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Question Number 190032 by Shrinava last updated on 26/Mar/23

$$\mathrm{x}>0$$ $$\mathrm{xy}-18=0$$ $${(2\mathrm{x}+\mathrm{y})}_{\mathbf{min}}=?$$

Answered by cortano12 last updated on 26/Mar/23

$$\mathrm{z}=2\mathrm{x}+\frac{18}{\mathrm{x}}=2\mathrm{x}+{18\mathrm{x}}^{-1}$$ $$\frac{\mathrm{dz}}{\mathrm{dx}}=2-\frac{18}{{\mathrm{x}}^2}=0\Rightarrow \{\begin{array}{l}\mathrm{x}=3\\ \mathrm{x}=-3\end{array}$$ $$\frac{{\mathrm{d}}^2\mathrm{z}}{{\mathrm{dx}}^2}=\frac{36}{{\mathrm{x}}^3}>0\mathrm{for}\mathrm{x}=3$$ $${\mathrm{z}}_{\min }=12$$

Answered by SEKRET last updated on 26/Mar/23

$$\mathbf{y}=\frac{18}{\mathbf{x}}$$ $$2\mathbf{x}+\frac{18}{\mathbf{x}}⩾2\sqrt{2\mathbf{x}\cdot \frac{18}{\mathbf{x}}}=12$$

Commented byShrinava last updated on 01/Apr/23

$$\mathrm{thankyou}\mathrm{professor}$$

Answered by mehdee42 last updated on 26/Mar/23

$$y=\frac{18}{x}$$ $$A=2x+\frac{18}{x}\Rightarrow A^{{}^{\prime }}=2-\frac{18}{x^2}=0\Rightarrow x=3$$ $$\Rightarrow minA=12$$

Answered by manxsol last updated on 27/Mar/23

$$2x+\frac{18}{x}=z$$ $$2x^2-zx+18=0$$ $$\mathrm{\Delta }\gg 0\Rightarrow z^2-4\left(18\right)\left(2\right)\gg 0$$ $$zϵ<-\mathrm{\infty },-12]U[12,+\mathrm{\infty }>$$ $$min=12$$

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