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Find-the-smallest-value-of-x-2-y-2-among-all-values-of-x-amp-y-satisfying-3x-y-20-




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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