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

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

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

extreme points

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

Minimum (4, 2), Saddle (\frac{4}{3}, \frac{2}{3})

Solution steps

Find the critical points:

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

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

D (x, y) = 12 y - 16

Check the sign of at each critical point

Check critical point

(4, 2) :

Minimum

Check critical point

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

Saddle

Minimum (4, 2), Saddle (\frac{4}{3}, \frac{2}{3})

V, V = 13 π r 2 h

f (x) = x^{3} + 3 x y + y^{3}

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

f (x) = x x - 2

f (x, y) = x^{3} + y^{3} - 36 x y

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