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Find-the-minimum-surface-area-of-a-solid-circular-cylinder-if-its-volume-is-16pi-cm-3-leave-your-answer-in-terms-of-pi-




Question Number 23133 by tawa tawa last updated on 26/Oct/17
Find the minimum surface area of a solid circular cylinder ,  if its volume is  16π cm^3    (leave your answer in terms of π)
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{solid}\:\mathrm{circular}\:\mathrm{cylinder}\:,\:\:\mathrm{if}\:\mathrm{its}\:\mathrm{volume}\:\mathrm{is} \\ $$$$\mathrm{16}\pi\:\mathrm{cm}^{\mathrm{3}} \:\:\:\left(\mathrm{leave}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\pi\right) \\ $$
Answered by ajfour last updated on 26/Oct/17
S=2πrh+2πr^2   V=πr^2 h   ⇒  (dh/dr)=−((2V)/(πr^3 ))   ...(i)  (dS/dr)=2πh+2πr(dh/dr)+4πr =0  using (i), this ⇒     2πh+2πr(−((2V)/(πr^3 )))+4πr=0  ⇒   h=((2V)/(πr))−2r = ((2(πr^2 h))/(πr^2 ))−2r   ⇒   h=2h−2r  or h=2r  Now V=πr^2 h =πr^2 (2r)=16π cm^3   ⇒  r^3 =8 cm^3    or  r=2cm  and h=2r =4cm  so  S=2πrh+2πr^2             =2π(3r^2 ) =2π×12cm^2        S=24π cm^2  .
$${S}=\mathrm{2}\pi{rh}+\mathrm{2}\pi{r}^{\mathrm{2}} \\ $$$${V}=\pi{r}^{\mathrm{2}} {h}\:\:\:\Rightarrow\:\:\frac{{dh}}{{dr}}=−\frac{\mathrm{2}{V}}{\pi{r}^{\mathrm{3}} }\:\:\:…\left({i}\right) \\ $$$$\frac{{dS}}{{dr}}=\mathrm{2}\pi{h}+\mathrm{2}\pi{r}\frac{{dh}}{{dr}}+\mathrm{4}\pi{r}\:=\mathrm{0} \\ $$$${using}\:\left({i}\right),\:{this}\:\Rightarrow \\ $$$$\:\:\:\mathrm{2}\pi{h}+\mathrm{2}\pi{r}\left(−\frac{\mathrm{2}{V}}{\pi{r}^{\mathrm{3}} }\right)+\mathrm{4}\pi{r}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{h}=\frac{\mathrm{2}{V}}{\pi{r}}−\mathrm{2}{r}\:=\:\frac{\mathrm{2}\left(\pi{r}^{\mathrm{2}} {h}\right)}{\pi{r}^{\mathrm{2}} }−\mathrm{2}{r} \\ $$$$\:\Rightarrow\:\:\:{h}=\mathrm{2}{h}−\mathrm{2}{r} \\ $$$${or}\:{h}=\mathrm{2}{r} \\ $$$${Now}\:{V}=\pi{r}^{\mathrm{2}} {h}\:=\pi{r}^{\mathrm{2}} \left(\mathrm{2}{r}\right)=\mathrm{16}\pi\:{cm}^{\mathrm{3}} \\ $$$$\Rightarrow\:\:{r}^{\mathrm{3}} =\mathrm{8}\:{cm}^{\mathrm{3}} \:\:\:{or}\:\:{r}=\mathrm{2}{cm} \\ $$$${and}\:{h}=\mathrm{2}{r}\:=\mathrm{4}{cm} \\ $$$${so}\:\:{S}=\mathrm{2}\pi{rh}+\mathrm{2}\pi{r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\pi\left(\mathrm{3}{r}^{\mathrm{2}} \right)\:=\mathrm{2}\pi×\mathrm{12}{cm}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\boldsymbol{{S}}=\mathrm{24}\pi\:{cm}^{\mathrm{2}} \:. \\ $$
Commented by tawa tawa last updated on 26/Oct/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

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