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Find-maximum-and-minimum-values-of-f-x-y-xy-on-the-circle-g-x-y-x-2-y-2-1-0-




Question Number 122695 by liberty last updated on 19/Nov/20
Find maximum and minimum values  of f(x,y) = xy on the circle g(x,y)=  x^2 +y^2 −1=0 .
$${Find}\:{maximum}\:{and}\:{minimum}\:{values} \\ $$$${of}\:{f}\left({x},{y}\right)\:=\:{xy}\:{on}\:{the}\:{circle}\:{g}\left({x},{y}\right)= \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:. \\ $$
Answered by john santu last updated on 19/Nov/20
By Lagrange multiplier    ▽f(p) = λ ▽g(p) ,where P is any point    (((∂f/∂x)),((∂f/∂y)) ) = λ  (((∂g/∂x)),((∂g/∂y)) )    ((y),(x) ) =  (((2λx)),((2λy)) ) → { ((λ=(y/(2x)))),((λ=(x/(2y)))) :}   ⇒ (y/(2x)) = (x/(2y)) ⇒x^2 =y^2  ∧x^2 +y^2 =1   this gives four solutions     { ((((1/( (√2))),(1/( (√2)))) , ((1/( (√2))),−(1/( (√2)))))),(((−(1/( (√2))),(1/( (√2)))) ,(−(1/( (√2))),−(1/( (√2)))))) :}  Evaluating f at these four points  we get max value is (1/2) and min value  is −(1/2).
$${By}\:{Lagrange}\:{multiplier}\: \\ $$$$\:\bigtriangledown{f}\left({p}\right)\:=\:\lambda\:\bigtriangledown{g}\left({p}\right)\:,{where}\:{P}\:{is}\:{any}\:{point} \\ $$$$\:\begin{pmatrix}{\frac{\partial{f}}{\partial{x}}}\\{\frac{\partial{f}}{\partial{y}}}\end{pmatrix}\:=\:\lambda\:\begin{pmatrix}{\frac{\partial{g}}{\partial{x}}}\\{\frac{\partial{g}}{\partial{y}}}\end{pmatrix} \\ $$$$\:\begin{pmatrix}{{y}}\\{{x}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{2}\lambda{x}}\\{\mathrm{2}\lambda{y}}\end{pmatrix}\:\rightarrow\begin{cases}{\lambda=\frac{{y}}{\mathrm{2}{x}}}\\{\lambda=\frac{{x}}{\mathrm{2}{y}}}\end{cases} \\ $$$$\:\Rightarrow\:\frac{{y}}{\mathrm{2}{x}}\:=\:\frac{{x}}{\mathrm{2}{y}}\:\Rightarrow{x}^{\mathrm{2}} ={y}^{\mathrm{2}} \:\wedge{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\:{this}\:{gives}\:{four}\:{solutions}\: \\ $$$$\:\begin{cases}{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:,\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)}\\{\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:,\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)}\end{cases} \\ $$$${Evaluating}\:{f}\:{at}\:{these}\:{four}\:{points} \\ $$$${we}\:{get}\:{max}\:{value}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{min}\:{value} \\ $$$${is}\:−\frac{\mathrm{1}}{\mathrm{2}}. \\ $$
Answered by mr W last updated on 19/Nov/20
P=xy  y=(P/x)  x^2 +(P^2 /x^2 )−1=0  2x+((2PP′)/x^2 )−((2P^2 )/x^3 )=0  P′=0  x−(P^2 /x^3 )=0  x−((x^2 y^2 )/x^3 )=0  x^2 =y^2 =(1/2)  ⇒x=±(1/( (√2))), y=±(1/( (√2)))  (xy)_(max) =(1/2)  (xy)_(min) =−(1/2)
$${P}={xy} \\ $$$${y}=\frac{{P}}{{x}} \\ $$$${x}^{\mathrm{2}} +\frac{{P}^{\mathrm{2}} }{{x}^{\mathrm{2}} }−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}{x}+\frac{\mathrm{2}{PP}'}{{x}^{\mathrm{2}} }−\frac{\mathrm{2}{P}^{\mathrm{2}} }{{x}^{\mathrm{3}} }=\mathrm{0} \\ $$$${P}'=\mathrm{0} \\ $$$${x}−\frac{{P}^{\mathrm{2}} }{{x}^{\mathrm{3}} }=\mathrm{0} \\ $$$${x}−\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{x}^{\mathrm{3}} }=\mathrm{0} \\ $$$${x}^{\mathrm{2}} ={y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\:{y}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\left({xy}\right)_{{max}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({xy}\right)_{{min}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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