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

The extreme f(x,y)=x^4+y^4-xy is Saddle(0,0),Minimum(-1/2 ,-1/2),Minimum(1/2 , 1/2)

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

extreme points

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

Saddle (0, 0), Minimum (- \frac{1}{2}, - \frac{1}{2}), Minimum (\frac{1}{2}, \frac{1}{2})

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{2}, - \frac{1}{2}), (\frac{1}{2}, \frac{1}{2})

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

D (x, y) = 144 x^{2} y^{2} - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{1}{2}, - \frac{1}{2}) :

Minimum

Check critical point

(\frac{1}{2}, \frac{1}{2}) :

Minimum

Saddle (0, 0), Minimum (- \frac{1}{2}, - \frac{1}{2}), Minimum (\frac{1}{2}, \frac{1}{2})

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

f (x) = - 3 x^{4} + 8 x^{3} - 6 x^{2} - 2

f (x) = 14 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

\frac{4 x}{x ^{2} - 25}

f (x) = x^{2} + x y + y^{2} - 10 y + 33

What is the extreme f(x,y)=x^4+y^4-xy ?
The extreme f(x,y)=x^4+y^4-xy is Saddle(0,0),Minimum(-1/2 ,-1/2),Minimum(1/2 , 1/2)