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Question Number 141475 by bramlexs22 last updated on 19/May/21

Find the smallest value of (√(x^2 +y^2 )) among all values of x & y satisfying 3x−y = 20

$$\:{Find}\:{the}\:{smallest}\:{value}\:{of}\: \\ $$ $$\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:{among}\:{all}\:{values}\:{of} \\ $$ $$\:{x}\:\&\:{y}\:{satisfying}\:\mathrm{3}{x}−{y}\:=\:\mathrm{20}\: \\ $$

Answered by MJS_new last updated on 19/May/21

y=3x−20 f(x)=(√(10(x^2 −12x+40))) x=t+6 f(t)=(√(10(t^2 +4))) min (f(t)) =2(√(10))

$${y}=\mathrm{3}{x}−\mathrm{20} \\ $$ $${f}\left({x}\right)=\sqrt{\mathrm{10}\left({x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{40}\right)} \\ $$ $${x}={t}+\mathrm{6} \\ $$ $${f}\left({t}\right)=\sqrt{\mathrm{10}\left({t}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$ $$\mathrm{min}\:\left({f}\left({t}\right)\right)\:=\mathrm{2}\sqrt{\mathrm{10}} \\ $$

Answered by mathmax by abdo last updated on 19/May/21

ϕ(x)=(√(x^2 +y^2 )) and y=3x−20 ⇒ϕ(x)=(√(x^2 +(3x−20)^2 )) =(√(x^2 +9x^2 −120x +400))=(√(10x^2 −120x +400)) ϕ^′ (x)=((20x−120)/(2(√(10x^2 −120x))+400)) =((10x−60)/( (√(10x^2 −120x+400)))) =((10(x−6))/( (√(....)))) ϕ^′ (x)=0 ⇒x=6 and ϕ^′ (x)>0 ⇒x>6 ϕ^′ (x)<0 ⇒x<6 ⇒inf ϕ(x)=ϕ(6)=(√(6^2 +(18−20)))^2 =(√(36+4)) =(√(40))=2(√(10))

$$\varphi\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:\:\mathrm{and}\:\mathrm{y}=\mathrm{3x}−\mathrm{20}\:\Rightarrow\varphi\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} \:+\left(\mathrm{3x}−\mathrm{20}\right)^{\mathrm{2}} } \\ $$ $$=\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{9x}^{\mathrm{2}} −\mathrm{120x}\:+\mathrm{400}}=\sqrt{\mathrm{10x}^{\mathrm{2}} −\mathrm{120x}\:+\mathrm{400}} \\ $$ $$\varphi^{'} \left(\mathrm{x}\right)=\frac{\mathrm{20x}−\mathrm{120}}{\mathrm{2}\sqrt{\mathrm{10x}^{\mathrm{2}} −\mathrm{120x}}+\mathrm{400}}\:=\frac{\mathrm{10x}−\mathrm{60}}{\:\sqrt{\mathrm{10x}^{\mathrm{2}} −\mathrm{120x}+\mathrm{400}}}\:=\frac{\mathrm{10}\left(\mathrm{x}−\mathrm{6}\right)}{\:\sqrt{....}} \\ $$ $$\varphi^{'} \left(\mathrm{x}\right)=\mathrm{0}\:\Rightarrow\mathrm{x}=\mathrm{6}\:\:\mathrm{and}\:\varphi^{'} \left(\mathrm{x}\right)>\mathrm{0}\:\Rightarrow\mathrm{x}>\mathrm{6} \\ $$ $$\left.\varphi^{'} \left(\mathrm{x}\right)<\mathrm{0}\:\Rightarrow\mathrm{x}<\mathrm{6}\:\Rightarrow\mathrm{inf}\:\varphi\left(\mathrm{x}\right)=\varphi\left(\mathrm{6}\right)=\sqrt{\mathrm{6}^{\mathrm{2}} \:+\left(\mathrm{18}−\mathrm{20}\right.}\right)^{\mathrm{2}} =\sqrt{\mathrm{36}+\mathrm{4}} \\ $$ $$=\sqrt{\mathrm{40}}=\mathrm{2}\sqrt{\mathrm{10}} \\ $$

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