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

The extreme f(x,y)=2xy+2x-x^2-2y^2 is Maximum(2,1)

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

extreme points

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

Maximum (2, 1)

Solution steps

Find the critical points:

(2, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(2, 1) :

Maximum

Maximum (2, 1)

f (x) = x^{3} + 4 x^{2} + 5 x - 2

3 x^{2} y + y^{3} - 3 x^{2} - 3 y^{2} + 2

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

f (x, y) = 10 x y - 5 x^{2} - 7 y^{2} + 40 x

f (x) = ln (1 + 8 x^{3})

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