Question Number 54137 by ajfour last updated on 29/Jan/19
Commented by ajfour last updated on 29/Jan/19
$${Centres}\:{of}\:{two}\:{spheres}\:{of}\:{radii}\:{R}\: \\ $$$${r}\:{are}\:\mathrm{2}{a}\:{distance}\:{apart}.\:{Find}\:{a} \\ $$$${point}\:{on}\:{the}\:{circumference}\:{of}\:{the} \\ $$$${circle}\:{with}\:{AB}\:{as}\:{diameter}\:{from} \\ $$$${which}\:{maximum}\:{surface}\:{area}\:{is} \\ $$$${visible}. \\ $$
Commented by ajfour last updated on 29/Jan/19
$${mrW}\:{sir},\:{please}\:{help}.. \\ $$
Answered by ajfour last updated on 29/Jan/19
Commented by mr W last updated on 29/Jan/19
$${AP}=\mathrm{2}{a}\:\mathrm{cos}\:\theta \\ $$$${BP}=\mathrm{2}{a}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{cos}\:\alpha=\frac{{R}}{{BP}}=\frac{{R}}{\mathrm{2}{a}\:\mathrm{sin}\:\theta} \\ $$$$\mathrm{cos}\:\beta=\frac{{r}}{{AP}}=\frac{{r}}{\mathrm{2}{a}\:\mathrm{cos}\:\theta} \\ $$$${A}_{{R}} =\mathrm{2}\pi{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right)=\mathrm{2}\pi{R}^{\mathrm{2}} \left(\mathrm{1}−\frac{{R}}{\mathrm{2}{a}\:\mathrm{sin}\:\theta}\right) \\ $$$${A}_{{r}} =\mathrm{2}\pi{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\beta\right)=\mathrm{2}\pi{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{{r}}{\mathrm{2}{a}\:\mathrm{cos}\:\theta}\right) \\ $$$${A}={A}_{{R}} +{A}_{{r}} =\mathrm{2}\pi\left\{{R}^{\mathrm{2}} \left(\mathrm{1}−\frac{{R}}{\mathrm{2}{a}\:\mathrm{sin}\:\theta}\right)+{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{{r}}{\mathrm{2}{a}\:\mathrm{cos}\:\theta}\right)\right\} \\ $$$${A}=\mathrm{2}\pi\left\{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}{a}}\left(\frac{{R}^{\mathrm{3}} }{\mathrm{sin}\:\theta}+\frac{{r}^{\mathrm{3}} }{\mathrm{cos}\:\theta}\right)\right\} \\ $$$${let}\:{f}\left(\theta\right)=\frac{{R}^{\mathrm{3}} }{\mathrm{sin}\:\theta}+\frac{{r}^{\mathrm{3}} }{\mathrm{cos}\:\theta} \\ $$$$\frac{{df}\left(\theta\right)}{{d}\theta}=−\frac{{R}^{\mathrm{3}} \mathrm{cos}\:\theta}{\mathrm{sin}^{\mathrm{2}} \:\theta}+\frac{{r}^{\mathrm{3}} \mathrm{sin}\:\theta}{\mathrm{cos}^{\mathrm{2}} \:\theta}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{{R}}{{r}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{{R}}{\:\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{{r}}{\:\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} }} \\ $$$$\Rightarrow{A}_{{max}} =\mathrm{2}\pi\left({R}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} }}{\mathrm{2}{a}}\right) \\ $$
Commented by mr W last updated on 29/Jan/19
Commented by ajfour last updated on 30/Jan/19
$${Thank}\:{you}\:{Sir}.\:{Location}\:{of}\:{P}\:\:{is}\:{nicely}\: \\ $$$${depicted}. \\ $$