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

The extreme f(x,y)=2x^3+3y^2-4xy-4 is Minimum(4/9 , 8/27),Saddle(0,0)

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

extreme points

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

Minimum (\frac{4}{9}, \frac{8}{27}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{4}{9}, \frac{8}{27}), (0, 0)

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

D (x, y) = 72 x - 16

Check the sign of at each critical point

Check critical point

(\frac{4}{9}, \frac{8}{27}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{4}{9}, \frac{8}{27}), Saddle (0, 0)

f (x) = 14 + 4 x - x^{2}, 0 \leq x \leq 5

- 1.8 x^{2} + 7 x - 11

f (x, y) = 2 ln (x) + ln (y) - 4 x - y

f (x, y) = 3 x + y

f (t) = (1 + 2 t - 3 t^{2} + 4 t^{3}) u (t - 2)

What is the extreme f(x,y)=2x^3+3y^2-4xy-4 ?
The extreme f(x,y)=2x^3+3y^2-4xy-4 is Minimum(4/9 , 8/27),Saddle(0,0)