What is the extreme f(x,y)=-x^3+4xy-2y^2+1 ?

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

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extreme f(x,y)=-x^3+4xy-2y^2+1

extreme points

f (x, y) = - x^{3} + 4 x y - 2 y^{2} + 1

Saddle (0, 0), Maximum (\frac{4}{3}, \frac{4}{3})

Solution steps

Find the critical points:

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

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

D (x, y) = 24 x - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Maximum

Saddle (0, 0), Maximum (\frac{4}{3}, \frac{4}{3})

x^{2} - 2 x y + 2 y

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

\frac{3 x ^{2}}{x + 5}

f (x) = \frac{2}{3} x - 5, - 2 \leq x \leq 3

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