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Question Number 89805 by jagoll last updated on 19/Apr/20

$$\mathrm{find}\mathrm{minimum}\mathrm{and}\mathrm{maximum}$$$$\mathrm{value}\mathrm{of}\mathrm{f}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2-{\mathrm{y}}^2$$$$\mathrm{with}\mathrm{constraint}{\mathrm{x}}^2+{\mathrm{y}}^2=1$$$$\mathrm{with}\mathrm{Lagrange}\mathrm{method}$$

Commented by john santu last updated on 19/Apr/20

$$y^2=1-x^2$$$$\Rightarrow f\left(x\right)=x^2-(1-x^2)=2x^2-1$$$$f{}^{\prime }\left(x\right)=4x=0;x=0$$$$y=\pm 1\Rightarrow stationerpoint$$$$(0,1);(0,-1)$$$$f(0,1)=-1\leftarrow minimum$$$${}^{\prime }via{calculus}^{\prime }$$$$$$

Commented by mr W last updated on 19/Apr/20

$$x^2+y^2=1$$$$\Rightarrow -1⩽x⩽1$$$$f\left(x\right)=2x^2-1$$$$min.=-1atx=0$$$$max.=1atx=\pm 1$$

Commented by jagoll last updated on 19/Apr/20

$$\mathrm{if}\mathrm{use}\mathrm{Langrange}\mathrm{method},\mathrm{how}\mathrm{sir}?$$

Answered by mr W last updated on 19/Apr/20

$$usinglagrange$$$$$$$$F(x,y,\lambda )=x^2-y^2+\lambda (x^2+y^2-1)$$$$\frac{\partial F}{\partial x}=2x+2\lambda x=0\Rightarrow x(1+\lambda )=0...\left(i\right)$$$$\frac{\partial F}{\partial y}=-2y+2\lambda y=0\Rightarrow y(-1+\lambda )=0...\left(ii\right)$$$$\frac{\partial F}{\partial \lambda }=x^2+y^2-1=0\Rightarrow x^2+y^2=1...\left(iii\right)$$$$$$$$from\left(i\right):$$$$x=0or\lambda =-1$$$$withx=0:$$$$\Rightarrow y=\pm 1,\lambda =1$$$$\Rightarrow x^2-y^2=-1\Rightarrow \Rightarrow min.$$$$$$$$with\lambda =-1:$$$$\Rightarrow y=0,x=\pm 1$$$$\Rightarrow x^2-y^2=1\Rightarrow \Rightarrow max.$$

Commented by jagoll last updated on 19/Apr/20

$$\mathrm{may}\mathrm{way}$$$$▽\mathrm{f}=\lambda ▽\mathrm{g}$$$$\left(\begin{array}{c}2\mathrm{x}\\ -2\mathrm{y}\end{array}\right)=\lambda \left(\begin{array}{c}2\mathrm{x}\\ 2\mathrm{y}\end{array}\right)$$$$\Rightarrow 2\mathrm{x}=2\lambda \mathrm{x}\Rightarrow \{\begin{array}{l}\mathrm{x}=0\\ \lambda =1\end{array}$$$$\Rightarrow -2\mathrm{y}=2\lambda \mathrm{y}\Rightarrow \{\begin{array}{l}\mathrm{y}=0\\ \lambda =-1\end{array}$$$$\mathrm{why}\mathrm{the}\mathrm{value}\mathrm{of}\lambda \mathrm{not}\mathrm{same}\mathrm{sir}$$

Commented by mr W last updated on 19/Apr/20

$$formax.andmin.thevalueof\lambda is$$$$different.$$

Commented by mr W last updated on 19/Apr/20

$$yougotexactlythesamevaluesasi:$$$$\lambda =1,x=0\Rightarrow y=\pm 1\Rightarrow min.=-1$$$$\lambda =-1,y=0\Rightarrow x=\pm 1\Rightarrow max.=1$$

Commented by jagoll last updated on 19/Apr/20

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

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