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Find-among-all-right-circular-cylinders-of-fixed-volume-V-that-one-with-smallest-surface-area-counting-the-areas-of-the-faces-at-top-and-bottom-




Question Number 144200 by bramlexs22 last updated on 23/Jun/21
Find, among all right circular  cylinders of fixed volume V   that one with smallest surface area  (counting the areas of the faces   at top and bottom )
$$\mathrm{Find},\:\mathrm{among}\:\mathrm{all}\:\mathrm{right}\:\mathrm{circular} \\ $$$$\mathrm{cylinders}\:\mathrm{of}\:\mathrm{fixed}\:\mathrm{volume}\:\mathrm{V}\: \\ $$$$\mathrm{that}\:\mathrm{one}\:\mathrm{with}\:\mathrm{smallest}\:\mathrm{surface}\:\mathrm{area} \\ $$$$\left(\mathrm{counting}\:\mathrm{the}\:\mathrm{areas}\:\mathrm{of}\:\mathrm{the}\:\mathrm{faces}\:\right. \\ $$$$\left.\mathrm{at}\:\mathrm{top}\:\mathrm{and}\:\mathrm{bottom}\:\right) \\ $$
Answered by MJS_new last updated on 23/Jun/21
S=2πr^2 +2πrh  V=πr^2 h ⇒ h=(V/(πr^2 ))  ⇒  S=2πr^2 +((2V)/r)  (dS/dr)=4πr−((2V)/r^2 )=0 ⇒ r=((V/(2π)))^(1/3) ∧h=(((4V)/π))^(1/3)   or simply h=2r
$${S}=\mathrm{2}\pi{r}^{\mathrm{2}} +\mathrm{2}\pi{rh} \\ $$$${V}=\pi{r}^{\mathrm{2}} {h}\:\Rightarrow\:{h}=\frac{{V}}{\pi{r}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$${S}=\mathrm{2}\pi{r}^{\mathrm{2}} +\frac{\mathrm{2}{V}}{{r}} \\ $$$$\frac{{dS}}{{dr}}=\mathrm{4}\pi{r}−\frac{\mathrm{2}{V}}{{r}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{r}=\sqrt[{\mathrm{3}}]{\frac{{V}}{\mathrm{2}\pi}}\wedge{h}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{4}{V}}{\pi}} \\ $$$$\mathrm{or}\:\mathrm{simply}\:{h}=\mathrm{2}{r} \\ $$

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