What is the extreme f(x,y)=2x^2+x*y^3-4y ?

The extreme f(x,y)=2x^2+x*y^3-4y is Saddle(\sqrt[5]{4/27},-\sqrt[5]{16/3})

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extreme f(x,y)=2x^2+x*y^3-4y

extreme points

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

Solution steps

Find the critical points:

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 x y

D (x, y) = 24 x y - 9 y^{4}

Check the sign of at each critical point

Check critical point

Saddle

f (x) = (\frac{1}{2} - x^{2} + y^{2}) e^{1 - x^{2} - y^{2}}

f (y) = 2 x^{2} - 9 x + 5

f (x y) = 2 x^{2} + \frac{3}{2} y^{2} + 3 x y - 10 x - 9 y + 2

2 x + 1 - 2 x

f (x) = y = \frac{3 x ^{2}}{x + 1}, - 1 \leq x \leq 5

What is the extreme f(x,y)=2x^2+x*y^3-4y ?
The extreme f(x,y)=2x^2+x*y^3-4y is Saddle(\sqrt[5]{4/27},-\sqrt[5]{16/3})