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Global Extremum – Domain of a function with fixed negative powers – Exercise 6551

Exercise

Find the maximum value and the minimum value of the function

z(x,y)=1+\frac{1}{x}+\frac{1}{y}

In the domain:

D=\{ (x,y):\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{8}\}

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

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