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

The extreme f(x,y)=x-4x^2y+y^2 is Saddle(1/(4^{2/3)}, 1/(2\sqrt[3]{4)})

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

extreme points

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

Saddle (\frac{1}{4 ^{\frac{2}{3}}}, \frac{1}{2 3 4})

Solution steps

Find the critical points:

(\frac{1}{4 ^{\frac{2}{3}}}, \frac{1}{2 3 4})

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

D (x, y) = - 16 y - 64 x^{2}

Check the sign of at each critical point

Check critical point

(\frac{1}{4 ^{\frac{2}{3}}}, \frac{1}{2 3 4}) :

Saddle

Saddle (\frac{1}{4 ^{\frac{2}{3}}}, \frac{1}{2 3 4})

f (x) = x + \frac{25}{x}, 0.2 \leq x \leq 20

y = x^{3} - 30 x^{2} + 6000

f (x) = x^{2} + x y + y^{2} + y

f (x) - x^{4} - 8 x^{2} + 1

f (x) = x^{3} - 12 x + 17

What is the extreme f(x,y)=x-4x^2y+y^2 ?
The extreme f(x,y)=x-4x^2y+y^2 is Saddle(1/(4^{2/3)}, 1/(2\sqrt[3]{4)})