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Find the absolute maximum and minimum values of $$f$$ on the set $$D .$$
$$f(x, y)=x^{2}+y^{2}-2 x, \quad D$$ is the closed triangular region
with vertices $$(2,0),(0,2),$$ and $$(0,-2)$$

$$f(x, y)=x^{2}+y^{2}-2 x$$

$$f_{x}(x, y)=2 x-2=0 \quad x=1$$

$$f_{y}\left(x, y\right)=2 y=0 \quad y=0$$

$$(1,0)$$ Critical point

$$f(x, y)=1+(-2)=-1$$ Min

$$L_{1}\quad x=0$$

$$f(0, y)=y^{2} $$

$$y=[-2,2]$$

$$(0, \pm 2)$$

$$f(0, \pm 2)=y^{2}=( \pm 2)^{2}=4$$ Max

Max $$\Rightarrow 4$$ @ $$(0,\pm2)$$
Min $$\Rightarrow-1$$ @ $$(1,0)$$

 

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