What is the extreme f(x)=2xy-5x^2-2y^2+4x-4y+4 ?

The extreme f(x)=2xy-5x^2-2y^2+4x-4y+4 is Maximum(2/9 ,-8/9)

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extreme f(x)=2xy-5x^2-2y^2+4x-4y+4

extreme points

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

Maximum (\frac{2}{9}, - \frac{8}{9})

Solution steps

Find the critical points:

(\frac{2}{9}, - \frac{8}{9})

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

D (x, y) = 36

Check the sign of at each critical point

Check critical point

(\frac{2}{9}, - \frac{8}{9}) :

Maximum

Maximum (\frac{2}{9}, - \frac{8}{9})

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

f (x, y) = x^{3} - 3 x y + y^{2} + y - 5

f (x) = \frac{( 8 x ^{3} - x ^{2} - 108 )}{4 x ^{2}}

f (x) = 3 x - 4

- (x + 1)^{2} - 5 (y - 3)^{2}

What is the extreme f(x)=2xy-5x^2-2y^2+4x-4y+4 ?
The extreme f(x)=2xy-5x^2-2y^2+4x-4y+4 is Maximum(2/9 ,-8/9)