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Question Number 141135 by bobhans last updated on 16/May/21

$$Findtheminimumof$$ $$\frac{12}{x}+\frac{18}{y}+xyforall$$ $$positivenumberx\mathrm{\&}y.$$

Answered by mitica last updated on 16/May/21

$$\frac{12}{x}+\frac{18}{y}+xy⩾3\sqrt[3]{\frac{12}{x}\cdot \frac{18}{y}\cdot xy}=18$$ $$=for\frac{12}{x}=\frac{18}{y}=xy\iff x=2,y=3$$ $$\Rightarrow min=18$$

Commented bymitica last updated on 16/May/21

$$x=2,y=3$$

Commented byiloveisrael last updated on 16/May/21

$$x=\sqrt[3]{\frac{4}{3}};y=\sqrt[3]{\frac{9}{2}}$$ $$\Rightarrow \frac{12}{\sqrt[3]{\frac{4}{3}}}+\frac{18}{\sqrt[3]{\frac{9}{2}}}+\sqrt[3]{\frac{36}{6}}=23.62$$ $$howyouclaim23.62=18??$$

Answered by iloveisrael last updated on 16/May/21

$$Weknowthatthethreeterms$$ $$inthesumhavefixedproduct$$ $$12\times 18=216.Thesumistherefore$$ $$aminimumifwecanfindx\mathrm{\&}y$$ $$satisfy\frac{12}{x}=\frac{18}{y}=xy.Sincethe$$ $$productis{6}^{3}eachof\frac{12}{x},\frac{18}{y}\mathrm{\&}xy$$ $$mustequal6,hencetherequired$$ $$minimumis6+6+6=18,$$ $$occurringiffx=2\mathrm{\&}y=3.$$

Answered by EDWIN88 last updated on 16/May/21

$$\mathrm{z}=\mathrm{f}(\mathrm{x},\mathrm{y})={12\mathrm{x}}^{-1}+{18\mathrm{y}}^{-1}+\mathrm{xy}$$ $$\Rightarrow {\mathrm{z}}_{\mathrm{x}}=-\frac{12}{{\mathrm{x}}^2}+\mathrm{y}=0,\mathrm{y}=\frac{12}{{\mathrm{x}}^2}\dots \left(\mathrm{i}\right)$$ $$\Rightarrow {\mathrm{z}}_{\mathrm{y}}=-\frac{18}{{\mathrm{y}}^2}+\mathrm{x}=0,\mathrm{x}=\frac{18}{{\mathrm{y}}^2}\dots \left(\mathrm{ii}\right)$$ $$\Rightarrow \frac{12}{18}=\frac{{\mathrm{x}}^2\mathrm{y}}{{\mathrm{xy}}^2};\frac{2}{3}=\frac{\mathrm{x}}{\mathrm{y}}\mathrm{or}\mathrm{y}=\frac{3\mathrm{x}}{2}$$ $$\Rightarrow \frac{3\mathrm{x}}{2}=\frac{12}{{\mathrm{x}}^2}\Rightarrow \mathrm{x}=\sqrt[3]{8}=2\mathrm{\&}\mathrm{y}=3$$ $$\mathrm{minimum}\mathrm{z}=\mathrm{f}(2,3)=18\divideontimes$$

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