Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

$${S}=\mathrm{2}\pi{r}^{\mathrm{2}} +\mathrm{2}\pi{rh} \\ $$$${V}=\pi{r}^{\mathrm{2}} {h}\:\Rightarrow\:{h}=\frac{{V}}{\pi{r}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$${S}=\mathrm{2}\pi{r}^{\mathrm{2}} +\frac{\mathrm{2}{V}}{{r}} \\ $$$$\frac{{dS}}{{dr}}=\mathrm{4}\pi{r}−\frac{\mathrm{2}{V}}{{r}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{r}=\sqrt[{\mathrm{3}}]{\frac{{V}}{\mathrm{2}\pi}}\wedge{h}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{4}{V}}{\pi}} \\ $$$$\mathrm{or}\:\mathrm{simply}\:{h}=\mathrm{2}{r} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com