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Question Number 54137 by ajfour last updated on 29/Jan/19

Commented by ajfour last updated on 29/Jan/19

Centres of two spheres of radii R r are 2a distance apart. Find a point on the circumference of the circle with AB as diameter from which maximum surface area is visible.

$${Centres}\:{of}\:{two}\:{spheres}\:{of}\:{radii}\:{R}\: \\ $$$${r}\:{are}\:\mathrm{2}{a}\:{distance}\:{apart}.\:{Find}\:{a} \\ $$$${point}\:{on}\:{the}\:{circumference}\:{of}\:{the} \\ $$$${circle}\:{with}\:{AB}\:{as}\:{diameter}\:{from} \\ $$$${which}\:{maximum}\:{surface}\:{area}\:{is} \\ $$$${visible}. \\ $$

Commented by ajfour last updated on 29/Jan/19

mrW sir, please help..

$${mrW}\:{sir},\:{please}\:{help}.. \\ $$

Answered by ajfour last updated on 29/Jan/19

Commented by mr W last updated on 29/Jan/19

AP=2a cos θ BP=2a sin θ cos α=(R/(BP))=(R/(2a sin θ)) cos β=(r/(AP))=(r/(2a cos θ)) A_R =2πR^2 (1−cos α)=2πR^2 (1−(R/(2a sin θ))) A_r =2πr^2 (1−cos β)=2πr^2 (1−(r/(2a cos θ))) A=A_R +A_r =2π{R^2 (1−(R/(2a sin θ)))+r^2 (1−(r/(2a cos θ)))} A=2π{R^2 +r^2 −(1/(2a))((R^3 /(sin θ))+(r^3 /(cos θ)))} let f(θ)=(R^3 /(sin θ))+(r^3 /(cos θ)) ((df(θ))/dθ)=−((R^3 cos θ)/(sin^2 θ))+((r^3 sin θ)/(cos^2 θ))=0 ⇒tan θ=(R/r) ⇒sin θ=(R/(√(R^2 +r^2 ))) ⇒cos θ=(r/(√(R^2 +r^2 ))) ⇒A_(max) =2π(R^2 +r^2 )(1−((√(R^2 +r^2 ))/(2a)))

$${AP}=\mathrm{2}{a}\:\mathrm{cos}\:\theta \\ $$$${BP}=\mathrm{2}{a}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{cos}\:\alpha=\frac{{R}}{{BP}}=\frac{{R}}{\mathrm{2}{a}\:\mathrm{sin}\:\theta} \\ $$$$\mathrm{cos}\:\beta=\frac{{r}}{{AP}}=\frac{{r}}{\mathrm{2}{a}\:\mathrm{cos}\:\theta} \\ $$$${A}_{{R}} =\mathrm{2}\pi{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right)=\mathrm{2}\pi{R}^{\mathrm{2}} \left(\mathrm{1}−\frac{{R}}{\mathrm{2}{a}\:\mathrm{sin}\:\theta}\right) \\ $$$${A}_{{r}} =\mathrm{2}\pi{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\beta\right)=\mathrm{2}\pi{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{{r}}{\mathrm{2}{a}\:\mathrm{cos}\:\theta}\right) \\ $$$${A}={A}_{{R}} +{A}_{{r}} =\mathrm{2}\pi\left\{{R}^{\mathrm{2}} \left(\mathrm{1}−\frac{{R}}{\mathrm{2}{a}\:\mathrm{sin}\:\theta}\right)+{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{{r}}{\mathrm{2}{a}\:\mathrm{cos}\:\theta}\right)\right\} \\ $$$${A}=\mathrm{2}\pi\left\{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}{a}}\left(\frac{{R}^{\mathrm{3}} }{\mathrm{sin}\:\theta}+\frac{{r}^{\mathrm{3}} }{\mathrm{cos}\:\theta}\right)\right\} \\ $$$${let}\:{f}\left(\theta\right)=\frac{{R}^{\mathrm{3}} }{\mathrm{sin}\:\theta}+\frac{{r}^{\mathrm{3}} }{\mathrm{cos}\:\theta} \\ $$$$\frac{{df}\left(\theta\right)}{{d}\theta}=−\frac{{R}^{\mathrm{3}} \mathrm{cos}\:\theta}{\mathrm{sin}^{\mathrm{2}} \:\theta}+\frac{{r}^{\mathrm{3}} \mathrm{sin}\:\theta}{\mathrm{cos}^{\mathrm{2}} \:\theta}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{{R}}{{r}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{{R}}{\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{{r}}{\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} }} \\ $$$$\Rightarrow{A}_{{max}} =\mathrm{2}\pi\left({R}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} }}{\mathrm{2}{a}}\right) \\ $$

Commented by mr W last updated on 29/Jan/19

Commented by ajfour last updated on 30/Jan/19

Thank you Sir. Location of P is nicely depicted.

$${Thank}\:{you}\:{Sir}.\:{Location}\:{of}\:{P}\:\:{is}\:{nicely}\: \\ $$$${depicted}. \\ $$

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