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Question Number 150963 by EDWIN88 last updated on 17/Aug/21

$$Givenx,yrealnumbersuchthat$$$$0<\frac{y}{x}<\frac{1}{2}.Findminimumvalue$$$$of\frac{2y}{x-y}+\frac{3x}{x+2y}.$$

Answered by john_santu last updated on 17/Aug/21

$$\mathrm{let}\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{t}\Rightarrow \mathrm{f}\left(\mathrm{t}\right)=\frac{2\mathrm{t}}{1-\mathrm{t}}+\frac{3}{1+2\mathrm{t}}$$$$\mathrm{where}0<\mathrm{t}<\frac{1}{2}.$$$$\mathrm{taking}\mathrm{derivative}\mathrm{of}\mathrm{f}\left(\mathrm{t}\right)$$$$\mathrm{f}{}^{\prime }\left(\mathrm{t}\right)=\frac{2}{{(1-\mathrm{t})}^2}-\frac{6}{{(1+2\mathrm{t})}^2}=0$$$$\Rightarrow {(1+2\mathrm{t})}^2=3{(1-\mathrm{t})}^2$$$$\Rightarrow {4\mathrm{t}}^2+4\mathrm{t}+1=3-6\mathrm{t}+{3\mathrm{t}}^2$$$$\Rightarrow {\mathrm{t}}^2+10\mathrm{t}-2=0$$$$\Rightarrow {(\mathrm{t}+5)}^2-27=0\Rightarrow \mathrm{t}=-5\pm 3\sqrt{3}$$$$\mathrm{f}{\left(\mathrm{t}\right)}_{\min }\mathrm{when}\mathrm{t}=-5+3\sqrt{3}$$$$\mathrm{so}\min \mathrm{f}\left(\mathrm{t}\right)=\frac{2(-5+3\sqrt{3})}{1-(-5+3\sqrt{3})}+\frac{3}{1+2(-5+3\sqrt{3})}$$$$\min \mathrm{f}\left(\mathrm{t}\right)=\frac{6\sqrt{3}-10}{6-3\sqrt{3}}+\frac{1}{2\sqrt{3}-3}$$$$\min \mathrm{f}\left(\mathrm{t}\right)=\frac{(6\sqrt{3}-10)(6+3\sqrt{3})}{9}+\frac{2\sqrt{3}+3}{3}$$$$\min \mathrm{f}\left(\mathrm{t}\right)=\frac{6\sqrt{3}-6}{9}+\frac{2\sqrt{3}+3}{3}$$$$\min \mathrm{f}\left(\mathrm{t}\right)=\frac{2\sqrt{3}-2}{3}+\frac{2\sqrt{3}+3}{3}=\frac{4\sqrt{3}+1}{3}$$$$$$

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