What is the extreme f(x,y)=2x^3-6xy+3y^2 ?

The extreme f(x,y)=2x^3-6xy+3y^2 is Minimum(1,1),Saddle(0,0)

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extreme f(x,y)=2x^3-6xy+3y^2

extreme points

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

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

Solution steps

Find the critical points:

(1, 1), (0, 0)

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

D (x, y) = 72 x - 36

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) = ((45 - x) x - (10 + 24 x))

y = 7 + 5 x - 5 x^{2}

T (x, y) = 46 - 2 x^{2} - 3 y^{2}

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

f (x, y) = 2 y^{2} + 2 x y + x^{2} - 16 x - 20 y

What is the extreme f(x,y)=2x^3-6xy+3y^2 ?
The extreme f(x,y)=2x^3-6xy+3y^2 is Minimum(1,1),Saddle(0,0)