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Question Number 95924 by i jagooll last updated on 28/May/20

form a Lagrangian to maximize x^2 −y^2 subject to the constraint 2x+y = 3?

$$\mathrm{form}\:\mathrm{a}\:\mathrm{Lagrangian}\:\mathrm{to}\:\mathrm{maximize} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:\mathrm{subject}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{constraint}\:\mathrm{2x}+\mathrm{y}\:=\:\mathrm{3}? \\ $$

Commented by john santu last updated on 29/May/20

f(x,y,λ) = x^2 −y^2 +λ(2x+y−3) (∂f/∂x) = 2x+2λ=0 ⇒λ=−x (∂f/∂y) = −2y+λ=0 ⇒λ=2y (∂f/∂λ) = 2x+y−3=0 ⇒−2λ+(λ/2)=3 −3λ = 6 ⇒λ = −2⇒ { ((x=2)),((y= −1)) :} max f(2,−1) =2^2 −(−1)^2 =3

$$\mathrm{f}\left(\mathrm{x},\mathrm{y},\lambda\right)\:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} +\lambda\left(\mathrm{2x}+\mathrm{y}−\mathrm{3}\right) \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{x}}\:=\:\mathrm{2x}+\mathrm{2}\lambda=\mathrm{0}\:\Rightarrow\lambda=−\mathrm{x} \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\:=\:−\mathrm{2y}+\lambda=\mathrm{0}\:\Rightarrow\lambda=\mathrm{2y} \\ $$$$\frac{\partial\mathrm{f}}{\partial\lambda}\:=\:\mathrm{2x}+\mathrm{y}−\mathrm{3}=\mathrm{0}\:\Rightarrow−\mathrm{2}\lambda+\frac{\lambda}{\mathrm{2}}=\mathrm{3} \\ $$$$−\mathrm{3}\lambda\:=\:\mathrm{6}\:\Rightarrow\lambda\:=\:−\mathrm{2}\Rightarrow\begin{cases}{\mathrm{x}=\mathrm{2}}\\{\mathrm{y}=\:−\mathrm{1}}\end{cases}\: \\ $$$$\mathrm{max}\:\mathrm{f}\left(\mathrm{2},−\mathrm{1}\right)\:=\mathrm{2}^{\mathrm{2}} −\left(−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3}\: \\ $$$$ \\ $$

Answered by mr W last updated on 28/May/20

f(x,y)=x^2 −y^2 F(x,y,λ)=x^2 −y^2 +λ(2x+y−3) (∂F/∂x)=2x+2λ=0 ⇒x=−λ (∂F/∂y)=−2y+λ=0 ⇒y=(λ/2) (∂F/∂λ)=2x+y−3=0 ⇒−2λ+(λ/2)−3=0 ⇒λ=−2 ⇒x=2, y=−1 f_(max) =f(2,−1)=2^2 −(−1)^2 =3 or f(x)=x^2 −(3−2x)^2 (df/dx)=2x−2(3−2x)(−2)=0 ⇒x=2 f_(max) =f(2)=2^2 −(3−4)^2 =3

$${f}\left({x},{y}\right)={x}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$${F}\left({x},{y},\lambda\right)={x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\lambda\left(\mathrm{2}{x}+{y}−\mathrm{3}\right) \\ $$$$\frac{\partial{F}}{\partial{x}}=\mathrm{2}{x}+\mathrm{2}\lambda=\mathrm{0}\:\Rightarrow{x}=−\lambda \\ $$$$\frac{\partial{F}}{\partial{y}}=−\mathrm{2}{y}+\lambda=\mathrm{0}\:\Rightarrow{y}=\frac{\lambda}{\mathrm{2}} \\ $$$$\frac{\partial{F}}{\partial\lambda}=\mathrm{2}{x}+{y}−\mathrm{3}=\mathrm{0}\:\Rightarrow−\mathrm{2}\lambda+\frac{\lambda}{\mathrm{2}}−\mathrm{3}=\mathrm{0}\:\Rightarrow\lambda=−\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{2},\:{y}=−\mathrm{1} \\ $$$${f}_{{max}} ={f}\left(\mathrm{2},−\mathrm{1}\right)=\mathrm{2}^{\mathrm{2}} −\left(−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3} \\ $$$$ \\ $$$${or} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{2}} \\ $$$$\frac{{df}}{{dx}}=\mathrm{2}{x}−\mathrm{2}\left(\mathrm{3}−\mathrm{2}{x}\right)\left(−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$${f}_{{max}} ={f}\left(\mathrm{2}\right)=\mathrm{2}^{\mathrm{2}} −\left(\mathrm{3}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{3} \\ $$

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