Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

$$\mathrm{o}\:\mathrm{i}\:\mathrm{understand}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com