What is the extreme f(x,y)=x^2-2xy+1/3 y^3-4y ?

The extreme f(x,y)=x^2-2xy+1/3 y^3-4y is Minimum(1+sqrt(5),1+sqrt(5)),Saddle(1-sqrt(5),1-sqrt(5))

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extreme f(x,y)=x^2-2xy+1/3 y^3-4y

extreme points

f (x, y) = x^{2} - 2 x y + \frac{1}{3} y^{3} - 4 y

Minimum (1 + 5, 1 + 5), Saddle (1 - 5, 1 - 5)

Solution steps

Find the critical points:

(1 + 5, 1 + 5), (1 - 5, 1 - 5)

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

D (x, y) = 4 y - 4

Check the sign of at each critical point

Check critical point

(1 + 5, 1 + 5) :

Minimum

Check critical point

(1 - 5, 1 - 5) :

Saddle

Minimum (1 + 5, 1 + 5), Saddle (1 - 5, 1 - 5)

f (x) = 4 x^{5} - 25 x^{4} + 40 x^{3} - 4 [- 1.4]

f (x) = - \frac{1}{2} x^{2} + 15 x - 51

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

f (x) = x^{2} + y^{2} - 16 = 0

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

What is the extreme f(x,y)=x^2-2xy+1/3 y^3-4y ?
The extreme f(x,y)=x^2-2xy+1/3 y^3-4y is Minimum(1+sqrt(5),1+sqrt(5)),Saddle(1-sqrt(5),1-sqrt(5))