What is the extreme f(x,y)=x^2-12xy+2y^4 ?

The extreme f(x,y)=x^2-12xy+2y^4 is Saddle(0,0),Minimum(-18,-3),Minimum(18,3)

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extreme f(x,y)=x^2-12xy+2y^4

extreme points

f (x, y) = x^{2} - 12 x y + 2 y^{4}

Saddle (0, 0), Minimum (- 18, - 3), Minimum (18, 3)

Solution steps

Find the critical points:

(0, 0), (- 18, - 3), (18, 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = 24 y^{2}

D (x, y) = 48 y^{2} - 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 18, - 3) :

Minimum

Check critical point

(18, 3) :

Minimum

Saddle (0, 0), Minimum (- 18, - 3), Minimum (18, 3)

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

f (x, y) = x^{3} + y^{3} - 9 x y

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

x^{3} - 6 x^{2} + 5

f (x) = 3 - x, - 1 \leq x \leq 2

What is the extreme f(x,y)=x^2-12xy+2y^4 ?
The extreme f(x,y)=x^2-12xy+2y^4 is Saddle(0,0),Minimum(-18,-3),Minimum(18,3)