Question Number 190032 by Shrinava last updated on 26/Mar/23
$$\mathrm{x}\:>\:\mathrm{0} \\ $$$$\mathrm{xy}\:−\:\mathrm{18}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2x}\:+\:\mathrm{y}\right)_{\boldsymbol{\mathrm{min}}} \:=\:? \\ $$
Answered by cortano12 last updated on 26/Mar/23
$$\:\mathrm{z}\:=\mathrm{2x}+\frac{\mathrm{18}}{\mathrm{x}}\:=\:\mathrm{2x}+\mathrm{18x}^{−\mathrm{1}} \\ $$$$\:\frac{\mathrm{dz}}{\mathrm{dx}}\:=\:\mathrm{2}−\frac{\mathrm{18}}{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{0}\:\Rightarrow\begin{cases}{\mathrm{x}=\mathrm{3}}\\{\mathrm{x}=−\mathrm{3}}\end{cases} \\ $$$$\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{z}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\frac{\mathrm{36}}{\mathrm{x}^{\mathrm{3}} }\:>\mathrm{0}\:\mathrm{for}\:\mathrm{x}=\mathrm{3} \\ $$$$\:\mathrm{z}_{\mathrm{min}} \:=\:\mathrm{12} \\ $$
Answered by SEKRET last updated on 26/Mar/23
$$\:\boldsymbol{\mathrm{y}}\:=\:\frac{\mathrm{18}}{\boldsymbol{\mathrm{x}}} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{x}}+\frac{\mathrm{18}}{\boldsymbol{\mathrm{x}}}\:\geqslant\:\mathrm{2}\sqrt{\mathrm{2}\boldsymbol{\mathrm{x}}\centerdot\frac{\mathrm{18}}{\boldsymbol{\mathrm{x}}}\:\:}\:=\:\mathrm{12} \\ $$
Commented by Shrinava last updated on 01/Apr/23
$$\mathrm{thankyou}\:\mathrm{professor} \\ $$
Answered by mehdee42 last updated on 26/Mar/23
$${y}=\frac{\mathrm{18}}{{x}} \\ $$$${A}=\mathrm{2}{x}+\frac{\mathrm{18}}{{x}}\Rightarrow{A}^{'} =\mathrm{2}−\frac{\mathrm{18}}{{x}^{\mathrm{2}} }=\mathrm{0}\Rightarrow{x}=\mathrm{3} \\ $$$$\Rightarrow{minA}=\mathrm{12} \\ $$
Answered by manxsol last updated on 27/Mar/23
![2x+((18)/x)=z 2x^2 −zx+18=0 Δ≫0⇒z^2 −4(18)(2)≫0 zε <−∞,−12] U [12,+∞> min=12](https://www.tinkutara.com/question/Q190128.png)
$$\mathrm{2}{x}+\frac{\mathrm{18}}{{x}}={z} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −{zx}+\mathrm{18}=\mathrm{0} \\ $$$$\Delta\gg\mathrm{0}\Rightarrow{z}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{18}\right)\left(\mathrm{2}\right)\gg\mathrm{0} \\ $$$$\left.{z}\epsilon\:<−\infty,−\mathrm{12}\right]\:{U}\:\left[\mathrm{12},+\infty>\right. \\ $$$${min}=\mathrm{12} \\ $$