What is the extreme 1/(x^2)+1/y+2xy ?

The extreme 1/(x^2)+1/y+2xy is Minimum(2^{1/5}, 1/(\sqrt[5]{8)})

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extreme 1/(x^2)+1/y+2xy

extreme

Minimum (2^{\frac{1}{5}}, \frac{1}{5 8})

Solution steps

Find the critical points:

(2^{\frac{1}{5}}, \frac{1}{5 8})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2}{y ^{3}}

D (x, y) = \frac{12}{x ^{4} y ^{3}} - 4

Check the sign of at each critical point

Check critical point

(2^{\frac{1}{5}}, \frac{1}{5 8}) :

Minimum

Minimum (2^{\frac{1}{5}}, \frac{1}{5 8})

What is the extreme 1/(x^2)+1/y+2xy ?
The extreme 1/(x^2)+1/y+2xy is Minimum(2^{1/5}, 1/(\sqrt[5]{8)})