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Question Number 124924 by bemath last updated on 07/Dec/20
What should the dimensions be of a cylinder so that for a given volume v its total surface S minimum ?
$$\:{What}\:{should}\:{the}\:{dimensions}\:{be} \\ $$$${of}\:{a}\:{cylinder}\:{so}\:{that}\:{for}\:{a}\:{given}\:{volume}\: \\ $$$${v}\:{its}\:{total}\:{surface}\:{S}\:{minimum}\:?\: \\ $$
Commented by liberty last updated on 07/Dec/20
Denoting by r the radius of the base of the cylinder and by h the altitude , we have S = 2πr^2 + 2πrh and the volume is given v = πr^2 h ; whence h = (v/(πr^2 )) substituting this expression of h into the formula for S , give S=2(πr^2 +(v/r)) ; where 0<r<∞ (dS/dr) = 2(2πr−(v/r^2 )) = 0 ⇒ r_1 = ((v/(2π)))^(1/3) ((d^2 S/dr^2 ))_(r=r_1 ) = 2(2π+((2v)/r^3 ))_(r=r_1 ) > 0 Thus at the point r=r_1 the function S has a minimum . Noticing that lim_(r→0) S=∞ and lim_(r→∞) S = ∞, we arrive at conclusion that at r = r_1 = ((v/(2π)))^(1/3) then h = (v/(πr^2 ))= 2((v/(2π)))^(1/3) = 2r
$${Denoting}\:{by}\:{r}\:{the}\:{radius}\:{of}\:{the}\:{base}\:{of}\:{the} \\ $$$${cylinder}\:{and}\:{by}\:{h}\:{the}\:{altitude}\:,\:{we}\:{have}\: \\ $$$$\:{S}\:=\:\mathrm{2}\pi{r}^{\mathrm{2}} \:+\:\mathrm{2}\pi{rh}\:{and}\:{the}\:{volume}\:{is}\:{given}\: \\ $$$${v}\:=\:\pi{r}^{\mathrm{2}} {h}\:;\:{whence}\:{h}\:=\:\frac{{v}}{\pi{r}^{\mathrm{2}} } \\ $$$${substituting}\:{this}\:{expression}\:{of}\:{h}\:{into}\:{the} \\ $$$${formula}\:{for}\:{S}\:,\:{give}\:{S}=\mathrm{2}\left(\pi{r}^{\mathrm{2}} +\frac{{v}}{{r}}\right)\:;\:{where}\:\mathrm{0}<{r}<\infty \\ $$$$\:\frac{{dS}}{{dr}}\:=\:\mathrm{2}\left(\mathrm{2}\pi{r}−\frac{{v}}{{r}^{\mathrm{2}} }\right)\:=\:\mathrm{0}\:\Rightarrow\:{r}_{\mathrm{1}} \:=\:\sqrt[{\mathrm{3}}]{\frac{{v}}{\mathrm{2}\pi}} \\ $$$$\left(\frac{{d}^{\mathrm{2}} {S}}{{dr}^{\mathrm{2}} }\right)_{{r}={r}_{\mathrm{1}} } =\:\mathrm{2}\left(\mathrm{2}\pi+\frac{\mathrm{2}{v}}{{r}^{\mathrm{3}} }\right)_{{r}={r}_{\mathrm{1}} } >\:\mathrm{0}\: \\ $$$${Thus}\:{at}\:{the}\:{point}\:{r}={r}_{\mathrm{1}} \:{the}\:{function}\:{S}\:{has} \\ $$$${a}\:{minimum}\:.\:{Noticing}\:{that}\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{S}=\infty\:{and} \\ $$$$\underset{{r}\rightarrow\infty} {\mathrm{lim}}\:{S}\:=\:\infty,\:{we}\:{arrive}\:{at}\:{conclusion}\:{that} \\ $$$${at}\:{r}\:=\:{r}_{\mathrm{1}} =\:\sqrt[{\mathrm{3}}]{\frac{{v}}{\mathrm{2}\pi}}\:{then}\:{h}\:=\:\frac{{v}}{\pi{r}^{\mathrm{2}} }=\:\mathrm{2}\sqrt[{\mathrm{3}}]{\frac{{v}}{\mathrm{2}\pi}}\:=\:\mathrm{2}{r} \\ $$$$ \\ $$
Answered by som(math1967) last updated on 07/Dec/20
let radius=r height=h v=πr^2 h ⇒h=(v/(πr^2 )) S=2πrh+2πr^2 S=((2v)/r) +2πr^2 (dS/dr)=−((2v)/r^2 ) +4πr (d^2 S/dr^2 )=((4v)/r^3 ) +4π ∴(d^2 S/dr^2 )>0 ∴S minimum for (dS/dr)=0 ∴−((2v)/r^2 )+4πr=0 4πr^3 =2v 2πr^3 =πr^2 h 2r=h ∴S is minimum when Height is eqal to diameter or r:h=1:2
$$\mathrm{let}\:\mathrm{radius}=\mathrm{r}\:\mathrm{height}=\mathrm{h} \\ $$$${v}=\pi\mathrm{r}^{\mathrm{2}} \mathrm{h}\:\Rightarrow\mathrm{h}=\frac{{v}}{\pi\mathrm{r}^{\mathrm{2}} } \\ $$$$\mathrm{S}=\mathrm{2}\pi\mathrm{rh}+\mathrm{2}\pi\mathrm{r}^{\mathrm{2}} \\ $$$$\mathrm{S}=\frac{\mathrm{2}{v}}{\mathrm{r}}\:+\mathrm{2}\pi\mathrm{r}^{\mathrm{2}} \: \\ $$$$\frac{\mathrm{dS}}{\mathrm{dr}}=−\frac{\mathrm{2}{v}}{\mathrm{r}^{\mathrm{2}} }\:+\mathrm{4}\pi\mathrm{r} \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{S}}{\mathrm{dr}^{\mathrm{2}} }=\frac{\mathrm{4}{v}}{\mathrm{r}^{\mathrm{3}} }\:+\mathrm{4}\pi \\ $$$$\therefore\frac{\mathrm{d}^{\mathrm{2}} \mathrm{S}}{\mathrm{dr}^{\mathrm{2}} }>\mathrm{0}\:\therefore\mathrm{S}\:\mathrm{minimum}\:\mathrm{for} \\ $$$$\frac{\mathrm{dS}}{\mathrm{dr}}=\mathrm{0} \\ $$$$\therefore−\frac{\mathrm{2}{v}}{\mathrm{r}^{\mathrm{2}} }+\mathrm{4}\pi\mathrm{r}=\mathrm{0} \\ $$$$\mathrm{4}\pi\mathrm{r}^{\mathrm{3}} =\mathrm{2}{v} \\ $$$$\mathrm{2}\pi\mathrm{r}^{\mathrm{3}} =\pi\mathrm{r}^{\mathrm{2}} \mathrm{h} \\ $$$$\mathrm{2r}=\mathrm{h} \\ $$$$\therefore\mathrm{S}\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{when} \\ $$$$\mathrm{Height}\:\mathrm{is}\:\mathrm{eqal}\:\mathrm{to}\:\mathrm{diameter} \\ $$$$\mathrm{or}\:\:\mathrm{r}:\mathrm{h}=\mathrm{1}:\mathrm{2} \\ $$

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