What is the extreme f(x,y)=11x^2-2x^3+2y^2+4xy ?

The extreme f(x,y)=11x^2-2x^3+2y^2+4xy is Minimum(0,0),Saddle(3,-3)

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extreme f(x,y)=11x^2-2x^3+2y^2+4xy

extreme points

f (x, y) = 11 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

Minimum (0, 0), Saddle (3, - 3)

Solution steps

Find the critical points:

(0, 0), (3, - 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = - 48 x + 72

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(3, - 3) :

Saddle

Minimum (0, 0), Saddle (3, - 3)

y (x, t) = x (- 2 t + 2)

y = 2 x^{3} - 6 x

- \frac{6 x}{3 x ^{2} + 8}

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

6 x - 4 y - x^{2} - 2 y^{2}

What is the extreme f(x,y)=11x^2-2x^3+2y^2+4xy ?
The extreme f(x,y)=11x^2-2x^3+2y^2+4xy is Minimum(0,0),Saddle(3,-3)