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

$$Givenxy=\frac{1}{3}forx,y\in \mathbb{R}$$$$findminvalueof\frac{4}{x^6}+\frac{9}{y^6}$$

Answered by MJS_new last updated on 28/Sep/20

$$y=\frac{1}{3x}\Rightarrow \frac{4}{x^6}+\frac{9}{y^6}=\frac{4}{x^6}+6561x^6$$$$\frac{d}{dx}[\frac{4}{x^6}+6561x^6]=-\frac{24}{x^7}+39366x^5=0$$$$39366x^{12}=24$$$$x^{12}=\frac{24}{39366}=\frac{4}{6561}\Rightarrow x=\pm \frac{2^{1/6}}{3^{2/3}}$$$$\Rightarrow$$$$\min =324$$

Commented by bemath last updated on 28/Sep/20

$$thankyousir$$

Answered by Olaf last updated on 28/Sep/20

$$$$$$y=\frac{1}{3x}$$$$(x\ne 0\mathrm{otherwise}\frac{4}{x^6}+\frac{9}{y^6}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist})$$$$f\left(x\right)=\frac{4}{x^6}+\frac{9}{{\left(\frac{1}{3x}\right)}^6}=\frac{4}{x^6}+3^8{x}^6$$$$f\mathrm{is}\mathrm{a}\mathrm{even}\mathrm{function}.$$$$\mathrm{I}\mathrm{suppose}x>0.$$$$f^{\prime }\left(x\right)=-\frac{4.6}{x^7}+6.3^8{x}^5$$$$f^{\prime }\left(x\right)=0\iff \frac{6}{x^7}(3^8{x}^{12}-4)=0$$$$\iff x=\sqrt[12]{\frac{4}{3^8}}=x_1$$$$f^{\prime }\left(x\right)<0\mathrm{in}]0,x_1[.$$$$f^{\prime }\left(x\right)>0\mathrm{in}]x_1,+\mathrm{\infty }[.$$$$\Rightarrow {\mathrm{it}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{minimum}.$$$$x^6=\sqrt{\frac{4}{3^8}}=\frac{2}{3^4}=\frac{2}{81}$$$$y^6=\frac{1}{3^6{x}^6}=\frac{1}{3^6}.\frac{81}{2}=\frac{1}{18}$$$$f\left(x\right)=\frac{4}{\left(\frac{2}{81}\right)}+\frac{9}{\left(\frac{1}{18}\right)}=2.81+2.81=324$$$$$$$$$$

Commented by bemath last updated on 28/Sep/20

$$yessir.thankyou$$

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

$$\mathrm{We}\mathrm{have}\frac{4}{x^6}+\frac{9}{y^6}={(\frac{2}{{\mathrm{x}}^3}-\frac{3}{{\mathrm{y}}^3})}^2+\frac{12}{{\left(\mathrm{xy}\right)}^3}$$$$⩾\frac{12}{{\left(\mathrm{xy}\right)}^3}=12\times 3^3=324$$$$\mathrm{The}\mathrm{equality}\mathrm{ocurrs}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}$$$$\{\begin{array}{l}\frac{2}{{\mathrm{x}}^3}=\frac{3}{{\mathrm{y}}^3}\\ \mathrm{xy}=\frac{1}{3}\end{array}\iff \{\begin{array}{l}\mathrm{xy}=1/3\\ 2\times {27\mathrm{y}}^3=\frac{3}{{\mathrm{y}}^3}\end{array}$$$$\iff \{\begin{array}{l}{\mathrm{y}}^6=1/18\\ \mathrm{xy}=1/3\end{array}\iff \{\begin{array}{l}\mathrm{y}=\frac{1}{{}^6\sqrt{18}}\\ \mathrm{x}=\frac{{}^6\sqrt{18}}{3}\end{array}$$$$\mathrm{Thus},{(\frac{2}{{\mathrm{x}}^3}=\frac{3}{{\mathrm{y}}^3})}_{\min }=324$$$$\mathrm{when}\{\begin{array}{l}\mathrm{y}=\frac{1}{{}^6\sqrt{18}}\\ \mathrm{x}=\frac{{}^6\sqrt{18}}{3}\end{array}$$

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