What is the extreme f(x,y)=x^2y+2xy^2-12xy+2 ?

The extreme f(x,y)=x^2y+2xy^2-12xy+2 is Minimum(4,2),Saddle(0,6)

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extreme f(x,y)=x^2y+2xy^2-12xy+2

extreme points

f (x, y) = x^{2} y + 2 x y^{2} - 12 x y + 2

Minimum (4, 2), Saddle (0, 6)

Solution steps

Find the critical points:

(4, 2), (0, 6)

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

D (x, y) = - 4 x^{2} - 8 x y + 48 x - 16 y^{2} + 96 y - 144

Check the sign of at each critical point

Check critical point

(4, 2) :

Minimum

Check critical point

(0, 6) :

Saddle

Minimum (4, 2), Saddle (0, 6)

f (x, y) = 2 x^{3}

25 x^{2} - 260 x + 1700

x + \frac{31}{x}

f (x) = x^{4} + 4 x + 1

\frac{x ^{2} - 12}{x + 4}

What is the extreme f(x,y)=x^2y+2xy^2-12xy+2 ?
The extreme f(x,y)=x^2y+2xy^2-12xy+2 is Minimum(4,2),Saddle(0,6)