What is the extreme-2x^3+14x^2+2y^2+4xy ?

The extreme-2x^3+14x^2+2y^2+4xy is Minimum(0,0),Saddle(4,-4)

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extreme-2x^3+14x^2+2y^2+4xy

extreme

Minimum (0, 0), Saddle (4, - 4)

Solution steps

Find the critical points:

(0, 0), (4, - 4)

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

D (x, y) = - 48 x + 96

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(4, - 4) :

Saddle

Minimum (0, 0), Saddle (4, - 4)

What is the extreme-2x^3+14x^2+2y^2+4xy ?
The extreme-2x^3+14x^2+2y^2+4xy is Minimum(0,0),Saddle(4,-4)