What is the extreme f(x,y)=x^3+3/2 y^2-3xy-124 ?

The extreme f(x,y)=x^3+3/2 y^2-3xy-124 is Minimum(1,1),Saddle(0,0)

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extreme f(x,y)=x^3+3/2 y^2-3xy-124

extreme points

f (x, y) = x^{3} + \frac{3}{2} y^{2} - 3 x y - 124

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

Solution steps

Find the critical points:

(1, 1), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 3

D (x, y) = 18 x - 9

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)

3 x^{4} - 4 x^{3} + 1

f (x, y) = 8 x^{4} + y^{2} - 16 x y

f (x, y) = x^{2} - 2 x y + 4 y^{2} - 6 x - 6 y + 8

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

x 9 - x^{2}

What is the extreme f(x,y)=x^3+3/2 y^2-3xy-124 ?
The extreme f(x,y)=x^3+3/2 y^2-3xy-124 is Minimum(1,1),Saddle(0,0)