{81}{2} = \frac{1}{18}

f (x) = \frac{4}{(\frac{2}{81})} + \frac{9}{(\frac{1}{18})} = 2.81 + 2.81 = 324

Commented by bemath last updated on 28/Sep/20

y e s s i r . t h a n k y o u

Answered by 1549442205PVT last updated on 28/Sep/20

We have \frac{4}{x^{6}} + \frac{9}{y^{6}} = {(\frac{2}{x^{3}} - \frac{3}{y^{3}})}^{2} + \frac{12}{{(xy)}^{3}}

⩾ \frac{12}{{(xy)}^{3}} = 12 \times 3^{3} = 324

The equality ocurrs if and only if

{\begin{cases} \frac{2}{x^{3}} = \frac{3}{y^{3}} \\ xy = \frac{1}{3} \end{cases} \Leftrightarrow {\begin{cases} xy = 1 / 3 \\ 2 \times {27 y}^{3} = \frac{3}{y^{3}} \end{cases}

\Leftrightarrow {\begin{cases} y^{6} = 1 / 18 \\ xy = 1 / 3 \end{cases} \Leftrightarrow {\begin{cases} y = \frac{1}{^{6} \sqrt{18}} \\ x = \frac{^{6} \sqrt{18}}{3} \end{cases}

Thus, {(\frac{2}{x^{3}} = \frac{3}{y^{3}})}_{\min} = 324

when {\begin{cases} y = \frac{1}{^{6} \sqrt{18}} \\ x = \frac{^{6} \sqrt{18}}{3} \end{cases}