Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 6430 by sanusihammed last updated on 27/Jun/16

An oil can is to be made in form of a right circular cylinder that be inscribed in a sphere of radius R. obtain the maximum volume of the can.

$${An}\:{oil}\:{can}\:{is}\:{to}\:{be}\:{made}\:{in}\:{form}\:{of}\:{a}\:{right}\:{circular}\:{cylinder}\:{that}\:{be} \\ $$$${inscribed}\:{in}\:{a}\:{sphere}\:{of}\:{radius}\:{R}.\:{obtain}\:{the}\:{maximum}\: \\ $$$${volume}\:{of}\:{the}\:{can}. \\ $$

Commented by sanusihammed last updated on 27/Jun/16

Thanks so much sir.

$${Thanks}\:{so}\:{much}\:{sir}.\: \\ $$

Commented by Yozzii last updated on 27/Jun/16

V=πr^2 h (V=volume of can) Let θ=angle subtended between the horizontal and a line joining one point of contact, between the can and the sphere, and the centre of the sphere. sinθ=(h/(2R))⇒h=2Rsinθ 0<θ<90° cosθ=(r/R)⇒r=Rcosθ. V=πR^2 cos^2 θ×2×Rsinθ V=πR^3 cosθsin2θ V=((πR^3 )/2)(sin3θ+sinθ) (dV/dθ)=πR^3 (−sinθsin2θ+2cosθcos2θ) When (dV/dθ)=0 ⇒2cosθcos2θ=sinθsin2θ tanθtan2θ=2 ((2tan^2 θ)/(1−tan^2 θ))=2 tan^2 θ=1−tan^2 θ 2tan^2 θ=1⇒tanθ=(1/(√2))⇒θ=tan^(−1) (1/(√2)) 0<θ<90° (dV/dθ)=((πR^3 )/2)(3cos3θ+cosθ) (d^2 V/dθ^2 )=−((πR^3 )/2)(9sin3θ+sinθ) Substituting θ=tan^(−1) 2^(−0.5) gives (d^2 V/dθ^2 )<0 ⇒max(V) occurs at θ=tan^(−1) 2^(−0.5) . tanθ=(1/(√2))⇒cosθ=((√2)/(√3))=(√(2/3)) & sinθ=(1/(√3)). ∴max(V)=2πR^3 ×(2/3)×(1/(√3)) max(V)=((4πR^3 )/(3(√3)))=(1/(√3))V_(sphere)

$${V}=\pi{r}^{\mathrm{2}} {h}\:\left({V}={volume}\:{of}\:{can}\right) \\ $$$${Let}\:\theta={angle}\:{subtended}\:{between}\:{the} \\ $$$${horizontal}\:{and}\:{a}\:{line}\:{joining}\:{one}\:{point} \\ $$$${of}\:{contact},\:{between}\:{the}\:{can}\:{and}\:{the}\:{sphere}, \\ $$$${and}\:{the}\:{centre}\:{of}\:{the}\:{sphere}. \\ $$$${sin}\theta=\frac{{h}}{\mathrm{2}{R}}\Rightarrow{h}=\mathrm{2}{Rsin}\theta\:\:\:\:\mathrm{0}<\theta<\mathrm{90}° \\ $$$${cos}\theta=\frac{{r}}{{R}}\Rightarrow{r}={Rcos}\theta. \\ $$$${V}=\pi{R}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta×\mathrm{2}×{Rsin}\theta \\ $$$${V}=\pi{R}^{\mathrm{3}} {cos}\theta{sin}\mathrm{2}\theta \\ $$$${V}=\frac{\pi{R}^{\mathrm{3}} }{\mathrm{2}}\left({sin}\mathrm{3}\theta+{sin}\theta\right) \\ $$$$\frac{{dV}}{{d}\theta}=\pi{R}^{\mathrm{3}} \left(−{sin}\theta{sin}\mathrm{2}\theta+\mathrm{2}{cos}\theta{cos}\mathrm{2}\theta\right) \\ $$$${When}\:\frac{{dV}}{{d}\theta}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{cos}\theta{cos}\mathrm{2}\theta={sin}\theta{sin}\mathrm{2}\theta \\ $$$${tan}\theta{tan}\mathrm{2}\theta=\mathrm{2} \\ $$$$\frac{\mathrm{2}{tan}^{\mathrm{2}} \theta}{\mathrm{1}−{tan}^{\mathrm{2}} \theta}=\mathrm{2} \\ $$$${tan}^{\mathrm{2}} \theta=\mathrm{1}−{tan}^{\mathrm{2}} \theta \\ $$$$\mathrm{2}{tan}^{\mathrm{2}} \theta=\mathrm{1}\Rightarrow{tan}\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\Rightarrow\theta={tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\:\mathrm{0}<\theta<\mathrm{90}° \\ $$$$\frac{{dV}}{{d}\theta}=\frac{\pi{R}^{\mathrm{3}} }{\mathrm{2}}\left(\mathrm{3}{cos}\mathrm{3}\theta+{cos}\theta\right) \\ $$$$\frac{{d}^{\mathrm{2}} {V}}{{d}\theta^{\mathrm{2}} }=−\frac{\pi{R}^{\mathrm{3}} }{\mathrm{2}}\left(\mathrm{9}{sin}\mathrm{3}\theta+{sin}\theta\right) \\ $$$${Substituting}\:\theta={tan}^{−\mathrm{1}} \mathrm{2}^{−\mathrm{0}.\mathrm{5}} \:{gives}\:\frac{{d}^{\mathrm{2}} {V}}{{d}\theta^{\mathrm{2}} }<\mathrm{0} \\ $$$$\Rightarrow{max}\left({V}\right)\:{occurs}\:{at}\:\theta={tan}^{−\mathrm{1}} \mathrm{2}^{−\mathrm{0}.\mathrm{5}} . \\ $$$${tan}\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\Rightarrow{cos}\theta=\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}}=\sqrt{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\&\:{sin}\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}. \\ $$$$\therefore{max}\left({V}\right)=\mathrm{2}\pi{R}^{\mathrm{3}} ×\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$${max}\left({V}\right)=\frac{\mathrm{4}\pi{R}^{\mathrm{3}} }{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}{V}_{{sphere}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com