What is the extreme x^2y-2xy+2y^2x ?

The extreme x^2y-2xy+2y^2x is Minimum(2/3 , 1/3),Saddle(0,1)

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extreme x^2y-2xy+2y^2x

extreme points

x^{2} y - 2 x y + 2 y^{2} x

Minimum (\frac{2}{3}, \frac{1}{3}), Saddle (0, 1)

Solution steps

Find the critical points:

(\frac{2}{3}, \frac{1}{3}), (0, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 x

D (x, y) = - 8 x y - 4 x^{2} + 8 x + 16 y - 16 y^{2} - 4

Check the sign of at each critical point

Check critical point

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

Minimum

Check critical point

(0, 1) :

Saddle

Minimum (\frac{2}{3}, \frac{1}{3}), Saddle (0, 1)

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

f (x) = 5 (x - 3)^{2} (2 x + 1) (x + 4)^{3}

f (x) = 2 x - 1.75 x^{2} + 1. 1^{3} - 0.25 x^{4}

f (x y) = - x^{2} - y^{2} + 22 x + 18 y - 102

x^{2} + 2 x + 1

What is the extreme x^2y-2xy+2y^2x ?
The extreme x^2y-2xy+2y^2x is Minimum(2/3 , 1/3),Saddle(0,1)