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 points

\frac{1}{x ^{2}} + \frac{1}{y} + 2 x y

Solution steps

Find the critical points:

\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

Minimum

P (x, y) = 10 x + 12 y

f (x, y) = x^{4} - 2 x^{2} + y^{3} - 3 y

f (x) = 4 x y^{2} - 2 x^{2} y - x

f (x, y) = x^{4} y^{4} + 2 x^{2} y^{2} - 2 x^{2} - 2 y^{2}

f (x) = x^{2} - 2 x + 5

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)})