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

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

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

extreme points

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

Maximum (\frac{1}{4}, \frac{1}{4})

Solution steps

Find the critical points:

(\frac{1}{4}, \frac{1}{4})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(\frac{1}{4}, \frac{1}{4}) :

Maximum

Maximum (\frac{1}{4}, \frac{1}{4})

p (a, x) = a (x + 1) + x (x + 1) + 2 (x + 1)

\frac{4 \frac{x}{5 - 3}}{2 x ^{5} + 9 x}

f (x) = - 3 x^{2} + 6 x + 9

f (x) = x^{2} - x y + y^{2} - 9 x + 6 y + 8

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