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) = cos (3 x)

f (x) = 2 x^{3} - 3 x^{2} + 6

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

f (x) = 2 x + \frac{8}{x ^{2}} + 5, x > 0

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