What is the extreme f(x)=x^5y^2-3x^3+10+5y^3-xy ?

The extreme f(x)=x^5y^2-3x^3+10+5y^3-xy is

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extreme f(x)=x^5y^2-3x^3+10+5y^3-xy

extreme points

f (x) = x^{5} y^{2} - 3 x^{3} + 10 + 5 y^{3} - x y

Solution steps

Find the critical points:

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 x^{5} + 30 y

D (x, y) = - 60 x^{8} y^{2} - 36 x^{6} + 20 x^{4} y + 600 x^{3} y^{3} - 540 x y - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

Saddle

Check critical point

(0, 0) :

Saddle

Check critical point

Maximum

Check critical point

Saddle

f (x) = \frac{0.2 t}{t ^{2} + 59}

\frac{18}{x} + 3 x^{2}

f (x) = 3 x^{2} - 4 x + 12

f (x) = x (- \frac{2 x}{3} + 4)

f (x, y) = 2 x^{2} + 3 y^{2} + 2 x y - 6 x - 8 y + 11

What is the extreme f(x)=x^5y^2-3x^3+10+5y^3-xy ?
The extreme f(x)=x^5y^2-3x^3+10+5y^3-xy is