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Question Number 97266 by bobhans last updated on 07/Jun/20
If x &y satisfy the equation x^2 +y^2 −4x−6y−1 =0 find minimum value of x+y ?
$$\mathrm{If}\:\mathrm{x}\:\&\mathrm{y}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}−\mathrm{6y}−\mathrm{1}\:=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}\:? \\ $$
Commented by bobhans last updated on 07/Jun/20
f(x,y,λ) = x+y+λ(x^2 +y^2 −4x−6y−1) (∂f/∂x) = 1+λ(2x−4)=0 ⇒λ = (1/(4−2x)) (∂f/∂y) = 1+λ(2y−6)=0 ⇒λ=(1/(6−2y)) so 4−2x = 6−2y ; y=1+x substitute to constraint x^2 +(x+1)^2 −4x−6(1+x)−1=0 2x^2 −8x−6=0 ; x^2 −4x−3=0 x = ((4 ±(√(28)))/2) = 2 ±(√7) ⇒ { ((x=2+(√7) ∧y=3+(√7))),((x=2−(√7) ∧y=3−(√7))) :} so minimum value x+y = 5−2(√7) or 5 −(√(28))
$$\mathrm{f}\left(\mathrm{x},\mathrm{y},\lambda\right)\:=\:\mathrm{x}+\mathrm{y}+\lambda\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}−\mathrm{6y}−\mathrm{1}\right) \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{x}}\:=\:\mathrm{1}+\lambda\left(\mathrm{2x}−\mathrm{4}\right)=\mathrm{0}\:\Rightarrow\lambda\:=\:\frac{\mathrm{1}}{\mathrm{4}−\mathrm{2x}} \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\:=\:\mathrm{1}+\lambda\left(\mathrm{2y}−\mathrm{6}\right)=\mathrm{0}\:\Rightarrow\lambda=\frac{\mathrm{1}}{\mathrm{6}−\mathrm{2y}} \\ $$$$\mathrm{so}\:\mathrm{4}−\mathrm{2x}\:=\:\mathrm{6}−\mathrm{2y}\:;\:\mathrm{y}=\mathrm{1}+\mathrm{x}\: \\ $$$$\mathrm{substitute}\:\mathrm{to}\:\mathrm{constraint}\: \\ $$$$\mathrm{x}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4x}−\mathrm{6}\left(\mathrm{1}+\mathrm{x}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2x}^{\mathrm{2}} −\mathrm{8x}−\mathrm{6}=\mathrm{0}\:;\:\mathrm{x}^{\mathrm{2}} −\mathrm{4x}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{4}\:\pm\sqrt{\mathrm{28}}}{\mathrm{2}}\:=\:\mathrm{2}\:\pm\sqrt{\mathrm{7}}\:\Rightarrow\begin{cases}{\mathrm{x}=\mathrm{2}+\sqrt{\mathrm{7}}\:\wedge\mathrm{y}=\mathrm{3}+\sqrt{\mathrm{7}}}\\{\mathrm{x}=\mathrm{2}−\sqrt{\mathrm{7}}\:\wedge\mathrm{y}=\mathrm{3}−\sqrt{\mathrm{7}}}\end{cases} \\ $$$$\mathrm{so}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{x}+\mathrm{y}\:=\:\mathrm{5}−\mathrm{2}\sqrt{\mathrm{7}}\:\mathrm{or}\:\mathrm{5}\:−\sqrt{\mathrm{28}} \\ $$
Answered by john santu last updated on 07/Jun/20
let x+y = k . because (x,y) satisfy the circle x^2 +y^2 −4x−6y−1 = 0 then r = ((∣2+3−k∣)/( (√2) )) = (√(14)) ⇔∣k−5∣ = (√(28 )) { ((k_(max) = 5 + (√(28)))),((k_(min) = 5−(√(28)) )) :}
$$\mathrm{let}\:\mathrm{x}+\mathrm{y}\:=\:\mathrm{k}\:.\:\mathrm{because}\:\left(\mathrm{x},\mathrm{y}\right) \\ $$$$\mathrm{satisfy}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}−\mathrm{6y}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{r}\:=\:\frac{\mid\mathrm{2}+\mathrm{3}−\mathrm{k}\mid}{\:\sqrt{\mathrm{2}}\:}\:=\:\sqrt{\mathrm{14}} \\ $$$$\Leftrightarrow\mid\mathrm{k}−\mathrm{5}\mid\:=\:\sqrt{\mathrm{28}\:} \\ $$$$\begin{cases}{\mathrm{k}_{\mathrm{max}} \:=\:\mathrm{5}\:+\:\sqrt{\mathrm{28}}}\\{\mathrm{k}_{\mathrm{min}} \:=\:\mathrm{5}−\sqrt{\mathrm{28}}\:}\end{cases} \\ $$
Commented by bobhans last updated on 07/Jun/20
great sir your short cut
$$\mathrm{great}\:\mathrm{sir}\:\mathrm{your}\:\mathrm{short}\:\mathrm{cut} \\ $$
Answered by Farruxjano last updated on 07/Jun/20
x^2 +y^2 −4x−6y−1=0, min{x+y}=? x^2 +y^2 −4x−6y−1=0 ⇒(x−2)^2 +(y−3)^2 =14 (((x−2)/( (√(14)))))^2 +(((y−3)/( (√(14)))))^2 =1 ⇒ { ((((x−2)/( (√(14))))=sin𝛂 )),((((y−3)/( (√(14))))=cos𝛂)) :}⇒ ⇒ { ((x=(√(14))sin𝛂+2)),((y=(√(14))cos𝛂+3)) :}⇒x^2 +y^2 =14+ +4(√(14))sin𝛂+6(√(14))cos𝛂+4+9= =[using this inequality: asin𝛂+bcos𝛂≥ −(√(a^2 +b^2 ))]=27+[4(√(14))sin𝛂+6(√(14))cos𝛂]≥ ≥27+(√(16∙14+36∙14))=27−(√(52∙14))=27−(√(728))
$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}−\mathrm{6y}−\mathrm{1}=\mathrm{0},\:\boldsymbol{{min}}\left\{\boldsymbol{{x}}+\boldsymbol{{y}}\right\}=? \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}−\mathrm{6}\boldsymbol{{y}}−\mathrm{1}=\mathrm{0}\:\Rightarrow\left(\boldsymbol{{x}}−\mathrm{2}\right)^{\mathrm{2}} +\left(\boldsymbol{{y}}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{14} \\ $$$$\left(\frac{\boldsymbol{{x}}−\mathrm{2}}{\:\sqrt{\mathrm{14}}}\right)^{\mathrm{2}} +\left(\frac{\boldsymbol{{y}}−\mathrm{3}}{\:\sqrt{\mathrm{14}}}\right)^{\mathrm{2}} =\mathrm{1}\:\Rightarrow\:\begin{cases}{\frac{\boldsymbol{{x}}−\mathrm{2}}{\:\sqrt{\mathrm{14}}}=\boldsymbol{{sin}\alpha}\:\:\:\:}\\{\frac{\boldsymbol{{y}}−\mathrm{3}}{\:\sqrt{\mathrm{14}}}=\boldsymbol{{cos}\alpha}}\end{cases}\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\boldsymbol{{x}}=\sqrt{\mathrm{14}}\boldsymbol{{sin}\alpha}+\mathrm{2}}\\{\boldsymbol{{y}}=\sqrt{\mathrm{14}}\boldsymbol{{cos}\alpha}+\mathrm{3}}\end{cases}\Rightarrow\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} =\mathrm{14}+ \\ $$$$+\mathrm{4}\sqrt{\mathrm{14}}\boldsymbol{{sin}\alpha}+\mathrm{6}\sqrt{\mathrm{14}}\boldsymbol{{cos}\alpha}+\mathrm{4}+\mathrm{9}= \\ $$$$=\left[\boldsymbol{{using}}\:\boldsymbol{{this}}\:\boldsymbol{{inequality}}:\:\boldsymbol{{asin}\alpha}+\boldsymbol{{bcos}\alpha}\geqslant\right. \\ $$$$\left.−\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }\right]=\mathrm{27}+\left[\mathrm{4}\sqrt{\mathrm{14}}\boldsymbol{{sin}\alpha}+\mathrm{6}\sqrt{\mathrm{14}}\boldsymbol{{cos}\alpha}\right]\geqslant \\ $$$$\geqslant\mathrm{27}+\sqrt{\mathrm{16}\centerdot\mathrm{14}+\mathrm{36}\centerdot\mathrm{14}}=\mathrm{27}−\sqrt{\mathrm{52}\centerdot\mathrm{14}}=\mathrm{27}−\sqrt{\mathrm{728}} \\ $$
Commented by bobhans last updated on 07/Jun/20
sorry sir. it wrong
$$\mathrm{sorry}\:\mathrm{sir}.\:\mathrm{it}\:\mathrm{wrong} \\ $$
Commented by bobhans last updated on 07/Jun/20
it should be { ((x = 2+(√(14)) cos α)),((y = 3+(√(14)) sin α)) :} then x+y = 5 + (√(28)) cos (α−45^o ) ; so min { x+y } = 5−(√(28)) .
$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\begin{cases}{{x}\:=\:\mathrm{2}+\sqrt{\mathrm{14}}\:\mathrm{cos}\:\alpha}\\{\mathrm{y}\:=\:\mathrm{3}+\sqrt{\mathrm{14}}\:\mathrm{sin}\:\alpha}\end{cases} \\ $$$$\mathrm{then}\:{x}+{y}\:=\:\mathrm{5}\:+\:\sqrt{\mathrm{28}}\:\mathrm{cos}\:\left(\alpha−\mathrm{45}^{\mathrm{o}} \right)\:; \\ $$$$\mathrm{so}\:\mathrm{min}\:\left\{\:{x}+{y}\:\right\}\:=\:\mathrm{5}−\sqrt{\mathrm{28}}\:.\: \\ $$

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