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Question Number 177815 by infinityaction last updated on 09/Oct/22

Answered by mahdipoor last updated on 09/Oct/22

$$\Rightarrow \{\begin{array}{l}xy+yy+zy=2y\\ xx+yx+zx=2x\\ xy+yz+zx=1\end{array}\Rightarrow i+ii-iii\Rightarrow$$$$xy+x^2+y^2=2(y+x)-1\Rightarrow$$$$g(x,y)=xy+x^2+y^2-2(x+y)=-1\Rightarrow$$$$f(x,y)=x-y$$$$getmaxorminfinp=(x,y)$$$$\Rightarrow ▽\overrightarrow{f}\left(p\right)=\lambda ▽\overrightarrow{g}\left(p\right)$$$$\Rightarrow (1i-1j)=\lambda ((y+2x-2)i+(x+2y-2)j)$$$$\Rightarrow \{\begin{array}{l}1=\lambda (y+2x-2)\\ -1=\lambda (x+2y-2)\end{array}\Rightarrow \{\begin{array}{l}y=\frac{2}{3}-\frac{1}{\lambda }\\ x=\frac{2}{3}+\frac{1}{\lambda }\end{array}$$$$g(x,y)=0=\frac{1}{{\lambda }^2}-\frac{1}{3}\Rightarrow \lambda =\pm \sqrt{3}$$$$f(x,y)=\frac{2}{\lambda }\Rightarrow \{\begin{array}{l}maxf=2/\sqrt{3}\\ minf=-2/\sqrt{3}\end{array}$$

Commented by infinityaction last updated on 09/Oct/22

$$thankssir$$

Commented by Tawa11 last updated on 10/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 09/Oct/22

$$x+y=2-z$$$$xy+(x+y)z=1$$$$xy=1-z(2-z)$$$${(x-y)}^2={(x+y)}^2-4xy$$$$={(2-z)}^2-4[1-z(2-z)]$$$$=4z-3z^2$$$$=-3{(z-\frac{2}{3})}^2+\frac{4}{3}⩽\frac{4}{3}$$$$\Rightarrow \mid x-y\mid ⩽\frac{2}{\sqrt{3}}$$$$\Rightarrow -\frac{2}{\sqrt{3}}⩽x-y⩽\frac{2}{\sqrt{3}}$$$$\Rightarrow {(x-y)}_{max}=\frac{2}{\sqrt{3}}atz=\frac{2}{3}$$$$\Rightarrow {(x-y)}_{min}=-\frac{2}{\sqrt{3}}atz=\frac{2}{3}$$

Commented by infinityaction last updated on 09/Oct/22

$$thankyousir$$

Commented by Tawa11 last updated on 10/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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