What is the extreme f(x,y)=x^3-3xy+4y^2 ?

The extreme f(x,y)=x^3-3xy+4y^2 is Minimum(3/8 , 9/64),Saddle(0,0)

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extreme f(x,y)=x^3-3xy+4y^2

extreme points

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

Minimum (\frac{3}{8}, \frac{9}{64}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{3}{8}, \frac{9}{64}), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 48 x - 9

Check the sign of at each critical point

Check critical point

(\frac{3}{8}, \frac{9}{64}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{3}{8}, \frac{9}{64}), Saddle (0, 0)

f (x, y) = ∣ x - y ∣

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

f (x) = 10 x^{2} + 36 x + 36

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

f (x) = 4 x^{3} + 48 x^{2} + 180 x

What is the extreme f(x,y)=x^3-3xy+4y^2 ?
The extreme f(x,y)=x^3-3xy+4y^2 is Minimum(3/8 , 9/64),Saddle(0,0)