What is the extreme f(x)=(2x-x^2)(2y-y^2) ?

The extreme f(x)=(2x-x^2)(2y-y^2) is Saddle(0,0),Saddle(2,0),Saddle(0,2),Saddle(2,2),Maximum(1,1)

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extreme f(x)=(2x-x^2)(2y-y^2)

extreme points

f (x) = (2 x - x^{2}) (2 y - y^{2})

Saddle (0, 0), Saddle (2, 0), Saddle (0, 2), Saddle (2, 2), Maximum (1, 1)

Solution steps

Find the critical points:

(0, 0), (2, 0), (0, 2), (2, 2), (1, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2 (2 x - x^{2})

D (x, y) = - 12 x^{2} y^{2} + 24 x^{2} y - 16 x^{2} + 24 x y^{2} - 48 x y + 32 x - 16 y^{2} + 32 y - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, 0) :

Saddle

Check critical point

(0, 2) :

Saddle

Check critical point

(2, 2) :

Saddle

Check critical point

(1, 1) :

Maximum

Saddle (0, 0), Saddle (2, 0), Saddle (0, 2), Saddle (2, 2), Maximum (1, 1)

f (x, y) = 9 - 2 x + 8 y - x^{2} - 4 y^{2}

f (x) = - 4 x^{5} ln (2 x)

f (x) = \frac{x ^{3}}{x} - 3 x^{2} - 7 x

y = x^{3} + 3 x^{2} + 3 x + 1

f (x) = \frac{8 x - 3}{x}

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