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

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

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

extreme points

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

Minimum (1, 1), Saddle (0, 0)

Solution steps

Find the critical points:

(1, 1), (0, 0)

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

D (x, y) = 2 y - 1

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (1, 1), Saddle (0, 0)

f (x) = \frac{- ( \frac{1}{2} ) x ^{4} + 2}{3 x ^{3} - 4}

\frac{( x - 2 ) ^{3}}{( x - 4 ) ^{4}}

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

f (x) = ln (2 x) - 2 x^{2}

f (x) = (x - 6)^{\frac{2}{3}}

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