What is the extreme f(x,y)=4x^2y+3xy^2-12xy ?

The extreme f(x,y)=4x^2y+3xy^2-12xy is Minimum(1, 4/3),Saddle(0,4)

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extreme f(x,y)=4x^2y+3xy^2-12xy

extreme

Minimum (1, \frac{4}{3}), Saddle (0, 4)

Solution steps

Find the critical points:

(1, \frac{4}{3}), (0, 4)

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

D (x, y) = - 64 x^{2} - 48 x y + 192 x - 36 y^{2} + 144 y - 144

Check the sign of at each critical point

Check critical point

(1, \frac{4}{3}) :

Minimum

Check critical point

(0, 4) :

Saddle

Minimum (1, \frac{4}{3}), Saddle (0, 4)

What is the extreme f(x,y)=4x^2y+3xy^2-12xy ?
The extreme f(x,y)=4x^2y+3xy^2-12xy is Minimum(1, 4/3),Saddle(0,4)