I-want-to-make-cyllinders-of-maximum-volume-V-max-with-minimum-surface-area-S-min-because-they-are-more-profitable-for-me-I-need-the-help-of-mathematicians-of-this-forum-

Question Number 3996 by Rasheed Soomro last updated on 26/Dec/15

$${I}\:{want}\:{to}\:{make}\:{cyllinders}\:{of}\:\boldsymbol{{maximum}} \\ $$$$\boldsymbol{{volume}}\:\boldsymbol{\mathrm{V}}_{\boldsymbol{\mathrm{max}}} \:{with}\:{minimum}\:\boldsymbol{{surface}}\:\boldsymbol{{area}}\: \\ $$$$\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{min}}} \:,{because}\:{they}\:{are}\:{more}\:{profitable}\:{for}\:{me}. \\ $$$${I}\:{need}\:{the}\:{help}\:{of}\:{mathematicians}\:{of}\:{this} \\ $$$${forum}. \\ $$$$ \\ $$

Commented by Filup last updated on 26/Dec/15

V=πr^2 h S=2πr^2 +2πrh V_(max) when V ′=0 (dV/dr)=2πrh or (dV/dh)=πr^2 S_(min) when S ′=0 (dS/dr)=4πr+2πh or (dS/dh)=2πr im not sure what to do from here. I think I need more information. I′m unsure -continue-

$${V}=\pi{r}^{\mathrm{2}} {h} \\ $$$${S}=\mathrm{2}\pi{r}^{\mathrm{2}} +\mathrm{2}\pi{rh} \\ $$$$ \\ $$$${V}_{{max}} \:\mathrm{when}\:{V}\:'=\mathrm{0} \\ $$$$\frac{{dV}}{{dr}}=\mathrm{2}\pi{rh}\:\:\:\:\:\:{or}\:\:\:\:\:\:\:\:\frac{{dV}}{{dh}}=\pi{r}^{\mathrm{2}} \\ $$$$ \\ $$$${S}_{{min}} \:\mathrm{when}\:{S}\:'=\mathrm{0} \\ $$$$\frac{{dS}}{{dr}}=\mathrm{4}\pi{r}+\mathrm{2}\pi{h}\:\:\:\:\:\:{or}\:\:\:\:\:\:\:\frac{{dS}}{{dh}}=\mathrm{2}\pi{r} \\ $$$$ \\ $$$$\mathrm{im}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{what}\:\mathrm{to}\:\mathrm{do}\:\mathrm{from}\:\mathrm{here}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{need}\:\mathrm{more}\:\mathrm{information}. \\ $$$$\mathrm{I}'{m}\:\mathrm{unsure} \\ $$$$ \\ $$$$-{continue}- \\ $$

Answered by prakash jain last updated on 26/Dec/15

Ratio =(V/S) = ((πr^2 h)/(2πr(r+h))) = ((rh)/(r+h)) S=2πr^2 +2πrh⇒h=((S−2πr^2 )/(2πr)) Taking S constant find r so that V is maximum f(r)=((r((S−2πr^2 )/(2πr)))/(r+((S−2πr^2 )/(2πr))))=((rS−2πr^2 )/S) f(r)=(1/S)(rS−2πr^2 ) f ′(r)=1−((4πr)/S), f ′′(r)=−((4π)/S) 1−((4πr)/S)=0⇒r=(S/(4π))

$$\mathrm{Ratio}\:=\frac{{V}}{{S}}\:=\:\frac{\pi{r}^{\mathrm{2}} {h}}{\mathrm{2}\pi{r}\left({r}+{h}\right)}\:=\:\frac{{rh}}{{r}+{h}} \\ $$$${S}=\mathrm{2}\pi{r}^{\mathrm{2}} +\mathrm{2}\pi{rh}\Rightarrow{h}=\frac{{S}−\mathrm{2}\pi{r}^{\mathrm{2}} }{\mathrm{2}\pi{r}} \\ $$$$\mathrm{Taking}\:\mathrm{S}\:\mathrm{constant}\:\mathrm{find}\:{r}\:\mathrm{so}\:\mathrm{that}\:\mathrm{V}\:\mathrm{is}\:\mathrm{maximum} \\ $$$${f}\left({r}\right)=\frac{{r}\frac{{S}−\mathrm{2}\pi{r}^{\mathrm{2}} }{\mathrm{2}\pi{r}}}{{r}+\frac{{S}−\mathrm{2}\pi{r}^{\mathrm{2}} }{\mathrm{2}\pi{r}}}=\frac{{rS}−\mathrm{2}\pi{r}^{\mathrm{2}} }{{S}} \\ $$$${f}\left({r}\right)=\frac{\mathrm{1}}{{S}}\left({rS}−\mathrm{2}\pi{r}^{\mathrm{2}} \right) \\ $$$${f}\:'\left({r}\right)=\mathrm{1}−\frac{\mathrm{4}\pi{r}}{{S}},\:{f}\:''\left({r}\right)=−\frac{\mathrm{4}\pi}{{S}} \\ $$$$\mathrm{1}−\frac{\mathrm{4}\pi{r}}{{S}}=\mathrm{0}\Rightarrow{r}=\frac{{S}}{\mathrm{4}\pi} \\ $$

Commented by RasheedSindhi last updated on 26/Dec/15

$$\boldsymbol{{T}}{hanks}\:{a}\:{lot}! \\ $$$${Perhaps}\:{a}\:{comercial}\:{person} \\ $$$${prefer}\:{the}\:{answer}\:{in}\:{diameter} \\ $$$${and}\:{height}\:{relation}! \\ $$

Commented by Filup last updated on 27/Dec/15

$${r}=\frac{{S}}{\mathrm{4}\pi} \\ $$$${r}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}+\frac{{rh}}{\mathrm{2}} \\ $$$$\mathrm{2}−{r}={h} \\ $$

Commented by Rasheed Soomro last updated on 27/Dec/15

$$\mathrm{Thanks}! \\ $$

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