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Question Number 85241 by Power last updated on 20/Mar/20

Answered by jagoll last updated on 20/Mar/20

$$\mathrm{minimum}\frac{{4\mathrm{s}}^2+{\mathrm{t}}^2}{5\mathrm{st}}=\frac{{4\mathrm{s}}^2}{5\mathrm{st}}+\frac{{\mathrm{t}}^2}{5\mathrm{st}}$$$$=\frac{4}{5}\left(\frac{\mathrm{s}}{\mathrm{t}}\right)+\frac{1}{5}\left(\frac{\mathrm{t}}{\mathrm{s}}\right).$$$$\mathrm{let}\frac{\mathrm{s}}{\mathrm{t}}=\mathrm{u}\Rightarrow \mathrm{f}\left(\mathrm{u}\right)=\frac{4}{5}\mathrm{u}+\frac{1}{5}{\mathrm{u}}^{-1}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{u}\right)=\frac{4}{5}-\frac{1}{{5\mathrm{u}}^2}=0\Rightarrow \mathrm{u}=\frac{1}{2}$$$$\mathrm{minimum}\mathrm{f}\left(\frac{1}{2}\right)=\frac{4}{5}\times \frac{1}{2}+\frac{1}{5}\times 2$$$$=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}$$

Commented by mr W last updated on 20/Mar/20

$$wecangetthisresultwithoutcalculus.$$$$weknowforpositiveAandB:$$$$A^2+B^2-2AB={(A-B)}^2⩾0$$$$\Rightarrow A^2+B^2⩾2AB$$$$nowwithA=2s,B=t$$$$\frac{{4\mathrm{s}}^2+{\mathrm{t}}^2}{5\mathrm{st}}=\frac{{\left(2s\right)}^2+{\mathrm{t}}^2}{5\mathrm{st}}⩾\frac{2\left(2s\right)t}{5st}=\frac{4}{5}$$$$i.e.theminimumof\frac{{4\mathrm{s}}^2+{\mathrm{t}}^2}{5\mathrm{st}}is\frac{4}{5}.$$

Commented by jagoll last updated on 20/Mar/20

$$\mathrm{good}\mathrm{improvment}$$

Commented by Power last updated on 20/Mar/20

$$\mathrm{thanks}$$

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