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

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

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

extreme points

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

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)

\frac{9}{2} x^{2} - ln (x)

f (x) = 2 x - 0.3 y

f (x) = 1 - (x - 2)^{\frac{4}{5}}

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

f (x) = x^{4} - 98 x^{2} - 1, (- 8, 8)

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