Question-108864

Question Number 108864 by I want to learn more last updated on 19/Aug/20

Commented by I want to learn more last updated on 19/Aug/20

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{length}\:\mathrm{of}\:\:\mathrm{PQ} \\ $$

Answered by mr W last updated on 19/Aug/20

f(x)=x^3 −3x−2 g(x)=2x−2 f′(x)=3x^2 −3 such that PQ is maximum, tangent at Q must be parallel to g(x). f′(x)=3x^2 −3=g′(x)=2 ⇒x=±(√(5/3)) at x=−(√(5/3)): y_P =−2(√(5/3))−2 y_Q =−(5/3)(√(5/3))+3(√(5/3))−2 PQ=−(5/3)(√(5/3))+3(√(5/3))−2−(−2(√(5/3))−2) =((10(√(15)))/9) at x=(√(5/3)): y_P =2(√(5/3))−2 y_Q =(5/3)(√(5/3))−3(√(5/3))−2 PQ=2(√(5/3))−2−((5/3)(√(5/3))−3(√(5/3))−2) =((10(√(15)))/9) ⇒PQ_(max,local) =((10(√(15)))/9) PQ_(max) =∞ when x→±∞

$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{2} \\ $$$${g}\left({x}\right)=\mathrm{2}{x}−\mathrm{2} \\ $$$${f}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3} \\ $$$${such}\:{that}\:{PQ}\:{is}\:{maximum},\:{tangent} \\ $$$${at}\:{Q}\:{must}\:{be}\:{parallel}\:{to}\:{g}\left({x}\right). \\ $$$${f}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}={g}'\left({x}\right)=\mathrm{2} \\ $$$$\Rightarrow{x}=\pm\sqrt{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$${at}\:{x}=−\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}: \\ $$$${y}_{{P}} =−\mathrm{2}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}−\mathrm{2} \\ $$$${y}_{{Q}} =−\frac{\mathrm{5}}{\mathrm{3}}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}+\mathrm{3}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}−\mathrm{2} \\ $$$${PQ}=−\frac{\mathrm{5}}{\mathrm{3}}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}+\mathrm{3}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}−\mathrm{2}−\left(−\mathrm{2}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}−\mathrm{2}\right) \\ $$$$=\frac{\mathrm{10}\sqrt{\mathrm{15}}}{\mathrm{9}} \\ $$$${at}\:{x}=\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}: \\ $$$${y}_{{P}} =\mathrm{2}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}−\mathrm{2} \\ $$$${y}_{{Q}} =\frac{\mathrm{5}}{\mathrm{3}}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}−\mathrm{3}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}−\mathrm{2} \\ $$$${PQ}=\mathrm{2}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}−\mathrm{2}−\left(\frac{\mathrm{5}}{\mathrm{3}}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}−\mathrm{3}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}−\mathrm{2}\right) \\ $$$$=\frac{\mathrm{10}\sqrt{\mathrm{15}}}{\mathrm{9}} \\ $$$$ \\ $$$$\Rightarrow{PQ}_{{max},{local}} =\frac{\mathrm{10}\sqrt{\mathrm{15}}}{\mathrm{9}} \\ $$$$ \\ $$$${PQ}_{{max}} =\infty\:{when}\:{x}\rightarrow\pm\infty \\ $$

Commented by I want to learn more last updated on 25/Aug/20

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{appreciate} \\ $$

Answered by Aziztisffola last updated on 19/Aug/20

P(x ; 2x−2) and Q(x ; x^3 −3x−2) PQ(x)=(√(0^2 +( x^3 −3x−2−2x+2)^2 )) =x^3 −5x PQ(x)′=3x^2 −5 PQ(x)′=0⇒3x^2 −5=0⇒x=+_− (√(5/3)) min in x=(√(5/3)) Max local in x=−(√(5/3)) ; look at variation of PQ then PQ(−(√(5/3)))=(−(√(5/3)))^3 +5(√(5/3)) =((10(√(15)))/9)

$$\mathrm{P}\left(\mathrm{x}\:;\:\mathrm{2x}−\mathrm{2}\right)\:\mathrm{and}\:\mathrm{Q}\left(\mathrm{x}\:;\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}−\mathrm{2}\right) \\ $$$$\mathrm{PQ}\left(\mathrm{x}\right)=\sqrt{\mathrm{0}^{\mathrm{2}} +\left(\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}−\mathrm{2}−\mathrm{2x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{3}} −\mathrm{5x} \\ $$$$\mathrm{PQ}\left(\mathrm{x}\right)'=\mathrm{3x}^{\mathrm{2}} −\mathrm{5} \\ $$$$\mathrm{PQ}\left(\mathrm{x}\right)'=\mathrm{0}\Rightarrow\mathrm{3x}^{\mathrm{2}} −\mathrm{5}=\mathrm{0}\Rightarrow\mathrm{x}=\underset{−} {+}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$$\:\mathrm{min}\:\mathrm{in}\:\mathrm{x}=\sqrt{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$$\:\mathrm{Max}\:\mathrm{local}\:\mathrm{in}\:\mathrm{x}=−\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\:;\:\mathrm{look}\:\mathrm{at}\:\mathrm{variation}\:\mathrm{of}\:\mathrm{PQ} \\ $$$$\:\mathrm{then}\:\mathrm{PQ}\left(−\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)=\left(−\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)^{\mathrm{3}} +\mathrm{5}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$$\:\:=\frac{\mathrm{10}\sqrt{\mathrm{15}}}{\mathrm{9}} \\ $$

Commented by I want to learn more last updated on 25/Aug/20

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply