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# Global Extremum – Domain of a curve with absolute value – Exercise 3471

Exercise

Find the maximum value and the minimum value of the function

$$z(x,y)=x^2-xy+y^2$$

In the domain

$$D=\{ (x,y): |x|+|y|\leq 1\}$$

Final Answer

$$\max_D z(0,\pm 1)=\max_D z(\pm 1,0) =1$$

$$\min_D z(0,0) =0$$

Solution

Coming soon…

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