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

The extreme f(x,y)=4xy+x^4+y^4 is Saddle(0,0)

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y^{2}

D (x, y) = 144 x^{2} y^{2} - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x^{2} + 3 y^{2} - 2 x + 18 y

f (x) = - x^{2} - x + 1, - 1 \leq x \leq 1

f (x) = e^{0.5 x} + 25 6^{- 0.5 x}

f (x) = \frac{ln ( x )}{5 x ^{2}}

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

What is the extreme f(x,y)=4xy+x^4+y^4 ?
The extreme f(x,y)=4xy+x^4+y^4 is Saddle(0,0)