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Question Number 99003 by bramlex last updated on 18/Jun/20

Given 5x−3y=6 . find min value of (x−1)^2 +(y+1)^2 ?

$${Given}\:\mathrm{5}{x}−\mathrm{3}{y}=\mathrm{6}\:.\:{find}\:{min}\:{value} \\ $$$${of}\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} \:? \\ $$

Answered by bobhans last updated on 18/Jun/20

let (x−1)^2 +(y+1)^2 =R^2 min {(x−1)^2 +(y+1)^2 } if 5x−3y = 6 is tangent line of a circle . min value = R^2 =((∣5.1−3.(−1)−6∣^2 )/(25+9)) = (4/(34)) = (2/(17)) ■

$$\mathrm{let}\:\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} \\ $$$$\mathrm{min}\:\left\{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} \right\}\:\mathrm{if}\:\mathrm{5x}−\mathrm{3y}\:=\:\mathrm{6}\:\mathrm{is} \\ $$$$\mathrm{tangent}\:\mathrm{line}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:. \\ $$$$\mathrm{min}\:\mathrm{value}\:=\:\mathrm{R}^{\mathrm{2}} \:=\frac{\mid\mathrm{5}.\mathrm{1}−\mathrm{3}.\left(−\mathrm{1}\right)−\mathrm{6}\mid^{\mathrm{2}} }{\mathrm{25}+\mathrm{9}} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{34}}\:=\:\frac{\mathrm{2}}{\mathrm{17}}\:\blacksquare \\ $$

Commented by bramlex last updated on 18/Jun/20

thank you

$${thank}\:{you} \\ $$

Answered by 1549442205 last updated on 18/Jun/20

5(x−1)−3(y+1)=−2.ApplyingAM−GM we get (−2)^2 ≤(5^2 +(−3)^2 )[(x−1)^2 +(y+1)^2 ] ⇒(x−1)^2 +(y+1)^2 ≥(4/(34))=(2/(17)).The equality occurs when { (((5/(x−1))=((−3)/(y+1)))),((5x−3y=6)) :}⇔ { ((3x+5y=−2)),((5x−3y=6)) :}⇔ { ((x=((12)/(17)))),((y=((−14)/(17)))) :} Hence,S=(x−1)^2 +(y+1)^2 has the least value equal to (2/(34)) when { ((x=((12)/(17)))),((y=((−14)/(17)))) :}

$$\mathrm{5}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{3}\left(\mathrm{y}+\mathrm{1}\right)=−\mathrm{2}.\mathrm{ApplyingAM}−\mathrm{GM} \\ $$$$\mathrm{we}\:\mathrm{get}\:\left(−\mathrm{2}\right)^{\mathrm{2}} \leqslant\left(\mathrm{5}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} \right)\left[\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} \right] \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} \geqslant\frac{\mathrm{4}}{\mathrm{34}}=\frac{\mathrm{2}}{\mathrm{17}}.\mathrm{The}\:\mathrm{equality}\:\mathrm{occurs} \\ $$$$\mathrm{when}\:\begin{cases}{\frac{\mathrm{5}}{\mathrm{x}−\mathrm{1}}=\frac{−\mathrm{3}}{\mathrm{y}+\mathrm{1}}}\\{\mathrm{5x}−\mathrm{3y}=\mathrm{6}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{3x}+\mathrm{5y}=−\mathrm{2}}\\{\mathrm{5x}−\mathrm{3y}=\mathrm{6}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{12}}{\mathrm{17}}}\\{\mathrm{y}=\frac{−\mathrm{14}}{\mathrm{17}}}\end{cases} \\ $$$$\mathrm{Hence},\mathrm{S}=\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{has}\:\mathrm{the}\:\mathrm{least}\:\mathrm{value}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\frac{\mathrm{2}}{\mathrm{34}}\:\:\mathrm{when}\:\begin{cases}{\mathrm{x}=\frac{\mathrm{12}}{\mathrm{17}}}\\{\mathrm{y}=\frac{−\mathrm{14}}{\mathrm{17}}}\end{cases} \\ $$$$ \\ $$

Commented by bemath last updated on 18/Jun/20

the last line (2/(34)) or (2/(17)) or (4/(34)) ?

$$\mathrm{the}\:\mathrm{last}\:\mathrm{line}\:\frac{\mathrm{2}}{\mathrm{34}}\:\mathrm{or}\:\frac{\mathrm{2}}{\mathrm{17}}\:\mathrm{or}\:\frac{\mathrm{4}}{\mathrm{34}}\:? \\ $$

Commented by 1549442205 last updated on 18/Jun/20

Excuse me,thank sir you,I wrote a mistake it is (2/(17))

$$\mathrm{Excuse}\:\mathrm{me},\mathrm{thank}\:\mathrm{sir}\:\mathrm{you},\mathrm{I}\:\mathrm{wrote}\:\mathrm{a}\:\mathrm{mistake} \\ $$$$\mathrm{it}\:\mathrm{is}\:\frac{\mathrm{2}}{\mathrm{17}} \\ $$

Answered by MJS last updated on 18/Jun/20

why you folks do such complicated things? 5x−3y=6 ⇒ y=(5/3)x−2 (x−1)^2 +(y+1)^2 =((34)/9)x^2 −((16)/3)x+2 (d/dx)[((34)/9)x^2 −((16)/3)x+2]=0 ((68)/9)x−((16)/3)=0 x=((12)/(17)) (d/dx)[((68)/9)x−((16)/3)]>0 ⇒ minimum ((34)/9)x^2 −((16)/3)x+2=(2/(17))

$$\mathrm{why}\:\mathrm{you}\:\mathrm{folks}\:\mathrm{do}\:\mathrm{such}\:\mathrm{complicated}\:\mathrm{things}? \\ $$$$\mathrm{5}{x}−\mathrm{3}{y}=\mathrm{6}\:\Rightarrow\:{y}=\frac{\mathrm{5}}{\mathrm{3}}{x}−\mathrm{2} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{34}}{\mathrm{9}}{x}^{\mathrm{2}} −\frac{\mathrm{16}}{\mathrm{3}}{x}+\mathrm{2} \\ $$$$\frac{{d}}{{dx}}\left[\frac{\mathrm{34}}{\mathrm{9}}{x}^{\mathrm{2}} −\frac{\mathrm{16}}{\mathrm{3}}{x}+\mathrm{2}\right]=\mathrm{0} \\ $$$$\frac{\mathrm{68}}{\mathrm{9}}{x}−\frac{\mathrm{16}}{\mathrm{3}}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{12}}{\mathrm{17}} \\ $$$$\frac{{d}}{{dx}}\left[\frac{\mathrm{68}}{\mathrm{9}}{x}−\frac{\mathrm{16}}{\mathrm{3}}\right]>\mathrm{0}\:\Rightarrow\:\mathrm{minimum} \\ $$$$\frac{\mathrm{34}}{\mathrm{9}}{x}^{\mathrm{2}} −\frac{\mathrm{16}}{\mathrm{3}}{x}+\mathrm{2}=\frac{\mathrm{2}}{\mathrm{17}} \\ $$

Commented by bramlex last updated on 18/Jun/20

thank you

$${thank}\:{you} \\ $$

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