What is the extreme f(x,y)= 1/2 x^2+y^3-xy ?

The extreme f(x,y)= 1/2 x^2+y^3-xy is Minimum(1/3 , 1/3),Saddle(0,0)

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extreme f(x,y)= 1/2 x^2+y^3-xy

extreme points

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

Minimum (\frac{1}{3}, \frac{1}{3}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{1}{3}, \frac{1}{3}), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 6 y - 1

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, \frac{1}{3}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{1}{3}, \frac{1}{3}), Saddle (0, 0)

f (x) = 2 x^{2} - 8 x + 2 (0.5)

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

f (x, y) = x^{3} - 12 x + y^{2} + 6 y + 14

f (x) = \frac{1}{5} x^{5} - \frac{2}{3} x^{3} + x

f (x) = \frac{1}{16} x^{3} - 4

What is the extreme f(x,y)= 1/2 x^2+y^3-xy ?
The extreme f(x,y)= 1/2 x^2+y^3-xy is Minimum(1/3 , 1/3),Saddle(0,0)