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Question Number 141136 by bobhans last updated on 16/May/21

$$Findthesmallestvalue$$$$5x+\frac{16}{x}+21overpositive$$$$valueofx$$

Answered by iloveisrael last updated on 16/May/21

$$\Rightarrow 5x\mathrm{\&}\frac{16}{x}haveaconstantproduct$$$$80hencetheminimumisachieved$$$$byequatingthese5x=\frac{16}{x}or$$$$x=\sqrt{\frac{16}{5}}=\frac{4}{\sqrt{5}}.Ourconclusion$$$$isthatsmallestvalueof$$$$5x+\frac{16}{x}+21=10x+21$$$$=\frac{40}{\sqrt{5}}+21=8\sqrt{5}+21$$

Answered by EDWIN88 last updated on 16/May/21

$$\mathrm{say}\mathrm{f}\left(\mathrm{x}\right)=5\mathrm{x}+{16\mathrm{x}}^{-1}+21$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=5-\frac{16}{{\mathrm{x}}^2}=0\Rightarrow {\mathrm{x}}^2=\frac{16}{5}$$$$\mathrm{since}\mathrm{x}>0\mathrm{then}\mathrm{x}=\sqrt{\frac{16}{5}}=\frac{4\sqrt{5}}{5}$$$$\mathrm{minimum}\mathrm{f}\left(\frac{4\sqrt{5}}{5}\right)=4\sqrt{5}+\frac{\cancel{16}{}^4.5}{\cancel{4}\sqrt{5}}+21=8\sqrt{5}+21$$

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