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find-minimum-and-maximum-value-of-f-x-y-x-2-y-2-with-constraint-x-2-y-2-1-with-Lagrange-method-

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Question Number 89805 by jagoll last updated on 19/Apr/20
find minimum and maximum value of f(x,y) = x^2 −y^2 with constraint x^2 +y^2 = 1 with Lagrange method
$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{with}\:\mathrm{constraint}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\mathrm{with}\:\mathrm{Lagrange}\:\mathrm{method} \\ $$
Commented by john santu last updated on 19/Apr/20
y^2 = 1−x^2 ⇒f(x) = x^2 −(1−x^2 )=2x^2 −1 f ′(x) = 4x = 0 ; x = 0 y = ± 1 ⇒ stationer point (0,1) ; (0,−1) f(0,1) = −1 ← minimum ′via calculus′
$${y}^{\mathrm{2}} \:=\:\mathrm{1}−{x}^{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)\:=\:{x}^{\mathrm{2}} −\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1} \\ $$$${f}\:'\left({x}\right)\:=\:\mathrm{4}{x}\:=\:\mathrm{0}\:;\:{x}\:=\:\mathrm{0} \\ $$$${y}\:=\:\pm\:\mathrm{1}\:\Rightarrow\:{stationer}\:{point} \\ $$$$\left(\mathrm{0},\mathrm{1}\right)\:;\:\left(\mathrm{0},−\mathrm{1}\right)\: \\ $$$${f}\left(\mathrm{0},\mathrm{1}\right)\:=\:−\mathrm{1}\:\leftarrow\:{minimum}\: \\ $$$$'{via}\:{calculus}' \\ $$$$ \\ $$
Commented by mr W last updated on 19/Apr/20
x^2 +y^2 =1 ⇒−1≤x≤1 f(x)=2x^2 −1 min.=−1 at x=0 max.=1 at x=±1
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1} \\ $$$${min}.=−\mathrm{1}\:{at}\:{x}=\mathrm{0} \\ $$$${max}.=\mathrm{1}\:{at}\:{x}=\pm\mathrm{1} \\ $$
Commented by jagoll last updated on 19/Apr/20
if use Langrange method , how sir?
$$\mathrm{if}\:\mathrm{use}\:\mathrm{Langrange}\:\mathrm{method}\:,\:\mathrm{how}\:\mathrm{sir}? \\ $$
Answered by mr W last updated on 19/Apr/20
using lagrange F(x,y,λ)=x^2 −y^2 +λ(x^2 +y^2 −1) (∂F/∂x)=2x+2λx=0 ⇒x(1+λ)=0 ...(i) (∂F/∂y)=−2y+2λy=0 ⇒y(−1+λ)=0 ...(ii) (∂F/∂λ)=x^2 +y^2 −1=0 ⇒x^2 +y^2 =1 ...(iii) from (i): x=0 or λ=−1 with x=0: ⇒y=±1, λ=1 ⇒x^2 −y^2 =−1 ⇒⇒min. with λ=−1: ⇒y=0, x=±1 ⇒x^2 −y^2 =1 ⇒⇒max.
$${using}\:{lagrange} \\ $$$$ \\ $$$${F}\left({x},{y},\lambda\right)={x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\lambda\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\frac{\partial{F}}{\partial{x}}=\mathrm{2}{x}+\mathrm{2}\lambda{x}=\mathrm{0}\:\Rightarrow{x}\left(\mathrm{1}+\lambda\right)=\mathrm{0}\:\:\:\:…\left({i}\right) \\ $$$$\frac{\partial{F}}{\partial{y}}=−\mathrm{2}{y}+\mathrm{2}\lambda{y}=\mathrm{0}\:\Rightarrow{y}\left(−\mathrm{1}+\lambda\right)=\mathrm{0}\:\:\:\:…\left({ii}\right) \\ $$$$\frac{\partial{F}}{\partial\lambda}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\:\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right): \\ $$$${x}=\mathrm{0}\:{or}\:\lambda=−\mathrm{1} \\ $$$${with}\:{x}=\mathrm{0}: \\ $$$$\Rightarrow{y}=\pm\mathrm{1},\:\lambda=\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =−\mathrm{1}\:\Rightarrow\Rightarrow{min}. \\ $$$$ \\ $$$${with}\:\lambda=−\mathrm{1}: \\ $$$$\Rightarrow{y}=\mathrm{0},\:{x}=\pm\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{1}\:\:\Rightarrow\Rightarrow{max}. \\ $$
Commented by jagoll last updated on 19/Apr/20
may way ▽f = λ▽g ((( 2x)),((−2y)) ) = λ (((2x)),((2y)) ) ⇒2x = 2λx ⇒ { ((x=0)),((λ=1)) :} ⇒−2y=2λy ⇒ { ((y=0)),((λ=−1)) :} why the value of λ not same sir
$$\mathrm{may}\:\mathrm{way}\: \\ $$$$\bigtriangledown\mathrm{f}\:=\:\lambda\bigtriangledown\mathrm{g} \\ $$$$\begin{pmatrix}{\:\:\:\mathrm{2x}}\\{−\mathrm{2y}}\end{pmatrix}\:=\:\lambda\begin{pmatrix}{\mathrm{2x}}\\{\mathrm{2y}}\end{pmatrix} \\ $$$$\Rightarrow\mathrm{2x}\:=\:\mathrm{2}\lambda\mathrm{x}\:\Rightarrow\:\begin{cases}{\mathrm{x}=\mathrm{0}}\\{\lambda=\mathrm{1}}\end{cases} \\ $$$$\Rightarrow−\mathrm{2y}=\mathrm{2}\lambda\mathrm{y}\:\Rightarrow\:\begin{cases}{\mathrm{y}=\mathrm{0}}\\{\lambda=−\mathrm{1}}\end{cases} \\ $$$$\mathrm{why}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\lambda\:\mathrm{not}\:\mathrm{same}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 19/Apr/20
for max. and min. the value of λ is different.
$${for}\:{max}.\:{and}\:{min}.\:{the}\:{value}\:{of}\:\lambda\:{is} \\ $$$${different}. \\ $$
Commented by mr W last updated on 19/Apr/20
you got exactly the same values as i: λ=1, x=0 ⇒y=±1 ⇒min. =−1 λ=−1, y=0 ⇒x=±1 ⇒max. =1
$${you}\:{got}\:{exactly}\:{the}\:{same}\:{values}\:{as}\:{i}: \\ $$$$\lambda=\mathrm{1},\:{x}=\mathrm{0}\:\Rightarrow{y}=\pm\mathrm{1}\:\Rightarrow{min}.\:=−\mathrm{1} \\ $$$$\lambda=−\mathrm{1},\:{y}=\mathrm{0}\:\Rightarrow{x}=\pm\mathrm{1}\:\Rightarrow{max}.\:=\mathrm{1} \\ $$
Commented by jagoll last updated on 19/Apr/20
o i understand sir. thank you
$$\mathrm{o}\:\mathrm{i}\:\mathrm{understand}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$

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