Global Extremum – Domain of a circle – Exercise 6543

Exercise

Find the maximum value and the minimum value of the function

$z(x,y)=4x^2-4xy+y^2$

In the domain:

$D=\{ (x,y):x^2+y^2=25\}$

Final Answer

$\max_D z(-2\sqrt{5},\sqrt{5})=\max_D z(2\sqrt{5},-\sqrt{5})= 125$

$\min_D z(\sqrt{5},2\sqrt{5})=\min_D z(-\sqrt{5},-2\sqrt{5})=0$

Solution

Coming soon…

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