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:

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

$$\max_D z(-2\sqrt{5},\sqrt{5})=\max_D z(2\sqrt{5},-\sqrt{5})= 125$$
$$\min_D z(\sqrt{5},2\sqrt{5})=\min_D z(-\sqrt{5},-2\sqrt{5})=0$$